Vedanta Excel in Mathematics Book 10 TG Final

a solid cylindrical rod of the same length is formed. Find the diameter of the rod.

Solution:

Here, the internal diameter of an iron pipe (d) = 2.8 cm
122.m8 mcm==1110.4cmcm=
So, its internal radius (r) = 0.1 cm
thickness of pipe =

∴ the external radius of the pipe (R) = 1.4 cm + 0.1 cm = 1.5 cm

Length of pipe = h cm (say)

Now, Volume of the material (V) = πh(R + r) (R − r)

= πh(1.5 + 1.4) (1.5 − 1.4) = πh × 2.9 × 0.1 = 0.29πh

Again, Volume of cylindrical rod = Volume of material used to make the pipe

or, πr2h = 0.29πh [Lengths of pipe and rod are equal]

or, r2 = 0.29 ∴ r = 0.54 cm and d = 2r = 2 × 0.5 cm = 1.08 cm

Hence; the diameter of the rod is 1.08 cm.

10. A solid iron sphere of diameter 42 cm is dropped into a cylindrical drum partly

filled with water. If the radius of the drum is 1.4 m. by how much will the surface of

the water be raised ?

Solution: iron sphere (d) = 42 cm ∴ radius (r) = 42 cm = 21 cm
Here, the diameter of an = × 21 2 12348 π cm3
4 4
Now, volume of sphere 3 πr3 = 3 π × 21 × 21 =

Let height of water surface raised in the drum be h cm.

Then, volume of water displaced by the sphere = Volume of sphere

or, πr2h = 12348 π

or, 140 × 140 h = 12348 = 0.63 cm = 0.63 × 10 mm = 6.3 mm

Hence. the surface of water is raised by 6.3 mm.

11. A cylindrical jar of radius 6 cm contains water. How many iron solid spheres each

of radius 1.5 cm are required to immerse into the jar to raise the level of water by

2 cm.

Solution:

Here, radius of a cylindrical jar (r) = 6 cm

height of water level raised by iron spheres (h) = 2 cm

Now, Volume of water displaced by the iron sphere (V1) = πr2h = π × 62 ×2 = 72π cm3
Also, radius of each iron sphere (r1) = 1.5 cm = 4.5π cm3
4 4
∴ Volume of each iron sphere (V2) = 3 πr3 = 3 π × 1.5 × 1.5 × 1.5

∴ Required number of iron spheres (N) = Volume of displaced water (V1) = 72 π cm3 = 16
Volume of each sphere (V2) 4.5 π cm3

12. Given figure in a solid composed of a cylinder with hemisphere at

one end. If the total surface area and height of the solid are 770

sq.cm and 14 cm respectively, find the height of the cylinder. 14 cm
Solution:

Let r and h be the common radius of base and height of the cylinder

respectively.

Then, height of hemisphere = radius of hemisphere = r

Now, total height of combined solid = r + h = 14 cm

51 Vedanta Excel in Mathematics Teachers' Manual - 10

∴ h - (14 − r) cm ... (i)

Also

T.S.A. of solid = C.S.A of hemisphere + C.S.A of cylinder + Area of base

or, 770 = 2πr2 + 2πrh + πr2

or, 770 = πr (2r + 2h + r)

or, 770 = 22 r [3r + 2(14 − r)] [From (i)]
7
or, 245 = r (r + 28)

or, r2 + 28r − 245 = 0

or, r2 + 35r − 7r − 245 = 0

or, r(r + 35) − 7(r + 35) = 0

or, (r + 35) (r − 7) = 0

Eithere, r + 35 = 0 ∴ r = − 35 which is impossible as the radius can not be negative

or, r − 7 = 0 ∴ r = 7 and h = (14 − 7) cm = 7 cm

Hence, the height of the cylinder is 7 cm.

13. A combined solid made up of a cylinder of radius 3 cm and length h cm and a

hemisphere with the same radius as the cylinder has volume 792 cm3. Find the

value of h.

Solution:

Here, radius of cylinder = radius of hemisphere (r) = 3 cm

height of cylinder = h cm

Volume of combined solid = 792 cm3

Now, volume of combined solid = Volume of cylindr + Volume of hemisphere
ππ27rr222(×+h332+×π23r33r)(h
or, 792 cm3 = + 2 × 3) or, 792 × 7 = (h + 2)
or, 792 = 3 22 × 9

or, 792 =

or, 28 = h + 2

or, h = 26

Hence, the height of the cylindrical part is 26 cm.

14. A roller of diameter 112 cm and length 150 cm takes 550 complete revolution to

level a compound. Calculate the cost of levelling the compound at Rs 9 per square

meter.

Solution:

Here, diameter of the roller = 112 cm 56
1212 cm 100
∴ Radius of the roller (r) = 150 cm = 56 cm = = 0.56 m
length of the roller (h) = = 110500 m m
= 1.5

Now, area of covered by the roller in 1 resolution = C.S.A. of the roller
22
= 2πrh = 2 × 7 × 0.56 × 1.5 = 5.28 m2

∴ Area covered by the roller in 550 revolution = 550 × 5.28 m2 = 2904 m2

∴ Area of the compound = 2904 m2

Again, rate of levelling the compound (R) = Rs 9 per m2

∴ Cost of levelling the compound (T) = A × R = 2904 × Rs 9 = Rs 26,136

Hence, the required cost of levelling the compound is Rs 26,136.

15. The external and internal radii of hollow cylindrical metallic vessel 56 cm long are

10.5 cm and 10.1 cm respectively. Find the cost of metal contained by the vessel at

Rs 2 per cubic cm. Also, find the cost of polishing its outer surface at 20 paisa per

Vedanta Excel in Mathematics Teachers' Manual - 10 52

square cm.

Solution:

Here, the internal radius of teh cylindrical Vessel (r) = 10.1 cm

the external radius of the vesel (R) = 10.5 cm

the length of the vessel (h) = 56 cm

Now, Volume of metal contained by the vessel (V) = πh(R + r) (R − r)
22
= 7 × 56 (10.5 + 10.1) (10.5 − 10.1)

= 22 × 8 × 20.6 × 0.4 = 1450.24 cm3

∴ Cost of metal contained by the vessel (T) = V × R = 1450.24 × Rs 2 = Rs 2900.48
22
Again, the external curved surface area (A) = 2πRh = 2 × 7 × 10.5 × 56 = 3696 cm2

∴ Cost of polishing the outer surface (T) = A × R = 3696 × 20 paisa = 73920 paisa

= Rs 739.20

16. Agatehastwocylindricalpillarswithahemispherical

end on a top of each pillar. The height of each pillar

is 9.96 cm and the height of each cylinder is 9.75 m. 9.96 m
9.75 m
Find the cost of colouring the surface of both pillars 9.75 m

at Rs 500 per sq.m. 9.96 m

Solution:

Here, For a pillar height of cylindrical part (h) = 9.75 m

height of hemispherical part = radius of base (r) =

9.96 m − 9.75 m = 0.21 m

Now, C.S.A of a pillar = C.S.A of hemisphere + C.S.A of cylinder
22
= 2πr2 + 2πrh = 2πr(r + h) = 2 × 7 × 0.21(0.21 + 9.75) = 13.1472 m2

∴ Total surface area of two pillars = 2 × 13.1472 m2 = 26.2944 m2

Again, cost of colouring the surface of the both pillar (T) = A × R = 26.2944 × Rs 500

= Rs 13,147.20

17. The adjoining solid is a half portion of a cylinder obtained by 35cm

cutting vertically through its height. Find the curved surface

area, total surface area and volume of the solid. 7cm
Solution:

In the given half-cylinder; radius of base (r) = 7 cm and height (h) = 35 cm

Now,

Curved surface area = πrh + 2rh =πr2272=×2727××735×+352×+ 7 × 35 = 1260 cm2
2rh +
Total surface area = πrh + 2× 7 × 35 + 22 × 72 = 1414 cm2
7
Volume = 1 πr2h = 1 × 22 × 72 × 35 = 2695 cm3 35cm
2 2 7

18. The adjoining cylindrical vessel is 70 cm high and the radius 70cm 20cm
of its base is 35cm. If it contains some water up to the height
of 20 cm, how much water is required to fill the vessel
completely?

Solution:

53 Vedanta Excel in Mathematics Teachers' Manual - 10

In the given cylindrical vessel; radius of base (r) = 35 cm and height = 70 cm

Height of empty space of the vessel above the water surface (h) = 70 cm – 20 cm = 50 cm

Now,

Volume of empty space of the vessel = πr2h t=he27v2e×sse3l5c2 o×m5p0le=tel1y9=251091002c05m0003 litre = 192.5 l
Hence, the quantity of water required to fill

19. 50 circular plates, each of radius 7 cm and thickness 5 mm are placed one above

the other to form a cylindrical shape. Find the volume of the cylinder so formed.

Solution

Radius of base (r) = 7 cm and height of 50 plates = 50 × 5mm = 250 mm = 25 cm

Now,

Volume of the cylindrical shape = πr2h = 22 × 72 × 25 = 3850 cm3
7

20. A cylindrical tube of radius 16 cm contains water to certain depth. When a

spherical ball is dropped into the tube, the level of water is raised by 9 cm. Find

the radius of the ball.

Solution

Radius of cylindrical tube (R) = 16 cm

Height of water raised after dropping a spherical ball = 9 cm

Radius of ball (r) =?

Now,

Volume of the spherical ball = Volume of cylindrical tube

or, 4 πr3 = πR2h
3
4
or, 3 × r3 = 162 × 9

or, r3 = 1728

or, r3 = 123

or, r = 12

Hence, the radius of the ball is 12 cm.

21. A cylinder whose height is two-third of its diameter, has the same volume as a

sphere of radius 21 cm. Calculate the curved surface area of the cylinder.

Solution

Let the radius of the cylinder = r cm

Then, height (h) = 2 of diameter = 2 of 2r = 4r cm
3 3 3
Radius of sphere (R) =21 cm

Now, volume of the cylinder = Volume of sphere

or, πr2h = 4 πR3
3
4r 4
or, r2 × 3 = 3 × 213

or, r3 = 213

∴ r = 21 and h = 4r cm = 4 × 21 cm = 28 cm
3 3
Again,

Vedanta Excel in Mathematics Teachers' Manual - 10 54

Curved surface area of the cylinder = 2πrh = 2 ×272 × 21 × 28 = 3696 cm2
22. The surface area of a ball is twice the area of the curved surface area of a cylinder.

If the height and radius of base of the cylinder be 10.5 cm each, find the volume of

the sphere.

Solution

Let the radius of the ball = R cm

Radius of base of the cylinder (r) = 10.5 cm, height of cylinder (h) = 10.5 cm

According to the question

4πR2 = 2×2πrh or, R2 = 10.5 ×10.5 ∴ R = 10.5 cm

Now, volume of the sphere = 4 πR3 = 4 × 22 ×10.53 = 4851 cm3
3 3 7

23. A hemispherical bowl with radius 21 cm is full of water.

If the water is poured into a 1m tall jar with the diameter

of the base 28 cm, find the height of the water level in

the jar. h

Solution

In the given hemispherical bowl; radius of base (r) = 21 cm

Diameter of cylindrical jar = 28 cm ∴ radius of the jar (R) = 14 cm

Height of water level in the jar (h) =?

Now,

Volume of the cylindrical jar = Volume of hemispherical bowl

or, πR2h = 2 πr3
3
2
or, 142 ×h = 3 × 213

or, h = 31.5 cm

Hence, the height of the water level in the jar is 31.5 cm

Extra Questions

1. The total surface area and curved surface area of a cylindrical can are 748 cm2 and 440
cm2 respectively. Find its volume. [Ans: 1540 cm3]
2. Some circular plates of same size, each of diameter 14 cm and thickness 5 mm are
placed one over other to form a right cylinder having a volume of 7700 cm3, find the
number of plates. [Ans: 100]
3. The sum of the height and radius of base of a cylinder 14 cm. If the curved surface
area of the cylinder is 231 square centimetres, find the volume of the cylinder when its
height is less than the radius of base. [Ans: 1212.75 cm3]
4. The height and the total surface area of a solid log composed of a cylinder and a
hemisphere at one end are 19 cm and 2288 cm2 respectively, find the he radius of its
base. [Ans: Rs 55,440]
5. A roller of diameter 1.4 m and length 2m takes 600 complete revolutions to level a
playground. Calculate the cost of gravelling the ground at Rs 10.50 per square meter.
6. A water-well is composed of 28 identical cement rings with 1.4m diameter and 0.2m
height.
(i) If the labour cost for digging out the well is Rs. 500 per cubic metre and the cost of a
cement ring is Rs 2500, estimate the total cost of well construction.
(ii) If the water level is found up to 20 rings from the bottom, find how many litres of water

55 Vedanta Excel in Mathematics Teachers' Manual - 10

Unit Mensuration (II): Prism and Pyramid

7
were in the well? Find it. [Ans: Rs 74, 312, 6,160 lt.]

Competency Allocated teaching periods 11

- Solving the problems related to surface area and volume of regular solid objects (prism,

pyramid, cylinder, sphere, hemisphere and cone)

Learning Outcomes

- To solve the problems related to surface area and volume of regular solid objects

(prism, pyramid, cylinder, sphere, hemisphere and cone)

Level-wise learning objectives

S.N. LEVELS OBJECTIVES

- To define prism
- To recall the area formula of triangles
- To tell the formula of finding the volume, lateral surface
area and total surface area of triangular prism
- To define pyramid
- To relate the side of base (a), height (h) and slant height
(l) of square based pyramid
1. Knowledge (K) - To tell the formula of finding the volume and total

surface area of square based pyramid
- To label radius (r), height (h) and slant height (l) of cone
- To relate the radius of base (r), height (h) and slant
height (l) of cone
- To list the formula of finding the volume , C.S.A. and
T.S.A. of cone

- To find the volume of triangular prism
- To find the area of rectangular faces of prism
- To calculate the area of total surface area of prism
2. Understanding (U) - To find the volume, LSA and TSA of square based

pyramid
- To find the volume, LSA and TSA of cone with the

given measurements

- To apply the related formula to solve the problems of
triangular prism

3. Application (A) - To find the CSA, TSA and volume of combine d solid
figure composed of cone and hemisphere, cylinder and
cone, pyramid and cuboid etc.

