Vedanta Excel in Mathematics Book 8 TG Final (2079)

6. In the figure alongside, AB// CD and AB = CD. Prove that A B
D
i) DAOB @ DCOD ii) AO = OD and BO = OC.

Solution: C O
Here,

In ∆AOB and ∆COD;

(i) ∠OAB = ∠CDO (A) [AB//CD and alternate angles]

(ii) AB = CD (S) [Given]

(iii) ∠ABO = ∠OCD (A) [AB//CD and alternate angles]

(iv) ∆AOB ≅ ∆COD [By A.S.A. axiom]

(v) AO = CD and BO = OC [The corresponding sides of congruent triangles]

7. In the adjoining figure, PQ = SR and PQ // SR, show that S R
i) PS = QR ii) PS // QR P Q

Solution:

Here,

In ∆PQS and ∆QRS;

(i) PQ = SR (S) [Given]
[PQ//SR and alternate angles]
(ii) ∠PQS = ∠QSR (A) [Common side]
[By S.A.S. axiom]
(iii) QS = QS (S) [The corresponding sides of congruent triangles]
[The corresponding angles of congruent triangles]
(iv) ∆PQS ≅ ∆QRS [From (vi), being alternate angles equal]

(v) PS = QR

(vi) ∠PSQ = ∠SQR

(vii) PS//QR A

8. In the given figure, ABC is an isosceles triangle in which AB = AC.
If AP ^ BC, prove that

(i) DAPB @ DAPC (ii) ∠B = ∠C (iii) BP = PC B C

Solution: P

Here,

In ∆ABP and ∆ACP;

(i) ∠APB = ∠APC (R) [Both are right angles]

(ii) AB = AC (H) [Given]

(iii) AP = AP (S) [Common side]

(iv) ∆ABP ≅ ∆ACP [By R.H.S. axiom]

(v) ∠B = ∠C [The corresponding angles of congruent triangles]

(vi) BP = PC [The corresponding sides of congruent triangles]

9. In the adjoining figure, AP ^ CD, BQ ^ CD and A

OP = OQ. Prove that AP = BQ. CP Q D
Solution:
Here, O

In ∆AOP and ∆BOQ; [AP//BQ and alternate angles] B
(i) ∠APO = ∠BQO (A)

101 Vedanta Excel in Mathematics Teachers' Manual - 8

(ii) OP = OQ (S) [Given]
[The vertically opposite angles are equal]
(iii) ∠AOP = ∠BOQ (A) [By A.S.A. axiom]
[The corresponding sides of congruent triangles]
(iv) ∆AOP ≅ ∆BOQ

(v) AP = BQ

Extra Questions

1. Define congruent triangles. List out axioms of congruency of triangles.

2. By what axiom, ∆ABC and ∆PQR are congruent to each A P

other? Also, write a pair of corresponding sides.

[Ans: S.A.S. axiom, AC = PR] B CQ R

3. In the given figure, AP ⊥ BC and AB = AC. Prove that A
∆APB ≅ ∆APC. B PC

4. By what axiom, ∆ABC and ∆DEF are A D
congruent to each other? Find the value of x. 7 cm
(3x – 1) cm
[Ans: A.S.A., 4 cm]
B CE F

5. In the given figure, ∆PQR ≅ ∆XYZ, P X
find the value of two remaining 55°
angles and y. 60° Y 60° 65° Z
Q 2y cm R (y + 3) cm
[Ans: ∠R = 650, ∠X = 550, 3 cm]

Teaching Activities for Similar Triangles
1. Show the same photo of anyone/anything else with different sizes and make a discussion

whether they have same (i) shape or not (ii) size or not.

2. Draw two squares / circles of different sizes and discuss about similar figures. Similarly,
with various real life examples of similar objects, define similar figures.


3. Discuss about the similar triangle. Define similar triangles as the equiangular triangles.

Vedanta Excel in Mathematics Teachers' Manual - 8 102

4. Tell the students to draw a pair of similar triangles A
and measure the corresponding sides. Then tell to
P
fill up the following table.

B CQ R
In ∆PQR
In ∆ABC Corresponding sides

Length Length Length Length Length Length
of AB of BC of CA of PQ of QR of RP

… … … … … … … ……

After the experiment, conclude that the corresponding sides of similar triangles are

proportional.
5. Discuss about the following conditions of similarity of two triangles with necessary

figures.
(i) When two angles of one triangle are respectively equal to angles of another triangle,

the triangles are similar.
(ii) When the corresponding sides of two triangles are proportional, the triangles are

similar.
(iii) When two corresponding sides of triangles are proportional and the angles included

by them are equal then the triangles are similar.

6. Make the students to solve the questions related to the lesson in groups.

Solution of selected questions from Excel in Mathematics -Book 8

1. From these pairs of congruent triangles. Let's A P
tell and write the corresponding sides and

corresponding angles.

Solution:

Here, B CQ R

In ∆ABC ∼ ∆PQR, AB = 9 cm, AC = 12 cm, PR = 8 cm, PQ = ?

(i) Since, the corresponding angels of similar triangles are equal.

∠A = ∠P = 750 and ∠B = ∠Q = 550

(ii) Since, the corresponding sides of similar triangles are proportional.

AB = BC = AC
PQ QR PR

or, 9 cm = BC = 12 cm
PQ QR 8 cm

Taking the first and the last ratios, we get

9 cm = 12 cm
PQ 8 cm

or, 12PQ = 72

∴ PQ = 6 cm

103 Vedanta Excel in Mathematics Teachers' Manual - 8

2. In the figure alongside, ∠BAD = ∠ACD and ∆ABD ~ ∆ABC. 4 cmA 10 cm
Find the length of BD. B
Solution: D C
Here, 8 cm

In ∆ABC ∼ ∆ABD, AB = 4 cm, AC = 10 cm, BC = 8 cm and
∠BAD = ∠ACD. BD =?

(i) Since, the corresponding sides of similar triangles are proportional.

AB = AC = BC
BD AD AB

or, 4 cm = 10 cm = 8 cm
BD AD 4 cm

Taking the first and the last ratios, we get

4 cm = 8 cm
BD 4 cm

or, 8 BD = 16

∴ BD = 2 cm

P

3. In the given figure, ∆PQR ~ ∆XQY. If ∠Q = 40°, X
∠R = 60°, PR:XY = 5 :3 and PQ = 15 cm, find ∠QXY
40° 60°
and the length of QX. Q YR
Solution:

Here,

In ∆PQR ∼ ∆XQY, AB = 4 cm, ∠Q = 400, ∠R = 600, PR: XY = 5:3 and PQ = 15 cm. ∠QXY =?,

QX =?

(i) In ∆PQR; ∠P = 1800 – (400 + 600) = 800

∴∠QXY = ∠P = 800 [XY//PR, corresponding angles are equal]

(ii) Since, the corresponding sides of similar triangles are proportional.

PR = PQ = QR
XY QX QY

or, 5 = 15 cm = QR
3 QX QY

Taking the first and the second ratios, we get

5 = 15 cm = QR
3 QX QY

or, 5 QX = 45 A
4 cm
∴ QX = 9 cm
BX
4. In the given figure, ∆ABC ~ ∆AXC, AX ⊥ BC, 3 cm 5 cm
AB = 3 cm and AC = 4 cm. Find the length of CX.
C

Vedanta Excel in Mathematics Teachers' Manual - 8 104

Solution:

Here,

In ∆ABC ∼ ∆AXC, AX ⊥ BC, AB = 3 cm, BC = 5 cm and AC = 4 cm

CX =?