4. High Ability (HA) - To estimate the cost and quantity related the solid
objects (gate, compound wall, well etc.)

Required Teaching Materials/ Resources
Pencil, scissors, models of triangular prisms, nets of triangular prism, models of cone and
square based pyramid, model of combined solids, colourful chart-paper with definitions
and formulae etc.

Teaching learning strategies

Vedanta Excel in Mathematics Teachers' Manual - 10 56

Pre-knowledge: Perimeter and area of triangle, rectangle and circle.

A. PRISM

Teaching Activities

1. With manipulative models of prisms, discuss upon the prism

and its base and lateral surface.

2. Discuss with models on volume of prism as the product of cross-sectional area and the

height i.e., V = A × h Rectangle 1
3. Make clear on area of rectangular faces or lateral surface

area (L.S.A) as the product perimeter of base and height Rectangle 2
i.e., L.S.A = (a + b + c) × h = P × h and the total surface Rectangle 3
area = L.S.A + 2A = P × h + 2A by using models or nets

of prism. O

B. PYRAMID e
Teaching Activities

1. With manipulative models of prisms, hl

discuss upon the pyramid and its

base and lateral surface. P

2. Explain the relation among a, h and l as
a 2 + h2
l2 = 2 a 2 + l2 a
2
3. Explain the relation among a, l, e as e2 =

4. List the following formulae:

- Area of base (A) = a2 1 1 1 1 16d2h
3 3 3 2
- Volume of pyramid (V) = area of base (A) × height (h) = a2h or, × d2h =

- Area triangular faces or lateral surface area (L.S.A) = 4 × 1 al = 2al
2
- Total surface area (T.S.A.) = a2 + 2al

C. CONE

Teaching Activities

1. Using manipulative models of cone describe about the cone and its base and curved

surface.

2. Explain the relation among r, h and l

as l2 = r 2 + h 2

3. List the following formulae:

- Area of base (A) = πr2 1 1
3 3
- Volume of pyramid (V) = area of base (A) × height (h) = πr2h

- Curved surface area (C.S.A) = = πrl

- Total surface area (T.S.A.) = Area of base (A) + C.S.A. = πr2+ πrl = πr (r + l)

4. Using manipulative models of combined solid ask and explain the volume and surface

area of the solid.

Solution of selected questions from Vedanta Excel in Mathematics

1. The area of cross section of a triangular prism is 126 cm2 and its volume is 5040
cm3. If the perimeter of the base is 54 cm. find its

57 Vedanta Excel in Mathematics Teachers' Manual - 10

i) lateral surface area (ii) total surface area

Solution:

Here, Cross - sectional area (A) = 126 cm2

Volume (V) = 5040 cm3

Perimeter of base (P) = 54 cm

Now, Volume (V) = cross - sectional area (A) × height (h)

or, 5040 cm3 = 126 cm2 × h ∴ h = 40 cm

i) L.S.A. of triangular prism = Perimeter of base (P) × height (h)

= 54 cm × 40 cm = 2160 cm2

ii) T.S.A of triangular prism = L.S.A × 2A = 2160 cm2 + 2 × 126 cm2 = 2412 cm2

2. The volume of an isosceles right angled triangular prism is 1000 cm3. If the length

of the prism is 20 cm, find (i) its lateral surface area (ii) its total surface area

Solution:

Here, the triangular base is isosceles right angled triangle. Let the equal sides of base be

x cm. area of base (A) = 1 b × p = 1 × x×x = x2 cm2
Now, 2 2 2

We have, volume (V) = Area of base (A) × height (h)
x2
or, 1000 cm3 = 2 × 20 10 ∴ x = 10

∴ p = b = 10 cm and h = p2 + b2 = 102 + 102 = 10 2

Perimeter of base (P) = (p + b + h)

= (10 cm + 10 cm + 10 2 cm) = (20 + 10 2 ) cm = 34.14 cm

i) L.S.A. of the prism = P × h = 34.14 cm × 20 cm = 682.8 cm2

ii) T.S.A. of the prism = P × h + 2A 1
2
= 34.34 × 20 + 2 × × b × p = 682.8 cm2 + 10 × 10 = 782.8 cm2

3. The area and perimeter of the base of a triangular prism are 24 cm2 and 24 cm

respectively. If the total surface area of the prism is 408 cm2, find its lateral

surface area and volume.

Solution:

Here, in the triangular prism; area of base (A) = 24 cm2

perimeter of base (P) = 24 cm

T.S.A. = 408 cm2

Now, T.S.A of triangular prism = P × h + 2A
or, 408 = 24 × h + 2 × 24 or, h = 15 cm
Again, L.S.A. of prism = P × h = 24 cm × 15 cm = 360 cm2

Volume of prism = A × h = 24 cm2 × 15 cm = 360 cm2

4. Find the total surface area and volume of the given square - based pyramids.

OO

10 cm
Dh C D C
QP AP 24 cm B
25 cm
A 16 cm B
Solution:

Vedanta Excel in Mathematics Teachers' Manual - 10 58

a) Here, length of side of square base (a) = 16 cm

slant height (l) = 10 cm

Now, total surface area of the pyramid = a2 + 2al = 162 + 2 × 16 × 10 = 576 cm2

Let OQ be the height of the pyramid
1 1
Then, PQ = 2 AB = 2 × 16 cm = 8 cm

In right angled ∆POQ, OQ = OP2 − PQ2 = 102 − 82 = 6 cm
1 1
Also, volume of the pyramid (V) = 3 a2h = 3 × 16 × 16 × 6 = 512 cm3
Alternative method:
HWVoeelruehm,asvelea(,nVt)2ah=2e+i13ghah22th(=O=Pl2) 21=×2o14r6,c×m1261, 62+e×dhg62e
= 102 ∴ h = 6 cm

∴ = 512 cm3
b) (OB) = 25 cm

Now, in right angled ∆DPB; PB = OB2 − OP2 = 252 − 242 = 7 cm
∴ AB(a) = 2 × PB = 2 × 7 cm = 14 cm

Also, T.S.A. of pyramid = a2 + 2al = 142 + 2 × 14 × 24 = 868 cm2
We have, (2a)2 + h2 = l2 or, (124)2 + h2 = 242
or, h2 = 576 − 49 = 527 ∴ h = 22.96 cm
31a2h 1
Again, volume of the pyramid (V) = = 3 × 14 × 14 × 22.96 = 1500.05 cm3

5. The adjoining solid object is formed with the combination of a pyramid and a

square - based prism. If the volume of the solid object is 900 cu.cm, find the height

of the prism.

Solution:

Here, the given solid object is formed with the combination of a 13cm
pyramid and a square - based prism

For uppermost pyramid:

Length of side of squared base (a) = 10 cm

slant height (l) = 13 cm, height (h) = ?
a 2 10 2
We have 2 h2 = l2 or, 2 h2 = 132 10cm
+ +

or, h2 = 169 − 25 = 144 ∴ h = 12 cm 124
∴ForVololuwmeremoof sptypryamraimdi(dV:1) = 31a2h 1
= 3 × 10 × 10 × = 400 cm3

∴oHLArege,nVan9gioc0ntleh0u,, m=(tVlh1eo)e4l(=0uhV0me21i)+ge0=h1oct0fml1o0sf,×hobtl1bhird1eeoa×r(p,Vdhrht)ih11s=m=(=bVi151s)o0lc=5um×mc1m10e0co×mf ph, y1hrea=imghi1dt0(0(hVh11)1)=c+m?V3olume of prism (V2)

6. The solid given alongside is the combination of a square based pyramid and a

square -based prism. If the total surface area of the solid is 86cm2, find the volume

of the solid.

Solution:

Here

For uppermost square-based pyramid; 10 cm
length of side of base (a) = 12 cm

slant height = l cm 12 cm 12cm
∴For(Ll.oSw.Ae.r)1m=os2tapl r=ism2;× 12 × l = 24l cm2

59 Vedanta Excel in Mathematics Teachers' Manual - 10

∴LAerne(Lag.tSoh.fA(bl)1a2)s==e 12 cm, breadth 1(b01()1=2 +1212c)m=, h4e8ig0hctm(h21) = 10 cm
2h(l + b) = 2 ×

(A) = l2, = 12 cm × 12 cm = 144 cm2

Noro, w86, 4toctaml 2su=rf2a4cle+ar4e8a0ocfmth2e+s1o4li4dc=m2(L.S.A.)1 + (L.S.A.)2 + A

or, 24l = 240 ∴ l = 10 cm l2 – a2 = 102 – 12 2 = 8 cm
Also, height of pyramid (h) = 2 2

Again, volume of the solid (V) = Volume of pyramid + Volume of prism
1
= 3 a2h + l1 × b1 × h1 × 10 = 1824 cm3
= 1 × 124× 8 + 12 × 12
3

7. The gate lof stadium has two pillars, each of

height 10 ft. with four visible lateral faces 2 ft
2 ft
and 3 ft. × 3 ft. bases. The top of each pillar

has combined pyramid of height 2 ft. If the

combined structures of both pillars and 10 ft 10 ft

pyramid are painted at the rate Rs 80 per sq.ft,

calculate the cost of painting. 3 ft 3 ft

Solution: 3 ft 3 ft

Here,

For square - based pyramid;

length of side of base (a) = 3 ft

height (h) = 2ft a 2+ h2 = 3 2+ 22 = 2.25 + 4 = 6.25 = 2.5 ft
We have slant height (l) = 2 2

AN ∴∴losLL woaa,,tteeftorroaartllacssluuusbrruoffraaifdcca;eecleaaerrnaeergaaethaoo(ffol1pcf)uye=baraco3mhidfictd(,oLbm(.LrSbe..SaAin.dA.e)t2hd.)(1=sb=t12r)uh2=c(altlu3+r=febt )a2==n=×d(1L352h.S×+e×.iA21g1h.2)0510t((3+h= =+(1L)113.=S53) .5s1A=q0s).q2f1ft.t2ft0 sq.ft

∴Total surface area of both combined structure = 2 × 135 = 270

Again, rate of painting the structure (R) = Rs 80 per sq.ft

∴ Cost of painting = A × R = 270 × Rs 80 = Rs 21,600

8. 150 people can be accommodated inside a conical tent. If 3m2 space and 10 m3 air

is required for each person, find the height of the tent.

Solution:

Here,

Space required for 1 person on the ground = 3 m2

∴ Space required for 150 people on the ground = 150 × 3m2 = 450 m2

∴ Area of base of conical tent (A) = 450 m2

Also,

Volume of air required to breathe for 1 person = 10 m3

Volume of air required to breathe for 150 people = 150 × 10 m3 = 1500 m3

∴ Volume of conical tent (V) = Volume of air required for 150 people = 150 m3
1
Now, volume of conical tent (V) = 3 A × h

or, 1500 = 1 × 450150× h
∴h= 3
10

Vedanta Excel in Mathematics Teachers' Manual - 10 60

Hence the height of the tent is 10 m

9. A tent of height 33 m is in the form of a sight circular cylinder of

diameter 42 m and height 5 m surmounted by a right circular cone of

the same diameter. Find the total area of cloths required to cover the

tent. Also, find the cost of cloths at Rs 125 per sq.m. 5m
5m
Solution:

Here,

For conical Part ∴ radius (r) = 422m = 21 m
Diameter of base = 42 m = 28 m
Height (h) = 33 m − 5 m 42m

We have, slant height (l) = r2 + h2 = 212 + 282 = 35 cm
∴ Curved surface area (C.S.A)1 = πrl =272 × 21 × 35 = 2310m2
FNA∴ogorCawciu,nyrt,lvhirneeaddtterosiutocafarlflcaalpcroaeetrahats;orerf(aaRrde)(iCq=uu.sSi(Rr.rAes)d)=12=c22l5o12tpπhmerrh(Aamn)= 2=d2h(×Cei.2gS72h.At×.()2h111+)×(=C5.5Sm=.A6)26=02m3210 m2 + 660 m2 = 2970 m2

∴ Cost of required cloths (T) = A × R = 2970 × Rs 125 = Rs 3,71,250

10. The gate of a stadium has two cylindrical pillars, each of height 5 m and the

circumference of each base is 22 m. The top of each pillar has a combined cone of

height 1.2m. If the combined structures of both pillars and cones are painted at the

rate of Rs 100 per sq.m, find teh total cost of painting the pillars.

Solution: 1.2m
1.2m
Here, circumference of circular base = 22 m

or, 2πr = 22 m
22
or, 2× =7 ×r = 22 m
or, ∴r 3.5 m

Height of cylinder (h) = 5 m 5m
5m

Also, height of cone (h1) = 1.2 m, radius (r) = 3.5 m
∴ Slant height (l) = r2 + h12 = 3.52 + 1.22 = 3.7 cm

Now, the surface area of a tructure = C.S.A of cone + C.S.A. of cylinder

= πrl + 2πrh = πr(l + 2h)
22
= 7 × 3.5 (3.7 + 2 × 5) = 150.7 m2

∴ The surface area of two structures (A) = 2 × 150.7 m2 = 301.4 m2

Again, rate of painting the corbined structures (R) = Rs 100 per sq.m

∴ Cost of painting the pillar (T) = A × R = 301.4 × Rs 100 = Rs 30,140

11. The adjoining figure is a rectangular 25 m 12 cm
compound wall of a house. Estimate the cost 32 m
of building the wall by using bricks of size 20 5m
cm × 8 cm × 5 cm. The rate of the cost of the
brick is Rs 1,750 per 1,000 bricks. Labourer
wages is Rs 750 per labourer per day and 3
labourers are required for 4 days.

61 Vedanta Excel in Mathematics Teachers' Manual - 10

Solution:

Here, the outer length of the wall (l) = 32 m
the inner breadth of the wall = 25 m
the thickness of the wall (d) = 12 cm = 0.12 m
∴ the outer breadth of the wall (b) = 25 m + 2d = 25 m + 2 × 0.12 m = 25.24 m
the height of the wall (h) =5m

Now, the cross-sectional area of the wall (A) = 2d(l + b − 2d)

= 2 × 0.12 (32 + 25.24 − 2 × 0.12)

= 13.68 m2

∴ Volume of the wall (V) = cross - sectional area (A) × height(h)
= 13.68 m2 × 5 m = 68.4 cm3
= 68.4 × 100 × 100 × 100 cm3
= 68400000 cm3

Also Volume of each brick (v) = 20 cm × 8 cm × 5 cm = 800 cm3

∴ Number of bricks required (N) = Volume of wall (V)
Volume of each brick (v)

= 68400000 cm3
800 cm3

= 85,500

Again, the cost of 1,000 bricks = Rs 1,750

the cost of 1 brick = Rs 1,750
1000
1,750
∴ the cost of 85,500 brick = Rs 1000 × 85,500 = Rs 1,49,625

Also, labour wages for 3 labourers per day = 3 × Rs 750 = Rs 2,250

∴ labourer wages for 3 labourers for 4 days = 4 × Rs 2,250 = Rs 9,000

Hence, the total estimation of building the compound wall = Rs 1,49,625 + Rs 9,000

= Rs 1,58,625

12. The adjoining solid is a crystal formed by the combination of 16 cm

two equal square based pyramid of side 12 cm. If the height

of the crystal is 16 cm, find the total surface area and vol- 12 cm
ume.

Solution

In one of the equal square based pyramids,

Side length of square base (a) = 12 cm and height (h) = 8 cm

We have, slant height (l) = a 2+ h2 = 12 2+ 82 = 10 cm
2 2
Now,

L.S.A. = 2al = 2×12 cm × 10 cm = 240 cm2

∴Total surface area of the crystal = 2× L.S.A. = 2×240 cm2 = 480 cm2

Also,

Vedanta Excel in Mathematics Teachers' Manual - 10 62

Volume of a pyramid = 13a2h = 1 × 122 × 8 = 384 cm3
3

∴ Volume of the crystal (V) = 2 × 384 cm3 = 768 cm3

13. In the given solid there are two pyramids of the same size and 6cm
shape at the ends of a cubical solid. If the height of one pyramid
is 6 cm and that of the cubical solid is 16 cm, find the total sur- 16cm
face area of the whole solid.

Solution

In one of the equal square based pyramids,

side length base (a) = 16 cm and height (h) = 6 cm

Edge of cubical part (l) = 16 cm

Volume of solid (V) =?

Now,

Volume of combined solid = volume of two pyramids + volume of cube

=2× 1 a2h + l3 =2× 1 × 162 × 6 + 163 = 5,120 cm3
3 3

14. In the given solid, there are two pyramids of the same size 6cm
and shape at the ends of a cubical solid. If the height of one
pyramid is 6 cm and that of the cubical solid is 16 cm, find 16cm
the volume of the whole solid.

Solution 28cm

In one of the equal square based pyramids,

side length of base (a) = 16 cm and height (h) = 6 cm

We have, slant height (l) = a 2+ h2 = 16 2+ 62 = 10 cm
2 2
Now,

L.S.A. = 2al = 2×16 cm × 10 cm = 320 cm2

∴L.S.A. of two equal pyramids = 2× L.S.A. = 2×320 cm2 = 640 cm2

Also, side length of cube = 16 cm

∴L.S.A. of cube = 4l2 = 4× 162 = 1024 cm3

Hence, the total surface area of the solid = L.S.A. of (cube + 2 equal pyramids)

= 640 cm2 + 1024 cm2 = 1,664 cm2

15. In the figure, a right circular cylinder has two conical 42cm 56cm
ends with radius of the base 42 cm and height 56 cm.
If the solid is 2.1 m long, find its total surface area 2.1 m
and volume.

Solution
For conical end,
Radius of base (r) = 42 cm and height (h) = 56 cm

We have, slant height (l) = r2 + h2 = 422 + 562 = 70 cm

Now,

63 Vedanta Excel in Mathematics Teachers' Manual - 10

(i) C.S.A. = πrl = 22 × 42 cm × 70 cm = 9,240 cm2
7

∴C.S.A. of two equal conical ends = 2× C.S.A. = 2×9,240 cm2 = 18,480 cm2

(ii) Volume of a conical end = 31πr2h =13 ×272×422×56 = 103488 cm3

∴Volume of two ends = 2× 103488 cm3 = 206976 cm3

For cylindrical part

Radius of base (r) = 42 cm and height (h) = 210 cm – 56 cm – 56 cm = 98 cm

Now,

(i) C.S.A. = =2ππrrh2h==2×27227×2 4×224×29c8m=×549383c1m2 = 25,872 cm2
(ii) Volume cm3

Thus, T.S.A. of the combined solid = C.S.A. of (cylinder + two conical ends)

= 25,872 cm2 + 18,480 cm2 = 44,352 cm2

Volume of the combined solid = Volume of (cylinder + two conical ends)

= 543312 cm3+ 206976 cm3 = 7,50,288 cm3

42cm 72cm

16. In the adjoining solid, one of its ends is a cone and 143cm
the other opposite end is a hemisphere with diameter
42 cm. Calculate the total surface area and the
volume of the solid.

Solution

For hemispherical end,

Radius of base (r) = 21 cm

∴ C.S.A.1 = 2r2 = 2 × 22 × 212 = 2772 cm2
7
23πr3 2 22
Volume (V1) = = 3 × 7 × 213 = 19404 cm3

For conical end,

Radius of base (r) = 21 cm and height (h) = 72 cm

We have, slant height (l) = r2 + h2 = 212 + 722 = 75 cm

∴C.S.A.2 = πrl = 22 × 21×75 = 4950 cm2
7
31πr2h 1 ×272×212×72
Volume (V2) = = 3 = 33264 cm3

For cylindrical part

Radius of base (r) = 21 cm and height (h) = 143 cm – 21 cm – 72 cm = 50 cm

Now,

∴C.S.A.3 = 2πrh =2 × 22 × 21 cm × 50 cm = 6600 cm2
7
22
Volume (V3) = πr2h = 7 × 212 × 50 = 69300 cm3

Thus, T.S.A. of the combined solid = C.S.A.1 + C.S.A.2 + C.S.A.3
= 2772 cm2+ 4950 cm2 + 6600 cm2 = 14,322 cm2

Vedanta Excel in Mathematics Teachers' Manual - 10 64

Volume of the combined solid = V1 + V2 + V3 20 cm

= 19404 cm3 + 33264
cm3 + 69300 cm3 = 1,21,968 cm3

17. In the adjoining figure, a conical vessel with 30 cm ?

diameter of its circular base 20 cm is completely

filled with water. If the water is poured

completely into a cylindrical jar of the same diameter, find the height of water

level in the jar.

Solution

Radius of base of cone (r) = 10 cm

Height of cone (h) = 30 cm

Radius of base of cylindrical jar (R) = 10 cm

Height of water level in jar (H) =?

Now, volume of water in cone = Volume of water in jar or, H = 10 cm
or, 13πr2h = πR2H or, 13×102×30 = 102×H

Hence, the height of water level in the jar is 10 cm.

5 cm
18. A hemispherical bowl with diameter 5 cm is completely
10 cm
filled with water. If the water is completely poured ?
into a right circular conical vessel with diameter of
cirular level of water is 10cm, find the height of the
level of water in the vessel.

Solution

Radius of base of hemispherical bowl (r) = 2.5 cm

Radius of circular water surface (R) = 5 cm

Height of water level in jar (h) =?

Now,

Volume of water in hemispherical bowl = Volume of water in conical vessel
or, 23πr3 = 31πR2h
or, 2×2.53= 52×h or, h = 1.25 cm

Hence, the height of water level in the conical vessel is 1.25 cm.

Extra Questions

1. Mr. Rai builds a square based tent 24 feet high for the accommodation of the guests for
his daughter’s marriage ceremony. If each edge of the tent is 25 feet and each people
requires 16 cu feet of air to breathe. How many people can be arranged inside the tent?
[Ans: 49]

2. The roof of a temple is a square based pyramid. The vertical height of the roof is 3 m and
slant height is 5m. How much zinc plate is required to cover the roof of a temple? Estimate
the cost of zinc plates required at the rate of Rs 110 per sq. meter.
[Ans: 80 sq. m, Rs 8800]

3. Khushbu wants to fill the ice-cream cones with chocolate cream making hemispherical
tops. For this, she brings some cones each of having base diameter 4.2 cm and slant

65 Vedanta Excel in Mathematics Teachers' Manual - 10

Unit Highest Common Factor (H.C.F.) and
Lowest Common Multiples (L.C.M.)
8

Allocated teaching periods 6
Competency

- To find the H.C.F. and L.C.M. of algebraic expressions by factorization method

- To solve the contextual problems related to H.C.F. and L.C.M.

Learning Outcomes
- To find the H.C.F. and L.C.M. of algebraic expressions by factorization method

Level-wise learning objectives

S.N. LEVELS OBJECTIVES

- To define H.C.F. of given expressions
1. Knowledge (K)

- To define L.C.M. of given expressions

2. Understanding (U) - To factorise the simple expressions and determine the
H.C.F. and L.C.M.

3. Application (A) - To find the H.C.F. and L.C.M.

4. High Ability (HA) - To relate the contextual problems in to H.C.F. and L.C.M.
and solve them

Required Teaching Materials/ Resources
Charts with formulae, models of Venn-diagram, audio-video materials etc

Pre-knowledge: Factorisation of algebraic expressions
Teaching Activities
1. Warm up the class with recalling highest common factor and lowest common multiples

of the numbers as learnt in the arithmetic in basic level.
2. Recall the following formulae

S.N. Expressions Factorised form Expanded form
1. (a+ b)2 (a + b) (a +b) a2 + 2ab + b2 or (a – b)2 + 4ab
2. (a – b)2 (a - b) (a -b) a2 - 2ab + b2 or (a + b)2 – 4ab
3. a2 – b2 (a + b) (a – b)
4. a2 + b2 (a + b)2 - 2ab or (a – b)2 + 2ab
(a+ b)3 (a + b) (a +b)(a + b) a3 + 3a2b + 3ab2+ b3 or a3 + b3 + 3ab
5. (a + b)
a3 – 3a2b + 3ab2– b3 or a3 – b3 – 3ab
6. (a – b)3 (a – b) (a–b)(a – b) (a – b)

Vedanta Excel in Mathematics Teachers' Manual - 10 66

7. a3 + b3 (a + b) (a2 – ab+ b2) (a + b)3 – 3ab(a + b)
8. a3– b3 (a – b) (a2 + ab+ b2) (a – b)3 + 3ab(a – b)

3. Divide the students into four five groups and give some real life problems related to
factorisation

Group A: A students distributed x2 + 5x + 6 chocolates equally among some of his/her
friends, find the possible number of his/her friends and the number of chocolates
received by each

Group B: The area of a rectangular floor is (3x2 + 12xy) m2, find the length and breadth of
the room.

Group C: The area of a rectangular plot is (x2 – 9y2) ft2, find the length and breadth of the plot.

Group D: The volume of a rectangular room is (x3 – 8y3) m3, find the height and base area of
the room.

4. Discuss on following process of the factorisations

(i) Taking common (ii) Using proper formula

(iii) Splitting the middle terms (iv) Completing the perfect squares

5. Explain H.C.F. and L.C.M. with proper examples by using the models of intersection and
union of sets in Venn-diagrams. For example show the H.C.F. and L.C.M. of x2 – 4 and
x2 + 2x + 3 using chart paper

x2 – 4 x2 + 2x + 3 x2 – 4 x2 + 2x + 3

x–2 x+2 x+1 x–2 x+2 x+1

H.C.F. L.C.M.

From Venn-diagram, H.C.F. = x + 2 and L.C.M. = (x + 2) (x – 2) (x + 1)

Divide the students and ask to find the H.C.F. and L.C.M. of similar expressions by
showing in the Venn-diagram in chart paper.

6. Define H.C.F. as the algebraic term/s having highest degree and largest coefficient that
divides all the given expressions without remainders

7. Define L.C.M. as the algebraic term/s having lowest degree and least coefficient which
is exactly divisible by the given expressions.

8. While factorising the expressions, raise the questions like how?, why?, what comes
then? in every steps of solving problems.