Since, the corresponding sides of similar triangles are proportional.
BC AB AC
AC = AX = CX

or, 5 cm = 3cm = 4 cm
4 cm AX CX

Taking the first and the third ratios, we get
5 cm 4 cm
4 cm = CX

or, 5 CX = 16

∴ CX = 3.2 cm

A

5. a) In the adjoining figure, ∠ABC = ∠APQ = 55°. Show that P 55°
(ii) BC = AC . B 55°
(i) ∆ABC ~ ∆APQ PQ AQ Q
Solution: C

Here,

(i) In ∆ABC and ∆APQ

(1) ∠ABC = ∠APQ (A) [Given]

(2) ∠BAC= ∠PAQ (A) [Common angle]

(3) ∠ACB = ∠AQP (A) [Remaining angles of the triangles]

(4) ∆ABC ∼ ∆APQ [By A.A.A. axiom]

(ii) BC = AC [Being corresponding sides of similar triangles]
PQ AQ

6. In the given figure, ∠DEF = ∠DHG = 50°. Prove that, D

(i) ∆DHG ~ ∆DEF (ii) DG = GH 50° H F
Solution: DF EF G
Here, E 50°

(i) In ∆DHG and ∆DEF

(1) ∠DHG = ∠DEF (A) [Given]

(2) ∠HDG= ∠EDF (A) [Common angle]

(3) ∠GHD = ∠EFD (A) [Remaining angles of the triangles]

(4) ∆DHG ∼ ∆DEF [By A.A.A. axiom]

(ii) DG = GH [Being corresponding sides of similar triangles]
DF EF

7. In the figure alongside, PQ//WY. Show that, W

(i) ∆WXY ~ ∆PXQ (ii) WY.PX = WX.PQ P
Solution:

Here, X QY
(i) In ∆WXY and ∆PXQ

105 Vedanta Excel in Mathematics Teachers' Manual - 8

(1) ∠WXY= ∠PXQ (A) [Common angle]
[PQ//WY and corresponding angles]
(2) ∠XYW= ∠XQP (A) (A) [PQ//WY and corresponding angles]
[By A.A.A. axiom]
(3) ∠XWY = ∠XPQ
[Being corresponding sides of similar triangles]
(4) ∆WXY ∼ ∆PXQ
Proved
(ii) WY = WX
PQ PX

∴ WY.PX = WX.PQ

8. In the given figure, AB//CD. Prove that, A B
D
(i) ∆AOB ~ ∆COD (ii) AO.OC = BO.OD.
C
Solution: O

Here, C
(i) In ∆AOB and ∆COD

(1) ∠AOB= ∠COD (A) [Vertically opposite angles are equal]

(2) ∠OAB= ∠ODC (A) [AB//CD and alternate angles]

(3) ∠OBA = ∠OCD (A) [AB//CD and alternate angles]

(4) ∆AOB ∼ ∆COD [By A.A.A. axiom]

(ii) AO = BO [Being corresponding sides of similar triangles]
OD OC Proved

∴ AO.OC = BO.OD

9. In the adjoining figure, ABC is a right angled triangle A
where ∠A = 90°. If AD ⊥ BC, prove that,

(i) ∆ABC ~ ∆ABD (ii) ∆ABC ~ ∆ACD

(iii) ∆ABD ~ ∆ACD (iv) AB2 = BC.BD BD
(v) AC2 = BC.CD (vi) AC.BD = AB.AD

Solution: [Both are right angles]
[Common angle]
Here, [Remaining angles of triangles]
[By A.A.A. axiom]
(i) In ∆ABC and ∆ABD
(1) ∠BAC= ∠ADB (A) [Both are right angles]
(2) ∠ABC= ∠AD (A) [Common angle]
(3) ∠ACB = ∠BAD (A) [Remaining angles of triangles]
(4) ∆ABC ∼ ∆ABD [By A.A.A. axiom]
(ii) In ∆ABC and ∆ACD
(1) ∠BAC= ∠ADC (A) [Both are right angles]
(2) ∠ACB= ∠ACD (A) [From (ii), (3)]
(3) ∠ABC = ∠CAD (A) [Remaining angles of triangles]
(4) ∆ABC ∼ ∆ACD
(iii) In ∆ABD and ∆ACD
(1) ∠ADB= ∠ADC (A)
(2) ∠BAD= ∠ACB (A)
(3) ∠ABD = ∠CAD (A)

Vedanta Excel in Mathematics Teachers' Manual - 8 106

(4) ∆ABD ∼ ∆ACD [By A.A.A. axiom]

(iv) AB = BD ∴AB2 = BC. BD [Being corresponding sides of ∆ABC ∼ ∆ABD ]
BC AB

(v) AC = CD ∴AC2 = BC. CD [Being corresponding sides of ∆ABC ∼ ∆ADC ]
BC AC

(vi) AC = AD ∴AC.BD = AB.AD [Being corresponding sides of ∆ABD∼ ∆ACD]
AB BD

A 6 cm

10. In the adjoining figure, ∠BED = ∠BAC prove that, x cm 4 cm D 6 cm
(i) ∆ABC ~ ∆BDE

(ii) Find the value of x. CE 8 cm B
Solution:
9 cm A
C
Here,

(i) In ∆ABC and ∆BDE [Given]
(1) ∠BAC= ∠BED (A) [Common angle]
(2) ∠ABC= ∠DBE (A) [Remaining angles of triangles]
(3) ∠ACB = ∠BDE (A) [By A.A.A. axiom]
(4) ∆ABC ∼ ∆BDE

(ii) BC = AC = AB [Being corresponding sides of similar triangles]
BD DE BE
9 cm x 12 cm
or, 6 cm = 6 cm = 8 cm

Taking the first and second ratios, we get

6x = 36

∴x = 6 cm

11. In the figure given alongside, prove that, 8 cm 6 cm
(i) ∆ABC ~ ∆BDC D

(ii) Find the length of AC B 4 cm
Solution:

Here,

(i) In ∆ABC and ∆BDC

(1) ∠BAC= ∠BCD (A) [Given]

(2) ∠ABC= ∠DBC (A) [Common angle]

(3) ∠ACB = ∠BDC (A) [Remaining angles of triangles]

(4) ∆ABC ∼ ∆BDC [By A.A.A. axiom]
BC AC AB [Being corresponding sides of similar triangles]
(ii) BD = DC = BC

or, 4 cm = AC = 8 cm
2 cm 3 cm 4 cm

Taking the first and second ratios, we get

2AC = 12 cm
∴x = 6 cm

107 Vedanta Excel in Mathematics Teachers' Manual - 8

Extra Questions


1. In the adjoining figure, ∆ABC ∼ ∆EFG, find the value of x.

[Ans: 6 cm]
2. If ∆PMN ∼ ∆PQR, find the length of PQ.

[Ans: 10 cm]
3. In the given figure, ∆ABC ∼ ∆DEC, find the value of y.
[Ans: 2 cm]

4. In the figure given alongside, if QR = 6cm, SR
= 4cm, QS = 6cm and ∆PQR ∼ ∆QRS, find the
length of PQ.

5. In the given figure, AB // CD, show that: ∆AOB ∼ ∆COD.

Vedanta Excel in Mathematics Teachers' Manual - 8 108

Unit Geometry: Quadrilateral and Regular Polygon

16

Allocated teaching periods: 6

Competency
- Verifying experimentally the properties of parallelogram, rectangle and square
- Solving the problems based on regular polygon

Learning Outcomes
- To explore experimentally relationship between the opposite sides/angles of
parallelogram
- To verify that the diagonals of parallelogram bisect to each other.
- To explore experimentally the relationship of diagonals of rectangle/square
- To find the sum of interior angles of polygon, each angle of regular polygons.

Level-wise learning objectives

S.N. LEVELS OBJECTIVES
1. Knowledge (K)
2. Understanding (U) - To tell the relation of opposite sides/angles of parallelogram
- To recall the size of each angle of rectangle/square
3. Application (A) - To write the relation between the diagonals of rectangle/
4. High Ability (HA)
square
- To tell the formulae of finding the sum of interior angles of

polygons, each interior/exterior angle of regular polygons

- To solve the problems based on properties of parallelogram,
rectangle and square

- To find the sum of interior angles of n-gon
- To calculate the interior/exterior angle of regular polygons

- To explore experimentally the relationship between the
opposite sides/angles of parallelogram

- To verify experimentally that diagonals of parallelogram
bisect to each other.

- To explore experimentally the relationship between the each
angle of rectangle/square

- To establish the formulae of finding the sum of interior
angles of polygons and present in the classroom.

Required Teaching Materials/ Resources
Ruler, protractor, strips, sticks, chart paper, scissors, threads, ICT tools like GeoGebra etc.
Pre-knowledge: Quadrilateral, parallelogram, square etc.

109 Vedanta Excel in Mathematics Teachers' Manual - 8

Teaching Activities for Experimental Verification of Properties of
Parallelogram/Rectangle/ Square

1. Give the real life examples of representing parallelograms, rectangle and square.
2. Take a rectangular sheet of paper and fold it in to the diagonal. Ask about the relationships

of (i) opposite sides (ii) each angle (iii) diagonals of rectangle

3. With discussion, verify experimentally that the opposite sides of rectangle are equal, each
angle measures a right angle, diagonals are equal and do bisect to each other.

4. Fold the breadth of rectangular sheet of paper along the length and make a square. Make a
discussion on the relationships of (i) opposite sides (ii) each angle (iii) diagonals of square.

5. Experimentally verify that the opposite sides of square are equal, each angle measures a
right angle, diagonals are equal and do bisect to each other.