9. Guide the students to find the highest common factor (H.C.F.) and lowest common
multiples (L.C.M.) of the expressions from the book

10. Call the students in front of the class to solve the related problems on the board.

11. Give some contextual problems related to the lesson and make discussion in the group.

67 Vedanta Excel in Mathematics Teachers' Manual - 10

Solution of selected questions from Vedanta Excel in Mathematics

1. Find the HCF of 16x4 − 4x2 − 4x − 1 and 8x3 − 1

Solution:

Here, the 1st expression = 16x4 − 4x2 − 4x − 1

= 16x4 − (4x2 + 4x + 1) = 16x4 − [(2x)2 + 2.2x.1 + (1)2]

= (4x2)2 − (2x + 1)2 = (4x2 + 2x + 1) (4x2 − 2x − 1)

The 2nd expression = 8x3 − 1 =(2x)3 − (1)3 = (2x − 1) (4x2 + 2x + 1)

∴ H.C.F = 4x2 + 2x + 1

2. Find the HCF of 9x2 − 4y2 − 8yz − 4z2, 4z2 − 4y2 − 9x2 − 12xy, 9x2 + 12xz + 4z2 − 4y2

Solution:

Here, the 1st expression = 9x2 − 4y2 − 8yz − 4z2 = 9x2 − (4y2 + 8yz + 4z2)

= 9x2 − [(2y)2 + 2.2y.2z + (2z)2 ] = (3x)2 − (2y + 2z)2 = (3x + 2y + 2z) (3x − 2y − 2z)

the 2nd expression = 4z2 − 4y2 − 9x2 − 12xy = 4z2 − (4y2 + 9x2 + 12xy)

= 4z2 − [(2y)2 + 2.2y.3x + (3x)2] = (2z)2 − (2y + 3x)2

= (2z + 2y + 3x) (2z − 2y − 3x) = (3x + 2y + 2z) (2z − 2y − 3x)

the 3rd expression = 9x2 + 12xz + 4z2 − 4y2

= (3x)2 + 2.3x.2z + (2z)2 − (2y)2 = (3x + 2z)2 − (2y)2 = (3x + 2y + 2z) (3x − 2y + 2z)

∴ H.C.F = (3x + 2y + 2z)

3. Find the H.C.F and L.C.M of x2 − 10x − 11 + 12y − y2, x2 − y2 + 22y − 121
Solution:

Here, the 1st expression = x2 − 10x − 11 + 12y − y2

= x2 − 2.x.5 + 52 − 52 − 11 + 12y − y2

= (x − 5)2 − y2 + 12y − 36 = (x − 5)2 − (y2 − 12y + 36) = (x − 5)2 − (y2 − 2.y.6 + 62)

= (x − 5)2 − (y − 6)2 = (x − 5 + y − 6) [(x − 5) − (y − 6)] = (x + y − 11) (x − y + 1)

the 2nd expression = x2 − x2 + 22y − 121 = x2 − (y2 − 22y + 121)

= x2 − (y2 − 2.y.11 + 112) = x2 − (y − 11)2 = (x + y − 11) (x − y + 11)

∴ H.C.F =(x + y − 11) and L.C.M =(x + y − 11) (x − y + 1) (x − y + 11)

4. Find the H.C.F and L.C.M of 1 + 2x − 6x3 − 9x4, 9x4 + 2x2 + 1

Solution:

Here, 1st expression = 1 + 2x − 6x3 − 9x4

= 1 − 9x4 + 2x − 6x3 = (1)2 − (3x2)2 + 2x (1 − 3x2)

= (1 + 3x2) (1 − 3x2) + 2x(1− 3x2) = (1 − 3x2) (1 + 3x2 + 2x) = (1 − 3x2) (1 + 2x + 3x2)

2nd expression = 9x4 + 2x2 + 1 = (3x2)2 + (1)2 + 2x2 = (3x2 + 1)2 − 6x2 + 2x2

= (3x2 + 1)2 − (2x)2 = (3x2 + 2x + 1) (3x2 − 2x + 1)

∴ H.C.F = 3x2 + 2x + 1 and L.C.M = (1 − 3x2) (3x2 + 2x + 1) (3x2 − 2x + 1)

= (1 − 3x2) (9x4 + 2x2 + 1)

Extra questions for practice

A. Find the H.C.F and L.C.M of the following expressions

1. x3 + 1 + 2x2 + 2x, x3 − 1, x4 + x2 + 1 [Ans: x2 + x + 1, L.C.M. = (x2 – 1) (x4 + x + 1)]

2. (a + b)2 − 4ab, a3 − b3, a2 + ab − b2

[Ans: H.C.F. = (a – b), L.C.M. = (a + b)2 (a + 2b) (a2 + ab + b2)]

3. 16x4 − 4x2 − 4x − 1, 8x3 − 1, 16x4 + 4x2 + 1

[Ans: H.C.F. = 4x2 + 2x + 1, L.C.M. = (8x3 – 1) (4x2 – 2x + 1)]

4. p4 + 10p2 + 169, p3 + p (p + 13) + 3p2, 3p2 + 4(3p + 5) + 19

[Ans: H.C.F. = p2 + 4p + 13, L.C.M. = 3p(p4 + 10p2 + 169)]

5. x4 + (2b2 − a2) x2 + b4, x4 + 2ax3 + a2x2 − b4
[Ans: H.C.F. = x2 + ax + b2, L.C.M. = (x2 + ax + b2) (x2 – ax + b2) (x2 + ax – b2)]

Vedanta Excel in Mathematics Teachers' Manual - 10 68

Unit Simplification of Rational Expressions

9

Allocated teaching periods 5

Competency

- To simplify the algebraic fractions
Learning Outcomes

- To simplify the problems related to rational expressions
Level-wise learning objectives

S.N. LEVELS OBJECTIVES

- To define algebraic rational expressions

1. Knowledge (K) - To tell the condition under which the expression

become the rational expression

2. Understanding (U) - To simplify the simple rational expressions

3. Application (A) - To simplify the algebraic rational expressions

4. High Ability (HA) - To prove the rational expressions under the given
conditions

Required Teaching Materials/ Resources

Chart paper, scissors, colourful markers, ICT tools, audio-video materials etc.

Pre-knowledge: L.C.M. of expressions

Teaching Activities

1. Recall the lowest common multiples (L.C.M.) of the algebraic expressions
2. Explain about the existence of rational expressions
3. Simplify the expressions and give the same type problems as class-work in groups

Solution of selected questions from Vedanta Excel in Mathematics

1. Simplify a2 – (b – c)2 + b2 – (a – c)2 + c2 – (a – b)2
Solution: (a + c)2 – b2 (a + b)2 – c2 (b + c)2 – a2

a2 – (b – c)2 + b2 – (a – c)2 + c2 – (a – b)2
(a + c)2 – b2 (a + b)2 – c2 (b + c)2 – a2

= (a + b – c) (a – b + c) + (b + a – c) (b – a + c) + (c + a – b) (c – a + b)
(a + c + b) (a + c − b) (a + b + c) (a + b − c) (b + c + a) (b + c − a)

= a + b – c + b – a +c + c + a − b =aa + b + c =1
a + b +c + b + c

2. Simplify x+1 + x−1 − 1
Solution: 2x3 − 4x2 2x3 + 4x2 x2 − 4

x+1 x−1 1 = x+1 + x−1 1
2x3 − 4x2 + 2x3 + 4x2 − x2 − 4 2x2 (x − 2) 2x2 (x + 2) − x2 − 4

= (x + 1) (x + 2) + (x − 1) (x − 2) − (x 1 2) = x2 + 2x + x+ 2 + x2 − 2x − x − 2 − 2x2 = 0
2x2 (x − 2) (x + 1) + 2) (x 2x2 (x + 2) (x − 2)
−

69 Vedanta Excel in Mathematics Teachers' Manual - 10

3. Simplify 3x − 1 3x + 1 54x3
9x2 − 3x + 1 − 9x2 + 3x + 1 + 81x4 + 9x2 + 1
Sol3uxti−on1:
9x2 − 3x + 1 − 3x + 1 + 54x3 + 1
9x2 + 3x + 1 81x4 + 9x2

= (3x − 1) (9x2 + 3x + 1) − (3x + 1) (9x2 − 3x + 1) + 54x3
(9x2 − 3x + 1) (9x2 + 3x + 1) (9x2)2 + (1)2 + 9x2

= (3x)3 − (1)3 − [(3x)3 + (1)3] + (9x2 54x3 (3x)2
(9x2 − 3x + 1) (9x2 + 3x + 1) + 1)2 −

= 27x3 − 1 − 27x3 − 1 1) 54x3
(9x2 − 3x + 1) (9x2 + 3x + + (9x2 + 3x + 1) (9x2 − 3x + 1)

= 54x3 − 2 3x + 1) = 2 [(3x)3 − (1)3]
(9x2 + 3x + 1) (9x2 − (9x2 + 3x + 1) (9x2 − 3x + 1)

= 2(3x − 1) (9x2 + 3x + 1) = 2(3x − 1)
(9x2 + 3x + 1) (9x2 − 3x + 1) 9x2 − 3x + 1)

4. Simplify: 1 − 1 + 1 + 1
Solution: x−5 x−3 x+5 x+3

1111
x−5 − x−3 + x+5 + x+3

= x 1 5 − 1 − x 1 3 + x 1 3 = x + 5+x − 5 − x+3+x−3
− x+5 − + (x + 5) (x − 5) (x + 3) (x − 3)

= x2 2x 2x = 2x3 − 18x − 2x3 + 50x = (x2 − 32x − 9) = (x2 − 32 − 25)
− 25 − x2 − 9 (x2 − 25) (x2 − 9) 25) (x2 9) (x2
b 2ab a 4a3b
5. Simplify: 1 + a − b + a2 + b2 − a + b + a4 + b4
Solution:

1 + b + 2ab − a + 4a3b = a − b + b − a + 2ab + 4a3b
− a2 + b2 + a4 + b4 a − b + a2 + b2 a4 + b4
a b a b a b

= a2 + ab − a2 + ab + 2ab 4a3b = 2a3b + 2ab3 + 2a3b − 2ab3 4a3b
(a − b) (a + b) a2 + b2 + a4 + b4 + a4 + b4
(a2 − b2) (a2 + b2)

= 4a7b + 4a3b5 + 4a7b − 4a3b5 8a7b
(a4 − b4) (a4 + b4) = a8 − b8

Extra questions for practice

1. a2 + b2 − a2 − b2 b) [Ans: 1]
ab b(a + b) a(a +

2. a2 a−x x2 + a+x − a4 + 2x2 + x4 [Ans: 2(­a – x) ]
− ax + a2 + ax + x2 a2x2 a2 – ax + x2

3. p+q + q+r + (r − r+p − q) [Ans: 0]
(p − r) (q − r) (q − p) (r − p) q) (p

4. a+b + b−c + b2 c+ a a)2 [Ans: 0]
(a + b)2 − c2 a2 − (b − c)2 − (c +

5. 1 + 1 (x + y − z) + 1 + 1 (y + z − x) + 1 + 1 (z + x − y) [Ans: 6]
x y y z z x

Vedanta Excel in Mathematics Teachers' Manual - 10 70

Unit Indices

10

Allocated teaching periods 6

Competency
- To simplify the expressions involving indices and solve the exponential equations

Learning Outcomes
- To simplify the expressions by using the laws of indices related to negative and
fractional power
- To solve the exponential quadratic equations

Level-wise learning objectives

S.N. LEVELS OBJECTIVES

- To identify coefficient, base and power of expression
1. Knowledge (K) - To recall the laws of indices

- To express the product/quotient of expressions having
same base in terms of single base

- To evaluate the numerical problems by using laws of
2. Understanding (U) indices
- To simplify/prove the simple given expressions
- To solve the exponential equations

- To simplify/prove the given rational expressions
(involving roots as well) by applying the laws of indices

3. Application (A) - To solve the exponential equations of the quadratic form

4. High Ability (HA) - To prove the rational expression under the given
condition/s.

- To prepare the report about the use of indices

Required Teaching Materials/ Resources
Chart papers with laws of indices, scissors, ruler, glue-stick and computer/projector if
possible

Pre-knowledge: Laws of indices, basic operations

A. Indices

Teaching Activities
1. Give the practical examples of use of laws of indices.
For example
(i) The cost of 1 kg of apple is Rs 125. Find the cost of 5 kg of apples by using the product

law of indices. For, 5 × 125 = 51 × 53 = 51 + 3 = 54 = 625

(ii) Divide 64 copies are equally among 4 friends by using the quotient law of indices. For,
64 26
4 = 22 = 26 – 2 = 24 = 16

71 Vedanta Excel in Mathematics Teachers' Manual - 10

2. Recall the following laws of indices by presenting in chart paper with proper examples

(i) Product law am × an = am + n, where 'm' and 'n' are positive integers

(ii) Quotient law am ÷ an = am – n when m > n
am ÷ an = 1 when m < n

an – m

(iii) Power law of indices (am)n = am × n, (ab)m = ambm, am = am
b bm

(iv) Law of negative index 1 1
a– m = am or am = a–m

(v) Law of zero index a° = 1, b° = 1, x° = 1 and so on
(vi) Root law of indices
n am m

= an

3. Discuss, give the way of solving the various problems and involve the students in solving
the problems from exercise

4. Under given condition, prove the expressions and give the same type problems to the
students and tell them to prove in the class.

5. Call the students randomly to solve the problems on the board in order to make them
confident to solve the problems

Solution of selected questions from Vedanta Excel in Mathematics

1. Simplify: 11 11 11
Solution:
ax − y x − z . ay − z y − x . az − x z − y

11 11 11

ax − y x − z . ay − z y − x . az − x z − y

111

= a . a . a(x − y) (x − z) (y − z) (y − x) (z − x) (z − y)

1 1 1
(z − x) (z − y) (y − z) (x − y)
= a + +(x − y) (x − z)
1 1 1
(y − z) (y − x) (z − x) (y − z)
= a− − −(x − y) (z − x)

− (y − z) − (z − x) − (x − y) −y+z−z+x−x+y

= a (x − y) (y − z) (z − x) = a (x − y) (y − z) (z − x) = a0 = 1

11 1

2. Simplify: p + (pq2)3 + (p2q)3 × 1 – q3
p–q 1

Solution: p3

11 1

Here, p + (pq2)3 + (p2q)3 × 1– q3
p– q 1

p3

12 21 11

= p + p3q3 + p3q3 × p3 − q3
p–q
1

p3

1 2 2 11 11

p3 p3 + q3 + p3q3 p3 − q3
= p–q ×1
p3

Vedanta Excel in Mathematics Teachers' Manual - 10 72

11 12 1 1 12 13 13

p3 – q3 p3 + p3.q3 + q3 p3 – q3 = p – q =1
= p–q = p–q p – q

a2 – 1 a a– 1 b–a
b2 b
×

3. Simplify: b2 – 1 b b+ 1 a–b
Solution: a2 a
×

1 a 1 b–a a+ 1 a a– 1 a a– 1 b–a
b2 b b b b
a2 – × a– × ×

Here, 1 b 1 a–b = b+ 1 b b– 1 b b+ 1
a2 a a a a
b2 – × b + × ×

1 a 1 a+b−a 1 a 1 b
b b b b
a+ × a– a+ a−

= 1 b+a−b 1 b = 1 × 1
b+ a a a a
× b– b+ b −

= ab + 1 a a ab − 1 a b = a a a b a a+b
b ab + 1 b ab − b b b
× × × 1 × =

111
4. If a + b + c = 0, show that 1 + xa + x–b + 1 + xb + x–c + 1 + xc + x–a = 1
Solution:

Here, a + b + c = 0, ∴ a + b = – c
111

Now L.H.S = 1 + xa + x–b + 1 + xb + x–c + 1 + xc + x–a

xb 1 xa + b
= xb (1 + xa + x–b) + 1 + xb + xa + b + xa + b (1 + xc + x–a)

xb 1 xa + b
= xb + xa + b + 1 + 1 + xb + xa + b + xa + b + xa + b + c + xb

xb 1 xa + b
= 1 + xa + b + xb + 1 + xa + b + xb + xa + b + x0 + xb

xb 1 xa + b xa + b + xb + 1
= 1 + xa + b + xb + 1 + xa + b + xb + xa + b + 1 + xb = xa + b + xb + 1 = RHS Proved.

5. If abc + 1 = 0, prove that 1 b−1 + 1 − 1 c−1 + 1 − 1 a−1 = 1
Solution: = or 1−a− −1 b− c−
Here, abc + 1 0 1 c
= abc = − 1 ∴ ab = c−1 = −c−1
Now, L.H.S 1
1 + + 1
− a − b−1 1−b− 1 − c − a−1

= b(1 − b − b−1) + 1 + ab(1 ab − a−1)
a 1 − b + ab −c

= b − b − 1 + 1 + ab b
ab 1 − b + ab ab − abc −

73 Vedanta Excel in Mathematics Teachers' Manual - 10

= − 1 − b ab + 1 + ab b = ab −b+ 1 = R.H.S proved
b+ 1 − b + ab ab + 1 − ab −b+ 1

22

6. Simplify: x3 + x3 − 1 − 1

1 1 1 1

Solution: 2 x3 − 1 x3 + 1 x3 + 1 x3 − 1

2

Here, = x3 + x3 − 1 − 1

1 1 1 1

x3 − 1 x3 + 1 x3 + 1 x3 − 1

22 12 12

= x3 − 1 + x3 − 1 − x3 − (1)2 − x3 − (1)2
1 1
1 1

x3 − 1 x3 + 1 x3 − 1 x3 + 1

1 1 1 1 11 1

= x3 + 1 x3 − 1 + x3 + 1 x3 − 1 = x3 + 1 + x3 − 1 = 2x3

1 1

x3 − 1 x3 + 1

7. Simplify: p2 2p 1
Solution: (p − y)y − (p − y)y − 1 + (p − y)y − 2

Here, p2 2p 1
(p − y)y − (p − y)y − 1 + (p − y)y − 2

= p2 − 2p (p − y) + (p − y)2 = p2 − 2p(p − y) + (p − y)2 = {p − (p − y)}2 = y2
− y)y (p − y)y (p − y)y (p − y)y (p − y)y (p − y)y
(p

B. Exponential Equations

Teaching Activities
1. Ask the laws of indices
2. Discuss upon the exponential equations like 2x = 8, x =? etc.
3. With more examples, list he following ideas
(i) If ax = ap then x = p (ii) If xn = kn then x = k (iii) If ax = 1 then x = 0
4. Solved some equations and give same type of equations to solve in the class or at home
5. Discuss upon the problems given in the exercise

Solution of selected questions from Vedanta Excel in Mathematics

8. If xa.xb = (xa)b, prove that a + b = ab − 2.
Solution: b a

Here, xa . xb = (xa)b

or, xa + b = xab

or, a + b = ab

On squaring on both sides, we get

(a + b)2 = (ab)2

or, a2 + 2ab + b2 = a2b2

or, a2 + b2 = a2b2 − 2ab

Dividing both sides by ab, we get
a2a+bb2 a2b2 − 2ab
= ab

or, ab + b = ab − 2
a

Vedanta Excel in Mathematics Teachers' Manual - 10 74

2 −2

9. If x2 + 2 = 23 + 2 3 , show that 2x(x2 + 3) =3

Solution: 2 −2

Here, x2 + 2 = 23 + 2 3
2 −2

or, x2 = 23 − 2 + 2 3
2 1 −1
or, x2 = 1 −1 2
− 2 × 23 × 2 3 +
23 23
1 −1 2
or, x2 =
23 − 2 3
1 −1

or, x = 23 − 2 3

Cubing on both side. We get

x3 = 1 −1 3

23 − 2 3 31 −1
13
or, x3 = −1 − 3 × 23 × 23 1 −1 [∴(a − b)3 = a3 − b3 − 3ab(a − b)]
23 −
23 −1 23 − 2 3

−1

or, x3 = 2 − 2−1 − 3 × 1 × x [∴ 2 3 − 2 3 = x
1
or, x3 = 2 − 2 − 3x

or, 2x3 = 4 − 1 − 6x

or, 2x3 + 6x = 3

∴ 2x (x2 + 3) = 3 Proved

10. If 2x = 3y = 6−z, show that 1 + 1 + 1 = 0
x y z
Solution:
1 1 −1

Let 2x = 3y = 6−z = k then 2x = k ∴ 2 = kx, 3y = k ∴ 3 = ky and 6−z = k ∴ 6 = k z

Now, 2 × 3 = 6

11 −1 11 −1
or, kx + y = k z
or, kx× ky = k z or, 1 + 1 = −1 ∴ 1 + 1 + 1 = 0
x y z x y z

11. Solve: 23x − 5 ax − 2 = 2x a− 2 1−x

Solution:

Here, 23x − 5 ax − 2 = 2x − 2 a1−x
or, 223xx ax − 2
− 5 a1−x = 1 or, 23x − 5 − x + 2 ax − 2 − 1 + x = 1
− 2

or, 22x − 3 a2x − 3 = 1 or, (2a)2x − 3 = (2a)0

or, 2x − 3 = 0 ∴ x = 3
2
26
12. Solve: 51− x + 5x − 1 = 5
Solution:
26
51− x+ 5x − 1 = 5
or, 55x + 5x
5 = 26 ..... eqn (i)
5
L et's o5rx, = a2t5h5e+anae2qn=(i)256b ecomesoa5r, a 26
+ 5 = 5

a2 − 26a + 25 = 0

or, (a − 1) (a − 25) = 0

Either a − 1 = 0 or, a − 25 = 0

or, a = 1 or, a = 25

75 Vedanta Excel in Mathematics Teachers' Manual - 10

or, 5x = 50 or, 5x = 52

∴ x = 0 ∴x=2

Hence, x = 0, 2 xx

13. If xy = yx, prove that x y = xy
Solution: y
y

Here, xy = yx, ∴ x =xx

x x x x1− y x x −1
y y x y
Now, L.H.S = xy = = xy = R.H.S
=y

xx

Extra questions for practice

1. Simplify.
a. (1 − xm − n)−1 + (1 − xn − m)−1 b. 11n1+12n–×551.1161n – 1

c. 9p + 2 + 10 × 9p d. 3 (x + y)–8 × (x + 2
9p + 1 × 11 – 8 × 9p
y)3

2. Simplify. 11 1

a. a+b xa2 × b+c xb2 × c+a xc2 m + (m2n)3 + (mn2)3 × 1 – n3
c xb2 a xc2 b xa2 b. m – n 1

1 x 1 y–x m3
y2 y x–y
c. 1 1 1 1 1 c 1 x2 – 1 × x –
– b d. x2
xa – b a–c × xb–c b–a × xc – a y 1
y2 – x
× y+

3. Simplify. 111

a) If p+q+ r = 0, show t+haxt1+1 +y–1xp++1x+–q +1 1 + xq + x–r +1 1 + xr + x–p = 1
b) If xyz = 1, show that 1 y+ z–1 + 1 + z + x–1 = 1.

c) If a + b+ c = p, prove that x2a + x2a + xp − c + x2b + x2c x2c + xp − b =1
2 −2 xp − b x2b + xp − c + xp − a + xp − a
d) If x2 + 2 = 33 + 3 3, show that 3x (x2 + 3) = 8

4. Solve a. 7x + 7x + 1 = 56 b. 102x − 3 = 0.001

c. 2z + 3 × 3x + 4 = 18 d. 62x − 3 . ax − 2 = 65x − 4 . a4x − 3

5. Solve a. 4x + 4− x = 16116 b. 7x + 343 = 56
7x
c. 2x − 2 + 23 − a = 3
111 d. 5.4x + 1 − 16x = 64

6. a. If xp = yq = z r and xyz = 1, prove that p + q + r = 0

b. If ap . aq = (ap)q, prove that: (xq − 2)p × (xp − 2)q = 1

c. If x = ap.bq + r, y = aq.br + p and z = ar.bp + q, prove that xq − r yr − p zp − q = 1.

d. If 3x = 7y = 21−z, show that: 1 + 1 + 1 = 0
x y z

Answer key: 1 x x+y
+ y
1. a) 1 b) 1 c) 1 d) (x y)2 2. a) 1 b) 1 c) 1 d)

4. a) 1 b) 0 c) –2 d) 13 5. a) ±2 b) 1, 2 c) 2, 3 d) 1, 2

Vedanta Excel in Mathematics Teachers' Manual - 10 76

Unit Surds

11

Allocated teaching periods 5

Competency

To discover and present the simplifications of algebraic relations and solve the problems
Learning Outcomes

- To solve the problems related to radical and surds using four fundamental operations
Level-wise learning objectives

S.N. LEVELS OBJECTIVES

- To define surd
- To tell the order of the surd
1. Knowledge (K) - To state the rationalizing factor or conjugate of the surd

- To arrange the surds in ascending or descending order of
magnitudes

2. Understanding (U) - To simplify the surds by using the four fundamental
operations (+, - , × and ¸)

- To rationalize the denominator
- To solve the simple equations involving surds

3. Application (A) - To simplify the rational surds by rationalizing the
denominators

- To solve the equations involving surds

4. High Ability (HA) - To explain the laws of surds with examples

Required Teaching Materials/ Resources
Colourful chart paper, colourful markers, tape, scissors etc.

Pre-knowledge: Different sets of numbers like set of natural numbers, whole numbers,

A. Surds integers, rational numbers etc

Teaching Activities
1. Recall the sets of numbers like set of natural numbers, whole numbers, integers, rational
numbers and irrational number with examples by showing in chart paper
2. Discuss about rational number with the following properties
p
(i) Rational numbers can be expressed in the form q where p and q are integers, q ≠ 0.
For example 0 = any non 0 2
zero integer , 3 etc.

(ii) tWraeerrcehmuetreinrnrimanatgiernddaatetoiciorninmgnaoaldnle.n-ctuiemmrmableisnr,ai32tsin=egx0bp.ur6et6sr6see6cd.u..,rinr56ind=ge.cF0iom.8r3ael3,x3at.hm..epneltecth12aere=dne0co.in5m-t,ael85rmp=ainrt0a.tm6in2ag5ybebutect

3. To the contrary, discuss about the irrational numbers. For examples 2, 4 6 etc
4. With examples, note that the decimal part of irrational numbers is neither terminate nor
non-terminating recurring
5. Define surd as the irrational number whose exact root cannot be found. For example

2, 3, 3 6, 4 7 etc.

77 Vedanta Excel in Mathematics Teachers' Manual - 10

6. Discuss about the parts of a surd as shown below

Radical sign

Order n a Radicand
7. Explain the parts of surd n a as follows
(i) The order ‘n’ of the surd is a natural number
(ii) The radicand ‘a’ is a positive rational number
8. Discuss about the laws of surds

Laws of surds Examples

(i) n a = 1 So, n a n 1 35 33

n, =a 3 2 = 23, = 53 = 5

(ii) n ab = n a . n b 3×5= 3× 5

(iii) n a= na 4 4= 44
b nb 5 45

(iv) n a + n a = 2 n a 3 5 + 3 5 =23 5

m na n ma mn a 3 64 = 4 = 2 = 2×3 64 = 6 64

(v) = =

9. Explain about the pure surds, mixed surd, like surds, unlike surds, operations of surds
10. Divide the students into 5 groups and give the following activities and call for

presentation in the class
Group A: Find the orders of 3 4 and p a

Group B: Simplify 8 + 18 – 32
Group C: Simplify ( 5 – 3)2

Group D: Simplify (3 2 + 2 3) (3 2 – 2 3)

Group E: Simplify a2 – b2
a–b

11. Explain about rationalization and conjugate with examples
12. Divide the students into 5 groups and give the following activities and call for

presentation in the class

Group A: Rationalize the denominator and simplify 2 – 1
2+1

Group B: Rationalize the denominator and simplify 3 – 1
3+1

Group C: Rationalize the denominator and simplify 3 + 2
3– 2

Group D: Rationalize the denominator and simplify 5 + 2
5– 2

Vedanta Excel in Mathematics Teachers' Manual - 10 78

Group E: Rationalize the denominator and simplify 2 3 – 3 2

2 3+3 2

13. Explain the process of simplification of the form 3 2 + 6– 43
3+ 6 2+ 3 6+ 2
B. Simple surd equations

Teaching Activities
1. Recall the order and operations of surds
2. Explain about the equation involving surds with examples
3. Ask about the process of expressing the surds into rational number
4. Give some examples with discussion and divide the students into 5 groups and give the

following activities and call for presentation in the class
Group A: Solve 3x – 2 – 1 = 4
Group B: Solve x + 9 + x = 9

Group C: Solve 3 4x + 1 = 25

Group D: Solve x –9 = 2
x +3

Group E: Solve y +5 = 3
y–5

5. Discuss about the roots and extraneous roots of the equation
x–1 x –1
6. With discussion, solve the equations like x +1 =4+ 2 and give same type of

equation from textbook as class-work or homework

7. Solve the equations like x+ 2 + x– 2 = 6 and give same type of equation from
x– 2 x+ 2
textbook as class-work or homework

Solution of selected questions from Vedanta Excel in Mathematics

1. Simplify:

a. 3 2 + 4 2500 − 4 64 + 6 8
Solution:

Here, 3 2 + 4 2500 − 4 64 + 6 8 = 3 2 + 4 54 × 22 − 4 24 × 22 + 6 22 × 2
= 3 2 + 5 22 − 2 22 + 6 × 2 2 = 3 2 + 5 2 − 2 2 + 12 2 = 18 2

b. 3x − 16
4 + 3x

Solution:
3x − 16
= 4+ 3x =( 3x)2 − (4)2 =( 3x + 4) ( 3x − 4) = 3x − 4
4 + 3x (4 + 3x)

c. 3 3 81 − 3 3 24 + 2 3 375 = 33 81 − 3 3 24 + 2 3 375
13 3 192 = 3× 13 3 +2×
3 3 3 −3 1×9223 3 5 3 3 13 3 3 1
54 3 3 4
13 × 4 3 3 = =