6. Under discussion, very experimentally that the (i) opposite sides of parallelogram are
equal (ii) opposite angles of parallelogram are equal. (ii) Diagonals of parallelogram

bisect to each other.

7. Make the group of students and give classwork to solve in group.

Solution of selected questions from Excel in Mathematics -Book 8

1. Calculate the unknown sizes of angles in the given figure. 5x+5 y

Solution: 4x-5 z
Here,

(i) (5x + 5) + (4x – 5) = 1800 [Co-interior angles between parallel lines]

or, 9x = 1800 ∴x = 200

(ii) y =4x – 5 =( 4×20 – 5) 0 = 750 [Being opposite angles of parallelogram]

z = y = 750 [Being alternate angles within parallel lines]

Find the value of x. Also, find the length of sides of parallelogram. D (2x–3) cm C
2.

Solution:

Here,

(i) AB = CD [The opposite sides of parallelogram are equal] A (x+7) cm B

or, 2x – 3 = x + 7 ∴x = 10

(ii) AB = CD = (2x – 3) cm = (2×10 – 3)cm = 17 cm

3. In the adjoining quadrilateral ABCD, ∠A = ∠C and D C
∠B = ∠D. Prove that, quadrilateral ABCD is a B
parallelogram.
Hint: Sum of the angles of quad. ABCD is 360°
A
Solution:

Here,

Given: In the quad. ABCD; ∠A = ∠C and ∠B = ∠D

To prove: ABCD is a quadrilateral.

Proof:

(i) ∠A = ∠C and ∠B = ∠D [Given]

Vedanta Excel in Mathematics Teachers' Manual - 8 110

(ii) ∠A+∠B +∠C +∠D = 1800 [The sum of interior angles of quad. is 3600]
[From (i) and (ii)]
(iii) ∠C+∠D+∠C +∠D = 1800
[From (iii), the sum of co-interior angles is 1800]
or, 2∠C+2∠D = 1800 [From (i) and (ii)]

or, ∠C+∠D = 900 [From (v), the sum of co-interior angles is 1800]
[From (iv) and (v)]
(iv) AD//BC

(v) ∠A+∠D+∠A +∠D = 1800

or, 2∠A+2∠D = 1800

or, ∠A+∠D = 900

(vi) AB//DC

(vii) ABCD is a parallelogram

4. In the adjoining quadrilateral PQRS, PQ // SR and S R
PS // QR. Prove that,

(i) ∆PQR ≅ ∆PRS (ii) PQ = SR (iii) PS = QR

Solution: PQ
Here,

Given: In the quad. PQRS; PQ//SR and PS//QR

To prove: (i) ∆PQR ≅ ∆PRS (ii) PQ = SR (iii) PS = QR
Proof:

1. In ∆PQR and ∆PRS;

(i) ∠QPR = ∠PRS (A) [SR//PQ and alternate angles]
[Common side]
(ii) PR = PR (S) [SP//RQ and alternate angles]
[By A.S.A. axiom]
(iii) ∠PRQ = ∠SPR(A) [The corresponding sides of congruent triangles are equal]

2. ∆PQR ≅ ∆PRS

3. PQ = SR and PS = QR

5. In the given figure, ABCD is a parallelogram. D C
Diagonals AC and BD are intersecting at O. Prove that, A
O
(i) ∆AOB ≅ ∆COD B
(ii) AO = OC and BO = OD.

Solution:

Here,

Given: In the parallelogram ABCD; the diagonals AC and BD interest at O.

To prove: (i) ∆AOB ≅ ∆COD (ii) AO = OC (iii) BO = OD

Proof:

1. In ∆AOB and ∆COD;

(i) ∠BAO= ∠OCD (A) [AB//DC and alternate angles]

(ii) AB = CD (S) [The opposite sides of parallelogram are equal]

(iii) ∠ABO = ∠ODC (A) [AB//DC and alternate angles]

2. ∆AOB ≅ ∆COD [By A.S.A. axiom]

3. AO = OC and BO = OD [The corresponding sides of congruent triangles are equal]

111 Vedanta Excel in Mathematics Teachers' Manual - 8

6. In the adjoining figure, ABCD is a rectangle. Prove that, D C
B
(i) ∆BAD ≅ ∆ABC

(ii) AC = BD A
Solution:

Given: ABCD is a rectangle.

To prove: (i) ∆BAD ≅ ∆ABC (ii) AC = BD

Proof:

1. In ∆BAD and ∆ABC;

(i) AD = BC (S) [The opposite sides of rectangle are equal]

(ii) ∠DAB= ∠ABC (A) [Both are right angles]

(iii) AB = AB (S) [Common side]

2. ∆BAD ≅ ∆ABC [By S.A.S. axiom]

3. AC = BD [The corresponding sides of congruent triangles are equal]

7. In the given figure, PQRS is a square. Prove that, SR

(i) ∆POQ ≅ ∆QOR

(ii) QO ⊥ PR. O
Solution: PQ

Given: PQRS is a square.

To prove: (i) ∆POQ ≅ ∆QOR (ii) QO⊥PR

Proof:

1. In ∆POQ and ∆QOR;

(i) PQ = QR (S) [The opposite sides of square are equal]

(ii) OP = OR (S) [The diagonals of square bisect to each other]

(iii) OQ = OQ (S) [Common side]

2. ∆POQ ≅ ∆QOR [By S.S.S. axiom]

3. ∠POQ = ∠QOR [The corresponding angles of congruent triangles are equal]

4. ∠POQ +∠QOR = 1800 [Linear pair]

5. ∠POQ +∠POQ = 1800 [From (3) and (4)]

6. or, 2∠POQ = 1800

∴ ∠POQ = 900

7. QO⊥PR [From (6)]

Extra Questions

1. Verify experimentally that the opposite sides of a parallelogram are equal. (Two figures of
different measurement are necessary)

2. Experimentally explore the relationship between the opposite angles of a parallelogram.
(Draw two figures of different measurement)

3. Experimentally verify that the diagonals of parallelogram bisect to each other. (Two figures
of different measurement are required)

4. Verify experimentally that all angles of rectangle are equal and each angle is right angle.
(Two rectangels of different measurement are required)

Vedanta Excel in Mathematics Teachers' Manual - 8 112

5. Explore experimentally the relationship between the diagonals of rectangle. (Two figures of
different measurement are required)

6. Verify experimentally that the diagonals of a rectangle bisect to each other. (Two figures of
different measurement are necessary)

7. Verify experimentally that the diagonals of square bisect to each other. (Two figures of
different measurement are necessary)

Teaching Activities for Regular Polygons

1. Show the models of different types of polygons like triangle, quadrilateral, pentagon,
hexagon, heptagon, octagon, nonagon, decagon etc.

2. Ask the sum of interior angles of triangle and quadrilateral. Make a discussion on the
regular polygons and its interior angles.

3. With the following table, discuss upon the sum of interior angles of polygons.

Polygons No. of sides (n) No. of triangles Sum of the interior angles of
Triangle
formed polygon

3 1 1 × 180° = (3 – 2) × 180°

Quadrilateral

4 2 2 × 180° = (4 – 2) × 180°

Pentagon

5 3 3 × 180° = (5 – 2) × 180°

Hexagon

6 4 4 × 180° = (6 – 4) × 180°

n – gon

n n–2 (n – 2) × 180°

4. Under experiments and discussion, list out the following points.

(i) The sum of interior angles of polygons = (n – 2) ×1800

(ii) Each interior angle of a regular polygon = (n – 2) × 180°
n
(iii) Each exterior angle of a regular polygon =36n0°

5. Make the group of students and give classwork to solve in group.

113 Vedanta Excel in Mathematics Teachers' Manual - 8

Solution of selected questions from Excel in Mathematics -Book 8

1. Find the number of sides of polygon whose sum of interior angles is 5400.
Solution:
Let ‘n’ be the required number of sides of the polygon.
Then, the sum of interior angles of polygon = (n- 2) ×1800
According to question, the sum of interior angles of the polygon = 5400
or, (n – 2) ×1800 = 5400
or, n – 2 = 3
or, n = 5
Hence, the required number of sides of the polygon is 5.

2. Find the number of sides of polygon whose sum of interior angles is 14400.
Solution:
Let ‘n’ be the required number of sides of the polygon.
Then, the sum of interior angles of polygon = (n- 2) ×1800
According to question, the sum of interior angles of the polygon = 14400
or, (n – 2) ×1800 = 14400
or, n – 2 = 8
or, n = 10
Hence, the required number of sides of the polygon is 10.