79 Vedanta Excel in Mathematics Teachers' Manual - 10

2. Rationalize the denominator and simplify: a+b− a−b
Solution: a+b+ a−b

a+b − a−b = a+b − a−b × a+b − a−b = ( a+b − a − b)2
a+b + a−b a+b + a−b a+b − a−b a + b)2 + a − b)2
( (

= ( a + b)2 − 2 a + b . a − b + ( a− b )2 = 2a − 2 a2 − b2 = a − a2 − b2
a + b − (a − b) 2b b

3. Simplify: xx + x2 − 1 − x − x2 − 1
− x2 − 1 x + x2 − 1

Solution:

Here, x+ x2 − 1 − x − x2 − 1 = (x + x2 − 1)2 − (x − x2 − 1)2
x− x2 − 1 x + x2 − 1 (x − x2 − 1) (x + x2 − 1)

= x2 + 2.x x2 − 1 + ( x2 − 1)2 − x2 + 2x x2 − 1 − ( x2 − 1)2 = 4x x2 − 1) = 4x x2 − 1)
x2 − ( x2 − 1)2 1

4. Simplify: 5 5 2 1) − 85 2 + 3 10
( 2+ 10 + 2+ 5
Solution:

Here, 5 2 − 85 + 3 10
5( 2 + 1) 10 + 2 2+ 5

= 52 5 × 10 − 5 − 85 2 × 10 − 2 + 3 10 × 2− 5
10 + 10 − 5 10 + 10 − 2 2+ 5 2− 5

= 5( 20 − 10) − 8( 50 − 10) + 3( 20 − 50)
10 − 5 10 − 2 2−5

= 5( 20 − 10) − 8( 50 − 10) + 3( 20 − 50)
5 8 3

= 20 − 10 − 50 + 10 − 20 + 50 = 0

5. Simplify: 1 1 +x x + 1 1 −x x
+ 1+ − 1+

Solution:

Here, =1 1 +x x + 1 1 −x x
+ 1+ − 1+

= (1 + x) (1 − 1 + x) + (1 − x) (1 + 1 + x)
(1 + 1 + x) (1 − 1 + x)

= 1 − 1+x+x−x 1+x+1+ 1+x−x−x 1+x
1−1−x

= 2 − 2x 1 + x = 2(x 1 + x − 1)
−x x

6. Simplify: x2 + 2 + x2 – 2 + x2 + 2 – x2 – 2
Solution: x2 + 2 – x2 – 2 x2 + 2 + x2 – 2

Here, x2 + 2 + x2 – 2 + x2 + 2 − x2 – 2
x2 + 2 − x2 – 2 x2 + 2 + x2 – 2

Vedanta Excel in Mathematics Teachers' Manual - 10 80

= ( x2 + 2 + x2 − 2 )2 + ( x2 + 2 – x2 − 2 )2
( x2 + 2 )2 − ( x2 − 2 )2

= x2 + 2 + 2 x4 − 4 + x2 − 2 + x2 + 2 − 2 x4 − 4 + x2 − 2 = 4x2 = x2
4 4

7. Solve: x + 5 + x + 12 = 2x + 41
Solution:
Here, x + 5 + x + 12 = 2x + 41

On squaring both sides, we get

or, ( x + 5 + x + 12)2 = ( 2x + 41)2

or, x + 5 + 2 x + 5 . x + 12 + x + 12 = 2x + 41

or, 17 + 2 x2 + 17x + 60 = 41

or, 2 x2 + 17x + 60 = 24

or, x2 + 17x + 60 = 12
On squaring both sides, we get
x2 + 17x + 60 = 144

or, x2 + 17x − 84 = 0

or, x2 + 21x − 4x − 84 = 0

or (x + 21) (x − 4) = 0

Either x + 21 = 0 ∴ x = − 21

or, x − 4 = 0 ∴ x = 4

Substituting x = − 21 in the given equation

− 21 + 5 + − 21 + 12 = 2(− 21) + 41
or, − 16 + − 9 = − 1
or, 4 − 1 + 3 − 1 = − 1
or, 7 − 1 = − 1 (False)
Substituting x = 4 is the given equation
4 + 5 + 4 + 12 = 2 × 4 + 41
or, 3 + 4 = 7 (True)
Hence, required value of x is 4.

8. Solve: x + x − 1 − x = 1
Solution:

Here, x + x − 1 − x = 1

Squaring on both sides

or, ( x − 1 − x)2 = (1 − x)2

or, x − 1 − x = 1 − 2 x + x

or, 2 x − 1 = 1 − x

or, (2 x − 1)2 = 1 − x [squaring on both sides]

or, 4x − 4 x + 1 = 1 − x

81 Vedanta Excel in Mathematics Teachers' Manual - 10

or, 5x = 4 x

or, 25x2 = 16x [Squaring on both sides]

or, x(25x − 16) = 0 16
25
Either x = 0 or, 25x − 16 = 0 ∴ x =
Checking:

Substituting x = 0 in the given equation

0+ 0− 1−0=1

or, − 1 = 1 (False)
16
Substituting x = 25 in the given equation

or, 16 + 16 − 1 − 16
25 25 25

or, 4 + 16 − 3 =1
5 25 5

or, 4 + 1 =1 or, 1 = 1 (True)
5 5

Hence, the required value of x is 1.

9. Solve : 7x − 36 =9− 7x − 11
6+ 7x 3

Solution:

7x − 36 =9− 7x − 11
6 + 7x 3

or, ( 7x + 6) ( 7x − 6) = 27 − 7x + 11
(6 + 7x ) 3

or, 7x − 6 = 38 − 5 7x
3

or, 3 − 18 = 38 − 5 7x

or, 8 7x = 56

or, ( 7x )2 = (7)2 [Squaring on both sides]

∴x=7

Checking: Putting x = 7 in the given equation

7 × 7 − 36 =9−5 7 × 7 − 11
6+ 7×7 3

or, 13 = 9 − 24
13 3

or, 1 = 1 (True)

Hence, the required value of x is 7.

10. Solve: x + a = 4 2 + x − a
x− a x+ a

Vedanta Excel in Mathematics Teachers' Manual - 10 82

Solution:

Here, x+ a− x − a =4 2
x− a x+ a

or, x+2 ax + a − x + 2 ax − a =4 2
x−a

or, ax = 2 ax
or, x 2−(ax − a) =

or, { 2 (x − a)}2 = ( ax)2 [Squaring on both sides]

or, 2(x2 − 2ax + a2) = ax

or, 2x2 − 4ax − ax + 2a2 = 0

or, 2x(x − 2a) − a(x − 2a) = 0

or, (3 − 2a) (2x − a) = 0

Either x − 2a = 0 ∴ x = 2a
or, 2x − a = 0 a
∴ x = 2

Extra questions for practice

1. Simplify:

a. 5 − 45 + 125 b. 3 128 + 2 3 54 − 2 3 250 c. 4 81x4 − 3 8x3
x2

2. Simplify:

a. x+y − x−y b. 3− 2+ 3+ 2 c. 32 3 − 43 2 + 6
x+y + x−y 3+ 2 3− 3 6+ 6+ 3+ 2

3. Solve: c. 8 + 3 4x − 7 = 13
a. x + 8 − 2 = x b. 3x + 1 = 3 8

4. Solve:

a. 5x − 4 =4− 5x − 3 b. xx ++420 = 3 x− 4
5x + 2 2 15 +3 x

c. x+2 + x+7 = 15 7 d. xx +− 22 + x+ 2 =6
x+ x− 2

Answer key:

1. a) 3 5 b) 0 c) 1 2. a) x + x2 – y2 b) 10 c) 0
c) 33 y c) 2 d) 16

3. a) 1 b) 1 4. a) 5 b) 4

83 Vedanta Excel in Mathematics Teachers' Manual - 10

Unit Equations

12

Allocated teaching periods 9

Competency
- To search the algebraic equations and solve

Learning Outcomes

- To solve the simultaneous equations of two variables
- To solve the quadratic equations of two variables
Level-wise learning objectives

S.N. LEVELS OBJECTIVES

1. Knowledge (K) - To define simultaneous equations

- To tell the ages after x years ago or x years later when
present age is given

- To state the two digit number formed by the digits x at
tens and y at ones places

- To represent the fraction
2. Understanding (U) - To solve the simultaneous equations

- To find the value of two different items

- To identify the numbers under the given conditions

3. Application (A) - To find the present ages of two individual under the
given conditions

- To find out the two digit numbers under the given
conditions

- To find the fraction under the given conditions

4. High Ability (HA) - To make the simultaneous equations related to the
context and solve the problems

Required Teaching Materials/ Resources
Chart-paper, glue-stick, colourful markers etc

Pre-knowledge: Methods of solving the simultaneous equations, quadratic equations

A. Linear equations
Teaching Activities

1. Discuss about the simultaneous equations with real life examples
2. Recall the methods of solving simultaneous equations
3. Explain with discussion the important terminologies used in making equations
(a) Terminologies based on fundamental operations

Vedanta Excel in Mathematics Teachers' Manual - 10 84

(i) Addition: added, sum, exceed, altogether, all, more than, increased etc
(ii) Subtraction: subtracted, difference, left away, shorter than, less than, decreased by,

younger/cheaper than etc
(iii) Multiplication: product, times, multiplied by etc
(iv) Division: divided by, each, share, quotient etc.
(b) Terminologies based on mensuration
(i) If the length and breadth of a rectangle are x meter and y meter respectively then

perimeter = 2 (x + y) and area = xy
(ii) If the perpendicular, base and hypotenuse of a right angled triangle are p, b and h

respectively then p2 + b2 = h2
(c) Terminologies based on ages: If the present age of an individual is x years
(i) Age before/ago: (x – t) years
(ii) Age after/ hence/ later : (x + t) years
(d) Terminologies based on consecutive numbers
(i) Consecutive numbers: x, x+1, x+2, ... etc
(ii) Consecutive odd/even numbers: x, x+2, x+4, ... etc
(e) Terminologies based on two-digit number
(i) Two digit number = 10x + y, reversed number = 10y + x
(ii) Sum of digits = x + y, product of digits = xy
(f) Terminologies based on fraction
If x and y are the numerator and denominator of a fraction, the fraction = x/y
4. Divide the students into groups and give difference questions to make the corresponding

simultaneous equations and solve

Solution of selected questions from Vedanta Excel in Mathematics

1. If three times the sum of two numbers is 42 and five times their difference is 20, find

the numbers.

Solution:

Let the greater number be x and the smaller one be y.

Then, From the first given condition;

3(x + y) = 42

or, x + y = 14

∴ x = 14 − y ........(i)

From the second given condition

5(x − y) = 20

∴ x − y = 4

Now, substituting the value of x from equation (i) in equation (ii), we get

14 − y − y = 4

or − 2y = − 10

∴ y =5

Again, substituting the value of y in equation (i), we get

x = 14 − 5 = 9

Hence, the required numbers are 9 and 5.

2. The total cost of a watch and a radio is Rs 500. If the watch is cheaper than the
radio by Rs 150, find their cost.

Solution:
Let the cost of a watch be Rs x and that of a radio be Rs y.
From the first given condition;
x + y = 500 .... (i)

85 Vedanta Excel in Mathematics Teachers' Manual - 10

From the second given condition;
x = y − 150 ... (ii)
Now, substituting the value of x from equn (ii) in equn (i), we get
y − 150 + y = 500
or, 2y = 650
or, y = 325
Substituting the value of y in equation (ii), we get
x = 325 − 150 = 175
Hence, the cost of the watch is Rs 175 and that of the radio is Rs 325.

3. The cost of tickets of a comedian show of 'Gaijatra' is Rs 150 for an adult and Rs 50
for a child. If a family paid Rs 550 for 5 tickets, how many tickets were purchased
in each category?

Solution:
Let the number of tickets purchased for adult be x and for children be y.
Then,
From the first given condition;
x + y = 5 ∴ y = 5 − x .... (i)
From the second given conditon;
150x + 50y = 550
or, 50(3x + y) = 550
∴ 3x + y = 11
Now, Substituting the value of y from eqn (i) in eqn (ii), we get
3x + 5 − x = 11
or, 2x = 6 ∴ x = 3
Again substituting the value of x in eqn (i), we get
y = 5 − 3 = 2
Hence the number of tickets purchased for adult is 3 and that for children is 2.

4. The numerator 21o.f a fraction is 5 less than its denominator. If 1 is added to each, its
value becomes Find the original fraction.

Solution:

Lreeqtuthireednufrmacetriaotnorisanyxd. denominator of the original fraction be x and y respectively. Then the
From the first given condition;

x = y − 5 .... (i)

From the second given condition;
x +1 1
y +1 = 2

or, 2x − y = − 1 .... (ii)

Now substituting the value of x from equation (i) in equation (ii), we get

2 (y − 5) − y = − 1 ∴ y = 9

Again, putting the value of y in equation (i), we get

x= 9−5=4 is 4
Hence, the required fraction 9

5. The sum of present ages of a son and his son is 40 years. If they both live on till son
becomes as old as the father is now, the sum of their ages will be 96 years. Find their
present ages.

Solution:

Vedanta Excel in Mathematics Teachers' Manual - 10 86

Let the present age of the father be x years and that of his son be y years.
From the first given condition; x + y = 40 ∴ y = 40 − x ..... (i)
From the second given condition; [x +(x − y)] + [y + (x − y)] = 90
∴ 3x − y = 96 .... (ii)
Now, substituting the value of y from equation (i) in equation (ii), we get
3x − (40 − x) = 96 ∴ x = 34
Again, putting the value of x in equation (i), we get
y = 40 − 34 = 6
Hence, the present age of the father is 34 years and that of his son is 6 years.
Note:
If the present ages of the father and son are x years and y years respectively, then
(i) After (x − y) years, the son will be as old as his father is now.
i,e af ter (x − y) ye ars, the age of son = [y + (x − y)] = x, which is the present age of father
(ii) (x − y) years ago, the father was as old as his son is now.
i.e (x − y) years ago age of father = x − (x − y) = y, which is present age of his son.