3. The size of each interior angle of a regular polygon is 1080. Find the number of sides of

polygon.

Solution:

Let ‘n’ be the required number of sides of the regular polygon.
(n – 2) × 180°
Then, the each interior angle of regular polygon = n

According to question, the size of each interior angle = 1080
or, (n – 2)n× 180° = 1080
or, 180n – 360 = 108n

or, 72n = 360

or, n = 5

Hence, the required number of sides of the regular polygon is 5.

4. The size of each interior angle of a regular polygon is 1400. Find the number of sides of

polygon.

Solution:

Let ‘n’ be the required number of sides of the regular polygon.
(n – 2) × 180°
Then, the each interior angle of regular polygon = n

According to question, the size of each interior angle = 1400
or, (n – 2)n× 180° = 1400
or, 180n – 360 = 140n

or, 40n = 360

or, n = 9

Hence, the required number of sides of the regular polygon is 9.

Vedanta Excel in Mathematics Teachers' Manual - 8 114

5. Find the number of sides of regular polygon whose size of each exterior angle is 720.
Solution:

Let ‘n’ be the required number of sides of the regular polygon.
360°
Then, the each exterior angle of regular polygon = n

According to question, the size of each exterior angle = 720

or, 36n0° = 720

or, 72n = 360

or, n = 5

Hence, the required number of sides of the regular polygon is 5.

6. Find the number of sides of regular polygon whose size of each exterior angle is 400.
Solution:

Let ‘n’ be the required number of sides of the regular polygon.
360°
Then, the each exterior angle of regular polygon = n

According to question, the size of each exterior angle = 400

or, 36n0° = 400

or, 40n = 360

or, n = 9

Hence, the required number of sides of the regular polygon is 9.

7. Find the unknown sizes of angles of the following polygon.

Solution: x

Here, x is an exterior angle of regular hexagon. 360° 360°
n 6
Now, x = size of each exterior angle of hexagon = = = 600

8. Find the value of x from the following polygon. xx
165° 160°

Solution: (x+10)°
Here, the sum of interior angles of the octagon = (n – 2) ×1800 160°
50° 60°

or, x + 1650 + 1600 + (1800 – 500) + (1800 – 600) + (x + 100) + 1600 + x = (8 – 2) ×1800

or, 3x + 7450 = 10800

or, 3x = 3350

∴x = 111.670

9. Find the value of x in each case.

Solution: x

(a)

Here, 360° 360°
n 4
Interior angle of square (a) = = = 900

115 Vedanta Excel in Mathematics Teachers' Manual - 8

Interior angle of regular hexagon (b) = 360° = 360° = 600
n 6
∴x = a + b = 90° + 60° = 150°

(b)

Here, 360° 360°
n 4
a = each exterior angle of square = = = 900 x
x
Each exterior angle of regular pentagon= 360° = 360° = 720
n 5
(n – 2) × 180° (3 – 2) × 180°
Each interior angle of equilateral triangle = n = 3 = 60°

∴ b = 72° – 60° = 120

Hence, x = a + b = 90°+12° = 102°

(c)

Here, (n – 2) × 180°
n
a = each interior angle of regular pentagon =

= (5 – 2) × 180° = 108°
5

b = x. [ Beings base angles of isosceles triangle]

Now, a + b + x = 180°

or, 108° + x + x = 180°

or, 2x = 72°

∴ x = 36°

(d) Here, (n – 2) × 180°
n
a = each interior angle of regular hexagon =

= (6 – 2) × 180° = 120°
6

b = c [ Beings base angles of isosceles triangle] x
Now, a + b + c = 180°
or, 120° + b + b = 180°
or, 2b = 60°
∴ b = 30°
Again, x = 120° – b = 120° – 30° = 90°

Extra Questions [Ans: 12600]

1. Calculate the sum of interior angles of nonagon.

2. Find the number of sides of polygon whose sum of interior angles is 7200 [Ans: 6]

3. The size of each interior angle of a regular polygon is 1350. Find the number of sides of

polygon. [Ans: 8]

4. Find the number of sides of regular polygon whose size of each exterior angle is 360.

[Ans: 10]

Vedanta Excel in Mathematics Teachers' Manual - 8 116

Unit Geometry: Construction

17

Allocated teaching periods: 4

Competency

- Constructing rectangle and regular polygons (pentagon, hexagon and octagon)
Learning Outcomes

- To construct rectangle.
- To construct regular pentagon
- To construct regular hexagon
- To construct regular octagon
Level-wise learning objectives

S.N. LEVELS OBJECTIVES
1. Knowledge (K)
2. Understanding (U) - To tell the properties of rectangle
- To recall the formula to find each interior/exterior angle of
3. Application (A)
regular polygon

- To find the interior angle of regular polygon
- To find the exterior angle of regular polygon
- To sketch the rough figure for the construction

- To construct rectangle.
- To construct regular pentagon
- To construct regular hexagon
- To construct regular octagon

Required Teaching Materials/ Resources

Ruler, protractor, compass, chart paper, ICT tools like GeoGebra etc.

Pre-knowledge: Each interior/exterior angle of regular polygon

Teaching Activities for Construction

1. Discuss about the procedures of constructions of rectangle in various conditions and give
the different information to construct in groups.

2. Explain the constructions procedures of regular pentagon by both methods (i) finding the
interior angle (ii) inside the circle and give the different information to construct in groups.

3. Explain the constructions procedures of regular hexagon by both methods (i) finding the
interior angle (ii) inside the circle and give the different information to construct in groups.

4. Explain the constructions procedures of regular octagon by both methods (i) finding the
interior angle (ii) inside the circle and give the different information to construct in groups.

117 Vedanta Excel in Mathematics Teachers' Manual - 8

Solution of selected questions from Excel in Mathematics -Book 8

1. When the diagonals and an angle made by them are given X
Construct a rectangle ABCD in which diagonals D

AC = BD = 5cm and they bisect each other making an
angle of 60°.

Steps of construction.

(i) Draw AC = 5 cm and draw its perpendicular bisector A 60° C to
find its mid-point O. O of

(ii) At O, construct ∠AOX = 60° and produce XO

to Y. 1
2
(iii) With centre at O and radius 2.5 cm ( BY

BD) draw two arcs to cut OX at D and OY at B.

(iv) Join A, D; B, C; A, B and C, D. Thus, ABCD is the required rectangle.

2. When a side and an angle made by the diagonal with the side are given.

Construct a rectangle ABCD in whch AB = 5.4 cm and ∠BAC = 30°.

Steps of construction D X
(i) Draw AB = 5.4 cm. C
B
(ii) At A, construct ∠BAX = 30°

(iii) At B, construct an angle of 90°.

It intersects AX at C.

(iv) With centre at C and radius equal

to the length of AB draw an arc. 30°
A
(v) With centre at A and radius equal
to the length of BC draw another

arc to cut the previous arc at D.

(vi) Join A, D and C, D. Thus ABCD is the required rectangle.

3. When a diagonal and angle made by it with a side are given

Construct a rectangle PQRS in which diagonal X
S
PR = 6 cm and ∠RPS = 60° Y

Steps of construction 60° 30° R
(i) Draw a diagonal PR = 6 cm P 30° 6 cm Z
(ii) Construct ∠RPX = 60° at P.
(iii) Construct ∠PRY= 30° at R. Q
RY intersects PX at S
(iv) Construct ∠RPZ = 30° at P.
(v) Draw an arc with radius equals to RS from
P to cut PZ at Q.

Vedanta Excel in Mathematics Teachers' Manual - 8 118

(vi) Join R, Q.
PQRS is the required rectangle.

Construction of a regular pentagon

1. Construct a pentagon with the length of each side 4.5 cm.

In a pentagon, n = 5. So, each interior angle of pentagon

= n nC– o2ns×tru1c8t0io° n= 5 – 2 × 180° = 108°
Steps of 5
QD
(i) Draw AB = 4.5 cm. 4.5cm

(ii) At B, draw ∠ABP = 108° with the help of P

a protractor. E 108° C
(iii) Draw an arc of radius 4.5 cm from B to cut

BP at C. R

(iv) At C, draw BCQ = 108° with the help of 4.5cm

the protractor.

(v) Draw an arc of radius 4.5 cm from C to cut

CQ at D. A 4.5cm 108°
(vi) At D, draw ∠CDR = 108° with the help of B

the protractor.

(vii) Draw an arc of radius 4.5 cm from D to cut DR at E.

(viii) Join E, A

ABCDE is the required regular pentagon.