6. The sum of present ages of a father and his son is 80 years. When the father age was

equal to the present age of the son, the sum of their ages was 40 years. Find their

present ages.

Solution:

Let the present age of the father be x years and that of his son be y years.

From the first given condition; x + y = 80 ∴ x = 80 − y .... (i)

From the second given condition; [x − (x − y)] + [y − (x − y)] = 40

or, 3y − x = 40

Now, putting the value of x in equation (ii) from equation (i), we get

3y − (80 − y) = 40 ∴ y = 30

Again, putting the value of y in equation (i), we get

x = 80 − 30 = 50

Hence, the present ages of the father and his son are 50 years and 30 years respectively.

7. When the age of Rita was equal to the present age of shova; she was thrice as old

as shova was. If the sum of their ages is 40 years, find their ages.

Solution:

Let the present ages of Rita and Shova be x years and y years respectively.

From the first given equation,

[x − (x − y)] = 3[y − (x − y)]

or, y = 6y − 3x

or, 3x = 5y
thxe=se5c3yond...g..iv(ie)n
∴ equation;
From

x + y = 40 ..... (ii)

Now, putting the value of x in equation (ii) from (i), we get
5y
3 + y = 40 ∴ y = 15

Again, putting the value of y in equation (i), we get
5 + 15
x= 3 = 25

Hence, the present ages of the Rita and Shova are 25 years and 15 years respectively.

87 Vedanta Excel in Mathematics Teachers' Manual - 10

8. In 2010, Salim was two times as old as Kedar was. If the ratio of their ages will be

5:3 in 2015. Find the years of birth.

Solution:

Let the ages of Salim and Kedar in 2010 be x and y years respectively.

From the first given condition,

x = 2y ..... (i)

From the second given condition, in 2015, i.e, after (2015 -2010) = 5 years,
or, 3yxx 555y==351 0
+ .... (ii)
+
−

Now, putting the value of x in equation (ii) from (i), we get

x × 2y − 5y = 10 ∴ y = 10

Again, putting the value of y in equation (i), we get

x = 2 × 10 = 20

∴ In 2010, the ages of Salim and Kedar were 20 years and 10 years respectively.

Hence, the birth year of Salim = 2010 − 20 = 1990 and

the birth year of Kedar = 2010 − 10 = 2000

9. A number of two-digits exceeds four times the sum of its digits by 3. If 36 is added
to the number the digits are reversed. Find the number.

Solution:
Let the digits at tens and ones place of a number be x and y respectively. Then, the number
= 10x + y.
From the first given condition;
10 + y = 4(x + y) + 3
or, 6x = 3y + 3
∴ 2x − y = 1 ....... (i)
From the second given condition;
10x + y + 36 = 10y + x
or, 9x = 9y − 36
∴ x = y − 4 ...... (ii)
Now, putting the value of x in equation (i) from (ii), we get
2(y − 4) − y = 1 ∴ y = 9
Again, putting the value of y in equation (ii), we get
x=9−4=5
Hence, the required number = 10x + y = 10 × 5 + 9 = 59

10. A number consists of two digits. If the number formed by reversing its digits is
added to it, the sum is 143 and if the same number is subtracted from it the
remainder is 9. Find the number.

Solution:
Let the digits at tens and ones place of a number be x and y respectively. Then, the number
= 10x + y.
From the first given condition,
(10x + y) + (10y + x) = 143
or, 11(x + y) = 143
or, x + y = 13 ..... (i)
From the second given condition
(10x + y) − (10y + x) = 9
or, 9x − 9y = 9

Vedanta Excel in Mathematics Teachers' Manual - 10 88

∴ x − y = 1 ....... (ii)
Now, adding equation (i) and equation (ii), we get
x + y = 13
x ± y = 1
or 2x = 14
∴ x = 7
Again, putting the value of x in equation (i), we get
7 + y = 13
∴ y=6
Hence, the required number is 10 × 7 + 6 = 76

11. Five times the sum of the digit of a two digit number is 9 less the number formed by

reversing its digits. If four times the value of the digit at ones place is equal to half

of the place value of the digit at tens place, find the number.

Solution:

Let the digit at tens and ones places of a two digit number be x and y respectively. Then, the

numbers = 10x + y

From the first given condition,

5(x + y) = 10y + x − 9

or, 4x − 5y = − 9 ...... (i)

From the second given condition,
5 10x
4y = 2 [∴ Place value of x is 10x]

∴ y= 5x ....... (ii)
4
Now, putting the value of y from equation (ii) in equation (i), we get
5x
4x − 5 × 4 = − 9 ∴x=4

Again, putting the value of x in equation (ii), we get
5 4
y = × = 5
4
Hence, the required number is 10 × 4 + 5 = 45

12. In a city, the taxi charges consists of two types of charges: a fixed together with the

charges for the distance covered. If a person travels 10 km, he pays Rs 180 and for

travelling 12 km, he pays Rs 210. Find the fixed charges and the rate of charge per

km.

Solution:

Let the fixed charge of the taxi be Rs x and the rate of charge per km be Rs y.

Then,

From the first given condition, x + 10y = 180 ... (i)

From the second given condition, x + 12y = 210 .... (ii)

Now, subtracting equation (i) from equation (ii), we get

x + 12y = 210

x + 10y = 180

(−) (−) (−)

or 2y = 30 ∴ y = 15

Again, putting the value of y in equation (i), we get

x + 10 × 15 = 180 ∴ x = 30

Hence, the fixed charge is Rs 30 and rate of charge is Rs 15 per km.

89 Vedanta Excel in Mathematics Teachers' Manual - 10

13. A lending library has a fixed charge for the first four days and an additional charge
for each day thereafter. Dorje paid Rs 50 for a book kept for 9 days, while Kajal
paid Rs 35 for the book she kept for 6 days. Find the fixed charge and the charge for
each extra day.

Solution:
Let the fixed charge be Rs x and the additional charge for each extra days be Rs y.
Then,
From the first given condition,
x + 5y = 50 [ Extra days (9 − 4) = 5 days]
∴ x = 50 − 5y ....... (i)
From the second given condition
x + 2y = 35 [ 6 − 4 = 2 extra days]
50 − 5y + 2y = 35
or, − 3y = − 15
∴ y = 5
Again, putting the value of y in equation (i), we get x = 50 − 5 × 5 = 25
Hence, the fixed charge is Rs 25 and charge for each extra day is Rs 5.

14. When Gopal gives Rs 10 from his money to Laxmi her money becomes double than
that of remaining money of Gopal. When Laxmi gives Rs 10 to Gopal, his money
is still Rs 10 less than the remaining money of Laxmi. Find their original sum of
money.

Solution:
Let Gopal and Laxmi have Rs x and Rs y respectively.
From the first given condition,
2(x − 10) = y + 10
or, 2x − 30 = y ..... (i)
From the second given condition,
(y − 10) − 10 = x + 10
∴ y − x = 30 ..... (ii)
Now, putting the value of y from equation (i) in equation (ii), we get 2x − 30 − x = 30
∴ x = 60
Again, putting the value of x in equation (i), we get y = 2 × 60 − 30 = 90
Hence, Gopal had Rs 60 and Laxmi had Rs 90.

15. Janak started his bicycle Journey from Kohalpur to Dhangadi at 6:00 a.m. with an
average speed of 20 km/hr. Two hours later Ganesh also started his journey from
Kohalpur to Dhangadhi with an average speed of 30 km/hr. At what time would
they meet each other if both of their maintain non - stop journey.

Solution:
Suppose, Janak meets Ganesh after x hours
But, Ganesh meets Janak after ( x − 2) hours
Now, the distance travelled by Janak in x hours = 20x km
Also, the distance travelled by Ganesh in (x − 2) hours = 30(x − 2) km
When they travelled the equal distance, they meet each other.

So, 20x = 30(x − 2)
or, 2x = 3x − 6
∴x=6
Since, Janak started his journey at 6:00 am, they would meet at 12:00 noon.

Vedanta Excel in Mathematics Teachers' Manual - 10 90

16. Two buses were coming from two different places Situated just in the opposite di-

rection. The average speed of one bus is 5 km/hr more than that of another one and

they had started their journey in the same time. If the distance, between the places

is 500 km and they meet after 4 hours, find their speed.

Solution:

Let the speed of the faster bus be x km/hr

and the speed of the slower bus be y km/hr

From the first given condition,

x − y = 5 ...... (i)

From the second given condition,

4x + 4y = 500

or, 4(x + y) = 500

∴ x + y = 125 ..... (ii)

Adding equation (i) and (ii), we get

x − y = 5

x + y = 125

or 2x = 130

∴ x = 65

Now, putting the value of x in equation (i), we get

65 − y = 5

∴ y = 60

Hence, the speed of the faster bus is 65 km/hr and the slower bus is 60 km/hr.

17. When the length of a rectangular field is reduced by 5 m and breadth is increased

by 3 m, its area gets reduced by 9 sq.m. If the length is increased by 3 m and breadth

by 2 m the area increases by 67 sq.m. Find the length and breadth of the field.

Solution:

Let the length and breadth of the rectangular field be x m and y m respectively. Then area

field (A) = l × b

From the first given condition,

(l − 5) (b + 3) = lb − 9

or, lb + 3l − 5b − 15 = lb − 9

∴ 3l − 5b = 6 ...... (i)

From the second given condition,

(l + 3) (b + 2) = lb + 67

or, lb + 2l + 3b + 6 = lb + 67

or, 2l + 3b = 61
61 – 2l
or, b = 3 ..... (ii)

Now, putting the value of b in equation (i) from (ii), we get
5( 61 2l )
3l − − = 6
3
or, 9l − 305 + 10l = 18

or, 19l = 323

∴ l = 17

Again, putting the value of l in equation (ii), we get
61 2 17 61 34
b = − 3 × = × =9
3
Hence, the length and breadth of the field are 17 m and 9 m respectively.

91 Vedanta Excel in Mathematics Teachers' Manual - 10

Extra questions for practice

1. A year hence, a father will be 5 times as old as his son. Two years ago, he was three times
as old as his son will be four hence. Find their present ages. [Ans: 29 years, 5 years]

2. The age of two girls are in the ratio 5:7. Eight years ago, their ages were in the ratio of

7:13. Find their present ages. [Ans: 15 years, 21 years]

B. Quadratic equations

Teaching Activities
1. Make a proper discussion on the quadratic equations with real life examples

2. Recall the methods of solving quadratic equations

(i) Factorisation method (ii) Using formulae

3. With discussion, make the equations and solve the problems

4. Make the groups of students and encourage to make the correct equations and solve the
equation to get the required solution

5. Make contextual problems based on quadratic equations and ask to students to solve
them

Solution of selected questions from Vedanta Excel in Mathematics

1. The area of rectangular field is 720 sq. m. and perimeter is 108 m. By what percent

is the longer side of the field to be decreased to make it a square? Why?

Solution:

Let the length and breadth of the rectangular filed be l m and b m respectively.

From the first condition,

l × b = 720 … (i)

From the second condition,

2 (l + b) = 108 or, l + b = 54 ∴l = 54 – b … (ii)

Now, substituting the value of l from equation (ii) in equation (i), we get

(54 – b)× b = 720 or, b2 – 54b + 720 = 0 or, (b – 24) (b – 30) = 0

Either, b = 24 or, b = 30 (Considering that the breadth is shorter than length)

Putting the value of b in equation (ii), we get l = 54 – 24 = 30

Hence, length (l) = 30 m and breadth (b) = 24 m

Difference between the length and breadth = 30 m – 24 m = 6 m 6
30
Thus, to make the filed in to a square, the longer side is to be decreased by × 100% = 20%

2. A piece of cloth costs Rs 800. If the piece was 5m long and rate of cost of cloth per

meter was Rs 8 less, the cost h of the piece would have remained unchanged. How

long is the piece and what is the rate of cost of the cloth?

Solution:

Let the length of piece of the cloth be x m and the rate of cost of the cloth Rs y per meter.

From the first condition,

xy = 800 … (i)

From the second condition,

(x + 5) ( y – 8) = 800

Vedanta Excel in Mathematics Teachers' Manual - 10 92

or, xy – 8x + 5y – 40 = 800

or, 800 – 8x + 5y – 40 = 800 (From (i), xy = 800)
40 + 8x
or, y = 5 (ii)

Now, substituting the value of y from equation (ii) in equation (i), we get
x(40 + 8x)
5 = 800 or, x2 + 5x - 500 = 0 or, (x + 25) (x – 20) = 0

Either, x = -25 or, 20 But the length cannot be negative. So, length = 20 m

Again, putting the value of x in equation (i), we get y = 40

Hence, the length of the piece of cloth is 20 m and its rate of cost is Rs 40 per meter.

3. Mrs Magar bought a certain kilograms of vegetables for Rs 120 last week. This

week, the rate of cost of vegetable decreases by Rs 20 per kg and she can buy 1 kg

more vegetables for the same amount of money. By what percentage is the rate of

cost of vegetables decreased?

Solution:

Let the rate of cost of vegetable last week be Rs x per kg and the quantity of vegetable that

can be bought for Rs 120 be y kg.

From the first condition,

xy = 120 … (i)

From the second condition,

(x – 20) ( y + 1) = 120

or, xy + x – 20y – 20 = 120

or, 120 + x – 20y – 20 = 120 (From (i), xy = 120)

or, x = 20 (y + 1) ... (ii)

Now, substituting the value of x from equation (ii) in equation (i), we get

or, 20 (y + 1) y = 120 or, y2 + y – 6 = 0 or, (y + 3) (y – 2) = 0

Either, y = -3 or, 2 But the quantity of vegetable cannot be negative.

So, y = 2

Again, putting the value of y in equation (i), we get x = 60

Hence, rate of cost of vegetable last week was Rs 60 per kg. 3331
20
Again, percentage of decreased amount of rate of cost = 60 × 100% = %

Thus, the rate of cost of vegetable is decreased 3313 % by in this week.