2. Construct a regular pentagon inside a circle with diameter 6 cm.

Steps of construction R
(i) Draw a line segment PQ = 6 cm and construct its A

perpendicular bisector RS that intersects PQ at O.
(ii) With the centre at O and the radius OP = OQ, draw

a circle that intersects RS at A and X.

(iii) Construct the perpendicular bisector of OQ that
cuts OQ at T.
B E
(iv) From T draw an arc of radius equal to TA to cut OP TQ
at U.
P UO at
(v) From U take an arc of radius equal to UA and starting
from A draw arcs to cut the circumference of the circle D
B, C, D and E respectively.

(vi) Join A, B, C, D and E.

Now, ABCDE is the required regular pentagon. C

X

S

119 Vedanta Excel in Mathematics Teachers' Manual - 8

Construction of a regular hexagon RE Q of
1. Construct a regular hexagon with the length 120° P

each side 3.5 cm. 3.5cm

In a hexagon, n = 6.

So, each interior angle of hexagon
n – 2 6–2
= n × 180° = 6 × 180° = 120°. 120° C

Steps of construction

At every vertex, construct an angle of 120°. S 3.5cm
A 3.5cm B
Then, follow the similar process as in case of

construction of regular pentagon.

2. Construct a regular hexagon of a side 4.5 cm inside a circle.

Steps of construction E D
(i) Draw a straight line segment AB = 4.5 cm. O

(ii) From A and B draw two arcs each of radius B

equal to AB to interest at O.

(iii) With the centre at O and radius equal to OA

draw a circle. C

(iv) Starting from B draw arcs each of radius equal F

to OA to cut the circumference of the circle at

C, D, E and F respectively.

(v) Join B, C, D, E, F and A.

Now, ABCDEF is the required regular hexagon.

A

Construction of a regular octagon

Construct a regular octagon with the length of each side 3 cm.

In an octagon, n = 8. So, each interior angle of
of
octagon S F R Q
135° E
n – 2 8 – 2
n 8 135°
= × 180° = × 180° = 135° 135°
135° 3 cm
G 135° 3 cm135° D

Steps of construction

At every vertex, construct an angle of 135° T H 135° C
and follow the process as like in the case
construction of regular pentagon. U 135° 3 cm
A 3 cm B

Vedanta Excel in Mathematics Teachers' Manual - 8 120

1. Construct a regular octagon inside a circle with diameter 4.8 cm.

Steps of construction C
G
(i) Draw a straight line segment AE = 4.8 cm

and construct its perpendicular bisector. The B
perpendicular bisector intersects AE to O.

(ii) With centre at O and radius equal to OA, construct A E
a circle such that the circle cuts the perpendicular F

bisector at C and G.

(iii) Construct the angular bisector of ∠AOC that cuts the H

circumference of the circle at B and F.

(iv) Construct the angular bisector of ∠EOC that cuts the

circumference of the circle at D and H.

(v) Join A, B, C, D, E, F, G, H and A.

Now, ABCDEFGH is the required regular octagon.

Extra Questions

1. Construct a regular hexagon with side 4.8 cm by using compass.

2. Construct a regular hexagon in a circle measuring 5 cm diameter.

3. Find the interior angle and construct a regular pentagon with side 4 cm.

4. Construct a regular pentagon inside the circle of diameter 6 cm.

5. Construct a rectangle ABCD in which the sides AB = 5cm and BC = 4cm.

6. Construct a rectangle ABCD having diagonal AC = 6cm and angle between the diagonals
AC and BD is 600.

7. Construct a rectangle ABCD in which AB is 6cm and ∠CAB is 300

8. Construct a parallelogram PQRS in which QR = 4 cm, diagonal QS = 5 cm and ∠SQR is
300

9. Construct a parallelogram ABCD in which AB = 5 cm, diagonals AC is 6 cm and BD is
8 cm.

121 Vedanta Excel in Mathematics Teachers' Manual - 8

Unit Coordinates

18

Allocated teaching periods: 5
Competency

- Verifying and using Pythagoras theorem.
- Using formula to calculate the distance between two points
Learning Outcomes
- To verify the Pythagoras theorem
- To find the distance between any two points
Level-wise learning objectives

S.N. LEVELS OBJECTIVES

1. Knowledge (K) - To recall the axes and quadrants
- To state the Pythagoras theorem
- To tell the distance formula

- To verify the Pythagorean triples
2. Understanding (U) - To solve the simple problems using Pythagoras theorem

- To find the distance between two points

3. Application (A) - To apply Pythagoras theorem in solving problems
- To show that the points are vertices of certain polygons
- To show that the given points are collinear

- To derive the Pythagoras theorem
4. Higher Ability (HA) - To find the distance between any two points taken from

graph paper by measuring and using formula

Required Teaching Materials/ Resources
Graph board/ chart /paper, ruler, ICT tools like GeoGebra etc.
Pre-knowledge: Axes, quadrants, area of square etc.
Teaching Activities for Pythagoras theorem
1. Use graph board/chart and make a short discussion on the coordinate system: origin, axes

and quadrants.
2. Activity-I:
(i) Fold one corner of the paper to meet the opposite edge along

the length.
(ii) Line up these two edges and crease the fold. You should now

have a right triangle made from the folded sheet and an extra
rectangle of single-sheet paper.
(iii) Cut off the extra paper along the edge of the triangle.

Vedanta Excel in Mathematics Teachers' Manual - 8 122

(iv) Select a smallest right angled triangle in the middle of the diagram and draw its outline
as shown in the figure.

(v) Draw the outlines of squares standing on each side of right angled triangle.
(vi) Count the number of triangles formed inside each square and verify the relation

h2 = p2 + b2
3. Activity-II:
In graph paper/board, draw a right angled triangle with proper sides like 3, 4, 5 units and

draw the squares along base, perpendicular and hypotenuse. Then discuss upon the relation

among the areas.
4. Activity-III:
Use GeoGebra tool to visualize h2 = p2 + b2 by different ways.

5. Activity-IV:
Experimentally verify that the Pythagoras theorem.

6. Solve the problems related to Pythagoras theorem in group.

Solution of selected questions from Excel in Mathematics -Book 8

1. In the adjoining figure, ABC is a triangle. If BD ⊥ AC, B
AB = BC = 15 cm, BD = 9 cm, find the length of AC.
Solution: 15 cm 9 cm 15 cm

Here, in right angled triangle ABD; AB (h) = 15 cm, BD A D C
(p) = 9 cm, AD (b) =?

Now, by using Pythagoras theorem

h2 = p2 + b2

or, 152 = 92 + b2

or, 225 = 81 + b2

or, 144 = b2

or, 122 = b2

or, b = 12 ∴AD (b) = 12 cm

Also, in right angled triangle BCD; BC (h) = 15 cm, BD (p) = 9 cm, CD (b) =?

Now, by using Pythagoras theorem

123 Vedanta Excel in Mathematics Teachers' Manual - 8

h2 = p2 + b2
or, 152 = 92 + b2
or, 225 = 81 + b2

or, 144 = b2
or, 122 = b2
or, b = 12 ∴CD (b) = 12 cm
Hence, the length of AC = AD + CD = 12 cm + 12 cm = 24 cm

2. If the length and breadth of a rectangle are 12 cm and 5 cm 12 cm

respectively, what will be the length of its diagonal? 16 cm
Solution:
Here, in a right angled triangle of rectangle; length (b) = 16 cm,

breadth (p) = 12 cm, diagonal (h) =?
Now, by using Pythagoras theorem
h2 = p2 + b2
or, h2 = 162 + 122
or, h2 = 256 + 144

or, h2 = 400
or, h2 = 202
or, h = 20
∴The length of diagonal of the rectangle is 20 cm.

3. In the adjoining figure, find the values of x and y. S 13 cm
Solution: x cm
Here, in right angled triangle PQR; PR (h) = x cm, PQ y cm R
3 cm

(b) = 4 cm, QR (p) = 3 cm P 4 cm Q
Now, by using Pythagoras theorem

h2 = p2 + b2

or, x2 = 32 + 42

or, x2 = 9 + 16

or, x2 = 25

or, x2 = 52

or, x= 5 ∴x = 5 cm

Also, in right angled triangle PRS; RS (h) = 13 cm, SP (p) = y cm, PR (b) = x = 5 cm

Now, by using Pythagoras theorem

h2 = p2 + b2

or, 132 = y2 + 52

or, 169 = y2 + 25

or, 144 = y2

or, 122 = y2

or, y = 12 ∴y = 12 cm

Hence, x = 5 cm and y = 12 cm.

Vedanta Excel in Mathematics Teachers' Manual - 8 124

4. A ladder 10 m long rests against a vertical wall 6 m 10 m A 6m
above the ground. At what distance does the ladder C
touch the ground from the bottom of the wall?

Solution: B

Here, length of ladder = AB (h) = 10 m, height of wall =
AC (p) = 6 m and the distance between the foot of ladder
and bottom of the wall = BC (b) =?

Now, by using Pythagoras theorem
h2 = p2 + b2
or,102 = 62 + b2
or, 100 = 36 + b2

or, b2 = 64
or, b2 = 82
or, b= 8
∴The ladder touches 8 m far from the bottom of the wall on the ground.

5. The vertical height of a tree is 18 m. The uppermost part of A
the tree is broken by wind 5 m above the ground. At what
distance does the uppermost part of the tree touch the ground
from the foot of the tree?

Solution: BC
Here,

The vertical height of the tree before it was broken by the wind = AB = 18 m

The height of tree after it was broken = 5 m

The broken length of broken part of the tree = 18m – 5 m = 13 m

Now,

Hypotenuse (h) = 13 m, perpendicular (p) = 5 m and base (b) = ?

Now, by using Pythagoras theorem

h2 = p2 + b2

or,132 = 52 + b2

or, 169 = 25 + b2

or, b2 = 144

or, b2 = 122

or, b= 12

Hence, the top of the tree touches the ground at the distance of 12 m from the foot of

the tree.

6. In the figure alongside, ABCD is a rectangular ground B C

of length 40 m and breadth 30 m. Ramesh reached at C 30 m
walking from A to B and B to C. But Shyam reached to
C walking from A to C directly. Who walked the shorter
distance and by how much? A 40 m D

125 Vedanta Excel in Mathematics Teachers' Manual - 8

Solution:
Here,
Distance walked by Ramesh = AB + BC = 30 m + 40 m = 70 m
Also, AD (b) = 40 m, DC (p) = 30 m and AC (h) =?
Now, by using Pythagoras theorem
h2 = p2 + b2
or,h2 = 302 + 402
or, h2 = 900 + 1600
or, h2 = 2500
or, h2 = 502
or, h= 50
The distance walked by Shyam = AC = 50 m
Difference = 70 m – 50 m = 20 m
Hence, Shyam walked 20m the shorter distance.

Extra Questions

1. A 30 foot ladder is leaning against the side of a building and is positioned such that the

base of the ladder is 24 feet from the base of the building. How far above the ground is

the point where the ladder touches the building? [Ans: 18 ft]

2. Sudhir walked 56 m towards east and then 33 m towards south. How far is he from his
starting point? [Ans: 65 m]

3. In the figure, rectangle ABCD represents the
play-ground.

Eleena stands at the point E. Calculate the
distance of Eleena from

(i) the corner B of the play-ground.

(ii) the corner C of the play-ground.

[Ans: 10 m, 17 m]

4. A room is 12 m long, 4m broad and 3 m high. Find the length of longest rod that can be
placed in the room. [Ans: 13 m]

Teaching Activities for Distance between Two Points

1. Draw a line segment and discuss about its length.
2. Take any two points on a graph, join them and find its length.
3. Recall the coordinates of any two points, derive the distance between them using

Pythagoras theorem.

4. Note that the distance between the points (x1, x2) and (y1, y2) is given by d =

Vedanta Excel in Mathematics Teachers' Manual - 8 126

5. Take any two points and join them.
(i) Find the distance between them by measuring the length of line segment.
(ii) Find the distance between them by using formula
Then, verify the distance between the points.

6. Discuss upon the problems related to the distance formula
7. Recall the following properties of polygons.
(i) In equilateral triangle, the lengths of all the sides are equal.
(ii) In isosceles triangle, the lengths of only two sides are equal.
(iii) In scalene triangle, the lengths of all sides are unequal.
(iv) In parallelogram, the lengths of opposite sides are equal.
(v) In rectangle, the opposite sides are equal and the diagonals are equal.
(vi) In square, the lengths of all four sides are equal and the diagonals are equal.
(vii) In rhombus, the lengths of all four sides are equal and but the diagonals are unequal.

Solution of selected questions from Excel in Mathematics -Book 8

1. Find the distance between the following points.

(i) (4, 5) and (-3, 6) (ii) (5 + 3, 2 – 3) and (7 + 3, 2 + 3)

Solution:
(i) Here,
The given points are (4, 5) → (x1, y1­) and (-3, 6) → (x2, y2)
Now, by using distance formula

d = (x2 – x1)2 + (y2 – y1)2 = (–3 –4)2 + (6 – 5)2 = (–7)2 + (1)2
= 49 + 1 = 50 = 52 × 2 = 5 2units

(ii) Here, 3 ) → (x1, y1­) and (7 + 3, 2 + 3 )→ (x2, y2)
The given points are (5 + 3, 2 –

Now, by using distance formula

d = (x2 – x1)2 + (y2 – y1)2
= [7 + 3 – (5 + 3)]2 + [2 + 3 – (2 – 3)]2

= (7 + 3 – 5 – 3)2 + (2 + 3 – 2 – 3)2
= (2)2 + (2 3)2 = 4 + 12 = 16 = 4 units

2. A point P is on the x-axis, 6 units right from the origin and another point Q is on the
y-axis, 8 units above the origin. Find the distance between the points P and Q.
Solution:
Here,
The given points are P (6, 0) → (x1, y1­) and Q (0, 8) → (x2, y2)
Now, by using distance formula

127 Vedanta Excel in Mathematics Teachers' Manual - 8

d = (x2 – x1)2 + (y2 – y1)2 = (0 – 6)2 + (8 – 0)2 = (–6)2 + (8)2

= 36 + 64 = 100 = 10 units
Hence the distance between the points P and Q is 10 units.

3. A point A lies on the x-axis, 9 units left from the origin and another point B lies on the
y-axis 12 units below the origin. Find the distance between A and B.
Solution:
Here,
The given points are A (-9, 0) (x1, y1­) and Q (0, -12) → (x2, y2)
Now, by using distance formula
d = (x2 – x1)2 + (y2 – y1)2 = (0 + 9)2 + (–12 – 0)2 = (9)2 + (–12)2

= 81 + 144 = 225 = 15 units
Hence the distance between the points A and B is 15 units.

4. When a map of Nepal is presented in a coordinate plane, the coordinates of Kathmandu
is found to be (5 3) and that of PokharA (4, 2 3). Find the map distance between these
points. If 1 unit represents 100 km, find the actual distance of Pokhara and Kathmandu.
Solution:
Here,
The position of Kathmandu is (5 3 ) → (x1, y1­)
The position of Pokhara is (4, 2 3 ) → (x2, y2)
Now, by using distance formula

d = (x2 – x1)2 + (y2 – y1)2
= (4 – 5)2 + (2 3 – 3)2

= (– 1)2 + ( 3 )2

= 1 + 3 = 4 = 2 units

Given that 1 unit = 200 km. ∴2 units = 2×100 km = 200 km
Hence the actual distance between Pokhara and Kathmandu is 200 km.

5. The centre of circle lies at O (4, 6) and A (-5, 18) is any point in the circumference.

(i) Find the radius of the circle.

(ii) Find the diameter of the circle.

(iii) Show that B (13, -6) also lies on the circumference of the circle.

Solution:

Here,

The centre of circle is O (4, 6) → (x1, y1­) (–9)2 + (12)2
Pint on the circumference A (-5, 18) → (x2, y2)
(i) By using distance formula

OA (d) = (x2 – x1)2 + (y2 – y1)2 = (–5 – 4)2 + (18 – 6)2 =

Vedanta Excel in Mathematics Teachers' Manual - 8 128

= 81 + 144 = 225 = 15 units
Hence, the radius of the circle is 15 units.
(ii) The diameter of the circle = 2×OA = 2×15 units = 30 units

(iii) Again, finding the distance between the points O (4, 6) and B (13, -6)

OB (d) = (x2 – x1)2 + (y2 – y1)2 = (13 – 4)2 + (–6 – 6)2 = (9)2 + (–12)2
= 81 + 144 = 225 = 15 units

Since, OA = OB. So, OB is also the radius of the circle and hence the point
(13, -6) lies on the circumference.

6. Show that the points A (3, 4), B (7, 8) and C (11, 4) are the vertices of an isosceles
triangle.
Solution:
Here, the given vertices of triangle are A (3, 4), B (7, 8) and C (11, 4).
Now, by using distance formula
AB = (x2 – x1)2 + (y2 – y1)2 = (7 – 3)2 + (8 – 4)2 = 42 +42 = 4 2 units
BC = (x2 – x1)2 + (y2 – y1)2 = (11 – 7)2 + (7 – 4)2 = (–3)2 + 32 = 4 2 units
CA = (x2 – x1)2 + (y2 – y1)2 = (11 – 3)2 + (4 – 4)2 = (8)2 + 02 = 8 units

Since, AB = BC.
Thus, the given points are the vertices of an isosceles triangle.