Extra questions

1. A two digit number is three times the product of digits and four times the sum of digits.

Finds the number. [Ans: 24]

2. Bhawani was 25 years old when her daughter was born. Now, the product of their ages

is 600. Find their present ages. [Ans: 40 years, 15 years]

3. Rs120 is equally distributed among certain number of students. If there were 3 students

more, each would have received Rs 2 less. Find the number of students. [Ans: 10]

93 Vedanta Excel in Mathematics Teachers' Manual - 10

Unit Area of Triangles and Quadrilaterals

13

Allocated teaching periods 12

Competency
- To prove logically and experimentally that the properties on the area of triangle
and quadrilateral

Learning Outcomes
- To verify/prove the interrelationship between the areas of parallelograms standing
on the same base and between the same parallels.
- To verify/prove the interrelationship between the areas of triangle and
parallelogram standing on the same base and between the same parallels.

- To show the interrelationship between the areas of triangles standing on the same
base and between the same parallels.

Level-wise learning objectives

S.N. LEVELS OBJECTIVES

1. Knowledge (K) - To tell the diagonal property of parallelogram
- To tell the median property of triangle
- To write the interrelationship between the areas of

parallelograms standing on the same base and between
the same parallels
- To state the interrelationship between the areas of triangle
and parallelogram standing on the same base and between
the same parallels
- To recall the types of parallelogram
- To state the interrelationship between the areas of triangles
standing on the same base and between the same parallels

2. Understanding (U) - To area of triangle or parallelogram based on the properties
of area of triangle and parallelogram

3. Application (A) - To verify/prove the theorems on properties of area of
triangle and parallelogram

4. High Ability (HA) - To prove theorems by drawing required diagrams/ figures

Required Teaching Materials/ Resources

Geo-board, graph board, circular shaped colourful chart-paper, scale, scissors, pencil, ,
marker, thread, rubber-band ICT tools etc.

Pre-knowledge: Parts of circle

Teaching Activities
1. Recall the properties of parallelogram
2. Recall the formula to find the area of parallelograms and triangles
3. Draw the parallelogram on the graph board and tell to students to draw the parallelogram

on the graph paper and discuss about the area of parallelogram by counting the squares

Vedanta Excel in Mathematics Teachers' Manual - 10 94

4. Divide the students in to groups and tell to draw the parallelograms under the following
cases on the graph paper
(i) Parallelograms standing on the same base and between the same parallels
(ii) Parallelograms standing on different bases but lying between the same parallels
(iii) Parallelograms standing on the same base but not lying between the same parallels

After discussion about the areas of the parallelograms among the members of the groups,
ask the group leader about the conclusion on the area.

At last, with discussion draw out the conclusion and state the theorem
5. If possible visualize the above experiments using geo-gebra tool
6. Ask the criterion of congruency of triangles
7. Verify experimentally/prove logically the theorem with discussion
8. Draw a parallelogram and a triangle under the following cases on the graph paper

(i) Parallelogram and triangle standing on the same base and between the same parallels
(ii) Parallelogram and triangle standing on different bases but lying between the same

parallels
(iii) Parallelogram and triangle standing on the same base but not lying between the same

parallels
After discussion about the areas of the parallelogram and triangle among the members of

the groups, ask the group leader about the relationship between the area of parallelogram
and the triangle.
At last, with discussion draw out the conclusion and state the theorem
9. Ask the statement of the previous theorem
10. Verify experimentally/prove logically the theorem with discussion
11. Draw triangles under the following cases on the graph paper
(iv) Triangles standing on the same base and between the same parallels
(v) Triangles standing on different bases but lying between the same parallels
(vi) Triangles standing on the same base but not lying between the same parallels
Let the students make to explore the relationship between the areas of triangles
At last, with discussion draw out the conclusion and state the theorem
12. Ask the statement of the previous theorems
13. Verify experimentally/prove logically the theorem with discussion
14. Recall the properties on area of parallelogram and triangle and discuss upon the problems
given in exercise
15. Discuss about the nature of questions asked as higher ability level as shown below

Theorem

Diagramatic Word statements

Conditional Non-conditional
(construction is not required) (construction is usually required) Condition given Condition not given

95 Vedanta Excel in Mathematics Teachers' Manual - 10

Solution of selected questions from Vedanta Excel in Mathematics

1. In the figure, the area of trapezium ABCD is 100 sq.cm. A D
and the area of ∆ADC is 40 sq.cm. Find the area of ∆DEC.
Solution:

Here, Area of trapezium ABCD = 100 cm2

Area of ∆ADC = 40 cm2 BE C
Area of ∆DEC =?

Now,
i) Area of ABED = 2 area of ∆ADC [Both are stunding on the same base AD

2 × 40 cm2 = 80cm2 and between AD//BC]

ii) Area of trapezium ABCD = Area of ( ABED + ∆DEC) [By whole part axion]

or, 100 cm2 = 80 cm2 + area of ∆DEC

∴ Area of ∆DEC = 20 cm2

2. In the given figure, ABCD is a square whose perimeter is 40 cm and AB is

produced to the point F. If F is the mid-point of DC, find the area of ∆EFC.

Solution:

Construction : D and F are joined AB
i) Perimeter of square ABCD = 40 cm F

or, 4l = 40 cm

∴ l = 10 cm

ii) Area of square ABCD = l2 = (10 cm)2 = 100 cm2

iii) Area of ∆DCF = 1 × area of square ABCD D EC
2 AB
1 F
= 2 × 100 cm2 = 50 cm2

iv) Area of ∆EFC = 1 × Area of ∆DCF bisects the ∆DCF]
2 [Median EF

= 1 × 50 cm2 = 25 cm2 D EC
2 DC

So, area of ∆EFC = 25 cm2

3. In the adjoining figure, AE//DC, ED//FC and ABCD is a square.
If AC = 5√2 cm, find the area of parallelogram DEFC.
Solution:
1 1 cm)2 A E BF
i) Area of square ABCD = 2 (AC)2 = 2 × (5√2 = 25 cm2
Q
ii) Area of parallelogram DEFC= Area of square ABCD = 25 cm2 M

Hence, area of parallelogram DEFC = 25 cm2 P
4. In the adjoining figure, PS = 5 cm and SM = 8 cm. Calculate

the area of ∆PQN.

Solution:

i) Area of PQRS = QR × SM N SR

= 5 cm × 8 cm = 40 cm2 [ QR = PS] ∴

ii) Area of ∆PQN = 1 × Area of PQRS
2
1
= 2 × 40 cm2 = 20 cm2

Vedanta Excel in Mathematics Teachers' Manual - 10 96

5. Find the area of quadrilateral PQRS given in the adjoining Q
A
figure in which 3 RB = 2 PA = QS = 12 cm.
B
Solution: R P
i) 3 RB = 2 PA = QS = 12 cm

∴ RB = 4 cm, PA = 6 cm and QS = 12 cm
1
ii) Area of quadrilateral PQRS = 2 × d(p1 + p2) S
1 × QS
= 2 (RB + PA)

= 1 × 12 cm (4 cm + 6 cm) = 60 cm2
2
16 cm
6. In the given figure, ABCD is a trapezium. If AB = 10 cm, BC = 22 A D
cm, AD = 16 cm, AD//BC and DC⊥BC, calculate the area of ∆ADC.
10 cm
Solution:

Construction: AP⊥BC is drawn where P is on BC. B 22 cm C

∴ PC = AD = 16 cm, BP = 22 − 16 cm = 6 cm
i) In right angled ∆ABP; AP = √AB2−BP2 = √102−62 = 8 cm
A 16 cm
D
1
ii) Area of ∆ADC = 2 AD × DC [∆ = 1 b × h] 10 cm
2
1
= 2 × 16 cm × 8 cm = 64 cm2 B P C
22 cm

Extra Short questions DE C

1. In the given figure, ABCD is a parallelogram. If the area of
∆ABE = 25 cm2, find the area of ∆BCE + ∆ADE. [Ans: 25 cm2]

A B
S
2. In the given figure, PQRS is a rhombus in which the diagonal P
T

PR = 16 cm and diagonal QS = 14 cm. If RS is produced to T, find

the area of ∆QRT. [Ans: 56 cm2]

Q R
X
W N
Y A
3. In the given figures, WXYZ is a square whose side WX is

produced to N. If M is mid-point of YZ and area of ∆MYN is

25 cm2, find the perimeter of square WXYZ. [Ans: 40 cm]

Z M

4. In the adjoining figures, M is mid-point of side AB of ∆ABC, AP⊥BC. M 6 cm
If BC = 8 cm and AP = 6 cm, find the area of ∆ABC. [Ans: 12 cm2]

B 6 cm C P

97 Vedanta Excel in Mathematics Teachers' Manual - 10

PS TU

5. In the adjoining figures, PU//QR, PQ//SR, TQ//UR, SM⊥PQ and TN⊥UR. M N
If SM = 3 cm, PQ = 8 cm and TN = 4 cm, find the length of TQ. R

[Ans: 6 cm]

Q

Solution of selected questions from Vedanta Excel in Mathematics

DC

1. In the given figure, ABCD is a parallelogram. x is any point X
within it. Prove that the sum of area of ∆XCD and ∆XAB is
equal to half of the area of ABCD.

Solution: A D B
C
Given: In ABCD; X is any point within it.
1 Q
To prove: ∆XCD + ∆XAB = 2 Area of ABCD P X
Construction:
PQ//AB//DC is drawn where P is on DA and Q is on BC.

Proof AB

Statements Reasons

1 DPQC and PQBA are parallelograms. 1. By construction and given

2. ∆XCD = 1 DPQC 2. Both are standing on the same base
2 and between the same parallels

3. ∆XAB = 1 PQBA 3. Same as (2)
2

4. ∆XCD + ∆XAB = 1 ( DPQC + PQBA) 4. On adding statements (2) and (3)
2

5. ∆XCD + ∆XAB = 1 ABCD 5. From (4), by whole part axiom
2

2. In the given figure, ∆ABC and parallelogram MBCN are on the M Proved
same base BC and between the same parallels MN and BC.
AN

Prove that area of ∆ABC = area of rectangle APCN.

Solution:

Given: ∆ABC and parallelogram MBCN are on the same base BC and B C
between the same parallels MN and BC. P is mid-point of BC
P

and AP ⊥ BC.
To Prove: Area of ∆ABC = Area of rectangle APCN.

Proof

Statements Reasons
1. Median AP bisects ∆ABC
1. ∆APC = 1 ∆ABC
2

2. ∆APC = 1 Rectangle APCN 2. Diagonal AC bisects the rectangle
2

3. ∆ABC = Rectangle APCN 3. From statements (1) and (2)

Proved

Vedanta Excel in Mathematics Teachers' Manual - 10 98

D XC
B
3. In the given parallelogram ABCD, X and Y are any points on

CD and ADrespectively. Prove that:

area of ∆AXB = area of ∆BYC. Y

Solution:

Given: In ABCD; X and Y are any point on CD and AD A
respectively.

To Prove: Area of ∆AXB = Area of ∆BYC

Proof

Statements Reasons

1. Area of ∆AXB = 1 × Area of ABCD 1. Both are standing on the same base AB
2 and between DC // AB

2. Area of ∆BYC = 1 × Area of ABCD 2. Both are standing on the same base BC
2 and between AD // BC

3. Area of ∆AXB = Area of ∆BYC 3. From statements (1) and (2)

4. In the adjoining parallelogram ABCD, A is joined to any point Proved

E on BC. AE and DC produced meet at F. Prove that area of AD

∆BEF = area of ∆CDE. B EC
F
Solution:

Given: In ABCD; E is any point on BC. AE and DC produced meet
at F.

To prove: Area of ∆BEF = Area of ∆CDE

Proof

Statements Reasons
1. Both are standing on AD and
1. Area of ∆AED = 1 × Area of ABCD
2 between AD // BC

2. Area of (∆ABE + ∆CDE) = 1 × Area of ABCD 2. Remaing parts of ABCD
2

3. Area of ∆ABF = 1 × Area of ABCD 3. Both are standing on AB and
2 between AB // DF

4. Area of ∆ABF = Area of (∆ABE + ∆BEF) 4. By whole part axion

5. Area of (∆ABE + ∆CDE) = Area of (∆ABE + 5. From statements (2), (3) and

∆BEF) (4)

6. Area of ∆CDE = Area of ∆BEF 6. From statement (5)

Proved

5. In the adjoining parallelogram ABCD, PQ // AB and RS // BC. A P D
Prove that area of ROPA = Area of QCSO.
S
Solution: R C
O

BQ

99 Vedanta Excel in Mathematics Teachers' Manual - 10

Proof

Statements Reasons

1. Area of ∆ABD = Area of ∆BCD 1. Diagonal BD bisects ABCD

2. Area of ∆ROB = Area of ∆BOQ 2. Diagonal OB bisects RBQO

3. Area of ∆POD = Area of ∆SOD 3. Diagonal OD bisects POSD

4. Area of (∆ABD − ∆ROB − ∆POD) = 4. Subtracting statements (2) and (3)
Area of (∆BCD − ∆BOQ − ∆SOD) from statement (1)

5. Area of ROPA = Area of QCSO 5. Remaining parts of whole

6. In the given diagram, ABCD and PQRD are two A P Proved
parallelograms. Prove that ABCD = PQRD. D
Solution: R B
P Q
Construction: P and C are joined
Proof R C

Statements Reasons A A B
D Q
1. ∆BPC = 1 ABCD 1. Both are standing on BC and
2 between AD // BC C

2. ∆BPC = 1 PQRD 2. Both are standing on PB and D B
2 between PB // QR HC

3. ABCD = PQRD 3. From (1) and (2) G

Proved EF
7. In the given figure, if AB // DC // EF, AD // BE and AF // DE,
prove that parm DEFH = parm. ABCD.
Solution:

Proof

Statements Reasons

1. ABCD = AGED 1. Both are standing on AD and between AD // BE
2. AGED = DEFH 2. Both are standing on DE and between DE // AF

3. DEFH = ABCD 3. From (1) and (2)

Proved

8. In the adjoining ∆ABC, P and Q are the mid-points of the sides AB and AC
respectively and R be any point on BC. Prove 1
that area of ∆PQR = 4 area of ∆ABC.
Solution: A
A

Construction: P Q PQ
P and C are joined.

BR CB R C

Vedanta Excel in Mathematics Teachers' Manual - 10 100