7. Show that the points A (2, -2), B (8, 4), C (5, 7) and D (-1, 1) are the vertices of a rectangle.
Solution:
Here, the given vertices of quadrilateral are A (3, 4), B (7, 8) and C (11, 4).
Now, by using distance formula
AB = (x2 – x1)2 + (y2 – y1)2 = (8 – 2)2 + (4 + 2)2 = 62 +62 = 6 2 units
BC = (x2 – x1)2 + (y2 – y1)2 = (5 – 8)2 + (7 – 4)2 = (–3)2 + 32 = 3 2 units
CD = (x2 – x1)2 + (y2 – y1)2 = (–1 – 5)2 + (1 – 7)2 = (–6)2 + (–6)2 = 6 2 units
DA = (x2 – x1)2 + (y2 – y1)2 = (2 + 1)2 + (–2 – 1)2 = (3)2 + (–3)2 = 3 2 unit

Again,
AC = (x2 – x1)2 + (y2 – y1)2 = (5 – 2)2 + (7 + 2)2 = (3)2 + (9)2 = 90 unit
BD = (x2 – x1)2 + (y2 – y1)2 = (–1 – 8)2 + (1 – 4)2 = (–9)2+(–3)2 = 90 unit
Thus in quadrilateral ABCD, the sides AB = CD = 6 2 units, BC = DA = 3 2 units
Also, diagonal AC = diagonal BD = 90 units.
Since, in the quadrilateral ABCD; the opposite sides are equal and the diagonals are
also equal. Thus, the given points are the vertices of a rectangle.

129 Vedanta Excel in Mathematics Teachers' Manual - 8

8. Show that the points P (1, 1), Q (-1, -1) and R (– 3, 3) are the vertices of an equilateral
triangle.
Solution:
Here, the given vertices of triangle PQR are P (1, 1), Q (-1, -1) and R (– 3 , 3 )
Now, by using distance formula

PQ = (x2 – x1)2 + (y2 – y1)2 = (–1 – 1)2 + (–1 – 1)2 = (–2)2 + (–2)2 = 2 2 units
QR = (– 3 + 1)2 + [ 3 + 1)2 = 3 – 2 3 +1 + 3 + 2 3+ 1 = 2 2 units

PR = (– 3 – 1)2 + [ 3 – 1)2 = 3 + 2 3 +1 + 3 – 2 3+ 1 = 2 2 units

In triangle PQR; PQ = QR = PR. Thus, the given points are the vertices of an equilateral
triangle.

9. Show that the points (1, 2), (4, 5) and R (8, 9) are collinear.
Solution:
Let A (1, 2), B (4, 5) and C (8, 9) are the given points.
Now, by using distance formula
AB = (x2 – x1)2 + (y2 – y1)2 = (4 – 1)2 + (5 – 2)2 = 32 +32 = 3 2 units

BC = (x2 – x1)2 + (y2 – y1)2 = (8 – 4)2 + (9 – 5)2 = 42 +42 = 4 2 units

AC = (x2 – x1)2 + (y2 – y1)2 = (8 – 1)2 + (9 – 2)2 = 72 +72 = 7 2 units

Here, AB + BC = 3 2 units + 4 2 units = 7 2 units = AC.
Hence, the given points are collinear.

10. If the distance between the points (a, 3) and (8, 3) is 7 units, find the value of a.
Solution:
Let P (a, 3) and Q (8, 3) are two given points.
Now, by using distance formula

PQ = (x2 – x1)2 + (y2 – y1)2

or, 7 = (8 – a)2 + (3 – 3)2

or, 7 = (8 – a)2
or, 7 = 8 – a
∴a=1
Hence, the required value of ‘a’ is 1.

Extra Questions

1. If the distance between the points (3, 0) and (0, a) is 5 units, find the value of a. [Ans: 4]

2. The centre of circle lies at origin and A (-5, 12) is any point in the circumference. Find

the diameter of the circle. [Ans: 26 units]

3. Show that the points (1, 1), (6, 6) and (9, 9) are collinear.

Vedanta Excel in Mathematics Teachers' Manual - 8 130

Unit Circle

19

Allocated teaching periods: 2
Competency

- Solving the problems on perimeter and area of the circle
Learning Outcomes

- To find the circumference of circle.
- To find the area of circle.
Level-wise learning objectives

S.N. LEVELS OBJECTIVES

1. Knowledge (K) - To define circle
- To identify the parts of circle
- To tell the formulae of perimeter and area of circle

- To find the perimeter of the circle
2. Understanding (U) - To find the area of the circle.

3. Application (A) - To solve some word problems based on the circumference
of the circle.

- To solve some word problems based on the area of the
circle.

4. Higher Ability (HA) - To prepare a project work

Required Teaching Materials/ Resources
Wire/ thread/ rope, ruler, compass, chart paper, models of circle, ICT tools like GeoGebra etc.
Pre-knowledge: Circular objects, parts of circle
Teaching Activities
1. Discuss about the circular objects.
2. Draw a circle with its various parts and make a discussion on it.
3. To find the formula of perimeter of circle, do the following activities
Activity 1
Using the wire/thread of different lengths, make the circles separately. Measure the lengths of
each wire/thread separately by placing them on the ruler. Ask about the wire/thread used to
form the circles and their circumferences. Mark a point on the circular wire/thread and measure
the lengths diameter (d) in each successively then fill up the table given below.

Fig. (i) Fig. (ii) Fig. (iii)


131 Vedanta Excel in Mathematics Teachers' Manual - 8

Fig. no. Length of wire (C) Length of diameter (d) Value of C
d
(i) … …
(ii) … … …
(iii) … …
…

…

Check whether the ratio of circumference and diameter of the circles and introduce about
pie (π).

Activity 2

Make the group of students; provide 1/1 piece of chart paper, thread and compass with pencils

in each group. Tell them to follow the following steps.

(i) Draw three circles of different radii in chart paper.
(ii) Mark a point on the circumference of each figure and draw the diameter from the point.
(iii) Measure the length of diameter in each circle.
(iv) Use thread to measure the circumference of each figure.
(v) Complete the following table.

Fig. no. Circumference (C) Diameter (d) Value of C Remark
d

(i) … ……

(ii) … ……

(iii) … ……

(vi) Draw a conclusion and present in the class.

4. Discuss about the formula that measures the circumference of the circle as C = πd or 2πr.
5. For finding the area formula of circle, do the following activities.
(i) Take a circular piece of paper.

(ii) Cut it into equal pieces as small as possible and arrange the pieces as shown in the

figure.
DC

A B r (breadth)
1
2 of 2πr

(length)

Ask the following questions to show the formula of area of circle.

(i) In how many pieces, has the circle been divided equally?
(ii) What is the name the new shape formed by arranging the pieces?
(iii) What is the length of the rectangle?
(iv) What is the breadth of the rectangle?
(v) What is the area of the rectangle?
(vi) Is the area of rectangle equal to the area of the circle?
(vii) What is the formula to find the area of the circle?

Vedanta Excel in Mathematics Teachers' Manual - 8 132

6. If possible show the perimeter and area of circle by using GeoGebra.
7. List out the following points.
(i) The value of π is obtained by dividing the length of circumference of the circle by the

length of its diameter. The approximate value of π is 3.141592….
(ii) The formula of perimeter (circumference) of the circle = 2πr
(iii) The formula of area of the circle = πr2
8. Solve the problems related to Pythagoras theorem in group.

Solution of selected questions from Excel in Mathematics -Book 8

1. The radius of a circular play ground is 77 m. calculate the distance covered by a runner

in 5 complete rounds around the ground.

Solution:

Here, radius of the circular ground = 77 m

Now, covered in 1 round = perimeter of the ground = 2πr = 2 × 22 × 77 m = 484 m
Distance 7

∴Distance covered in 5 rounds = 5×484 m = 2420 m

2. When a marathon runner completes 3 rounds around a circular track, he covers a
distance of 66 km. find the radius of the track.

Solution:
Here,
The distance covered in 3 rounds = 66 km
The distance covered in 1 round = 22 km
We know, the perimeter of the tack = distance covered in 1 round

or, 2πr = 22 km

or, 2× 22 × r = 22 km
7
∴r = 3.5 km

Hence, the radius of the track is 3.5 km

3. If 3520 ft of wire is required to fence a circular garden with 4 rounds, find the diameter

of the garden.

Solution:

Here,

The required length of wire to fence a circular garden with 4 rounds =3520 ft

The length of wire to fence the garden with 1 round =880 ft

We know, the perimeter of the garden = length of wire to fence with 1 round

or, πd = 880 ft

or, 22 × d = 880 ft
7
∴d = 280 ft

Hence, the diameter of the garden is 280 ft.

133 Vedanta Excel in Mathematics Teachers' Manual - 8

4. The area of a circular pond is 1386 m2. Find the length of wire to fence around it with 1

round.

Solution:

Here,

The area of the pond =1386 m2

or, πr2 =1386 m2

or, 22 × r2 = 1386 m2
7

∴r = 21 m

Again, of the pond = 2πr = 2 × 22 × 21 m = 132 m
The perimeter 7

Hence, the length of wire to fence around the pond with 1 round is 132 m.

5. Samriddhi bent a rope to form a circle. If she got the circumference of the circle 30 cm

more than its diameter. What was the length of radius?

Solution:

Here,

Circumference (C) = 2πr and diameter (d) = 2r

According to question, C – d = 30 cm

or, 2πr – 2r = 30

or, 2 × 22 × r – 2r = 30
7
44r – 14 r
or, 7 = 30

or, 30r = 210

or, r = 7 cm

Hence, the radius of the circle is 7 cm.

6. The rope required to measure the diameter of a circular pond is 90 m shorter than
measuring the circumference. Find the diameter of the pond.

Solution:

Here,

Circumference (C) = 2πr and diameter (d) = 2r
According to question,

C – d = 90 m

or, 2πr – 2r = 90

or, 2× 22 × r – 2r = 90
7

or, 44r – 14 r = 90
7

or, 30r = 630

or, r = 21 m

Hence, the diameter of the pond is 2r = 2×21 m = 42 m.

Vedanta Excel in Mathematics Teachers' Manual - 8 134

7. Find the area of shaded region in the given figure.

Solution:

Here, 14 cm

(i) Area of square (A1) = l2 = 142 = 196 cm2
(ii) Diameter of circle (d) = 14 cm ∴radius (r) = 7 cm

∴ Area of circle (A2) = 22 × r2 = 22 × 72 = 154 cm2 14 cm
7 7
Hence, the area of shaded region (A) = A1 – A2 = 196 – 154 = 42 cm2

8. Find the area of shaded region in the given figure.

Solution: 42 cm

Here,

(i) Diameter of bigger circle (D) = 42 cm. ∴Radius (R) = 21 cm

∴ Area of bigger circle (A1) = 22 × R2 = 22 × 212 = 1386 cm2
7 7

(ii) Diameter of smaller circle (d) = 21 cm ∴Radius (r) = 10.5 cm

∴ Area of smaller circle (A2) = 22 × r2 = 22 × 10.52 = 346.5 cm2
7 7

Hence, the area of shaded region (A) = A1 – A2 = 1386cm2 – 346.5cm2 = 1039.5cm2

9. Find the area of shaded region in the given figure.
Solution:

Here, 14 cm

(i) Diameter of bigger semi-circle = 14 cm ∴radius (r) = 7 cm

Area of bigger semi-circle (A1) = 1 × 22 × r2 = 1 × 22 × 72 = 77 cm2
2 7 2 7

(ii) Diameter of each of two smaller semi-circles (d) = 7 cm ∴radius (r) = 3.5 cm

∴ Area of two semi-circles (A2) = 2 × 1 × 22 × r2 = 22 × 3.52 = 38.5 cm2
2 7 7

Hence, the area of shaded region (A) = A1 – A2 = 77 – 38.5 = 38.5 cm2

10. Find the area of shaded region in the given figure. 12 cm
Solution:

Here,

(i) Area of square (A1) = l2 = 122 = 144 cm2

(ii) Radius of each of two quadrant (r) = 6 cm 12 cm

∴ Area of a quadrant = 1 × 22 × r2 = 1 × 22 × 62 = 28.29cm2
4 7 4 7

Area of two quadrants (A2) = 2×28.285 = 56.57 cm2

Hence, the area of shaded region (A) = A1 – A2 = 144 – 56.57 = 87.43 cm2

135 Vedanta Excel in Mathematics Teachers' Manual - 8

11. Find the area of shaded region in the given figure.

Solution: 28 cm

Here,

(i) Area of square (A1) = l2 = 282 = 784 cm2
(ii) Radius of each of t41w×o q27u2a×drar2n=t (r14) = 1242 cm
∴Area of a quadrant = × 7 × 142 = 154 cm2 28 cm

Area of four quadrants (A2) = 4×154 cm2 = 616 cm2
Hence, the area of shaded region (A) = A1 – A2 = 784 – 616 = 168 cm2

12. Find the area of shaded region in the given figure.

Solution:

Here, 21 cm
60°
Radius of circle (r) = 21 cm

Angle at the centre of shaded sector (θ) = 1800 – 600 = 1200

Area of square (A1) = l2 = 122 = 144 cm2
We know,

∴311062T00mh00 aemmkaeaarskkeeea3ss61o130f 61csp0hoamar×dtpeo1ldfe2tt0sehe°ecc=itcroicrrplce=alret31ofpathrte ocfirtchlee.circle 2 πr2 = 1 × 22 × 212 = 462 cm2
3 3 7

Extra Questions

1. If the radius of a circular stadium is 84 m, find the perimeter and area of the stadium.
[Ans: 528 m, 22,176 m2]

2. The area of circular “Chapatti” is 154cm2, find its diameter. [Ans: 14 cm]

3. Mr. Gurung has a circular garden of diameter 28 m. A sprinkler at the centre of the garden
can spray the water in a circle with radius 13.3 cm.

(i) Find the area of the garden.

(ii) Find the area of the part of the garden which can be sprayed the water.

(iii) Find the part of the garden which cannot be sprayed with the water?
[Ans: 616 m2, 555.94 m2, 60.06 m2]

4. Suppose that you have a piece of wire of length 44 cm. If you bent it to form a square and
again a circle.

(i) Find the length of side of square and radius of circle.

(ii) Calculate the area of each shape.

(iii) Which shape covers more area and by how much? Find it.

[Ans: 10 cm, 7 cm, 100 cm2, 154cm2, area of circle is more by 54cm2]

Vedanta Excel in Mathematics Teachers' Manual - 8 136

Unit Area and Volume

20

Allocated teaching periods: 10

Competency
- Solving the problems on area of triangles and quadrilaterals
- Solving the problems related to area and volume of prisms (cube, cuboid, cylinder,
triangular prism)

Learning Outcomes
- To find the area of triangles and quadrilaterals
- To draw the net of prisms and pyramids
- To solve the problems related to area and volume of prisms (cube, cuboid, cylinder,
triangular prism)

Level-wise learning objectives

S.N. LEVELS OBJECTIVES

1. Knowledge (K) - To tell the formula to find the area of triangles and
quadrilaterals

- To identify the solid figure when its net is given
- To tell formula of finding the CSA, TSA and volume of cube,

cuboid, cylinder, triangular prism

- To find the area of triangles, quadrilaterals
2. Understanding (U) - To draw the net for the given prims and pyramids

- To calculate the area and volume of cube, cuboid, prisms
and cylinder.

- To solve some word problems based on the area of triangles
and quadrilaterals

3. Application (A) - To solve some word problems based on LSA/CSA, TSA and
volume of cuboid, cube, cylinder and triangular prism

Higher Ability (HA) - To compare the area of rectangular sheet of paper and the
4. curved surface area of the cylinder so formed from the same

sheet of paper.

Required Teaching Materials/ Resources
Cardboard paper, scissors, scale, ICT tools like GeoGebra etc
Pre-knowledge: area of rectangle and square

Teaching Activities for Area of Plane Figures
1. Discuss about the perimeter and area with proper examples.
2. Ask the formula of finding the area of rectangle and square.

137 Vedanta Excel in Mathematics Teachers' Manual - 8

3. Try to reduce the parallelogram, triangle, rhombus, trapezium and quadrilateral in to
rectangle by cutting the cardboard paper or visualizing the through GeoGebra.

(i) Activity for finding the formula of the area of triangle

-Cut a cardboard paper to form a triangular shape and name it as ABC.

-Fold the perpendicular line (height) from the vertex A to its base BC.

-Open and mark the perpendicular with a pen and mark it as h.

-Fold the height in half so that the vertex A sits on BC and mark half of the height as h/2.

- Cut the folded edges AX, PX and QX and place the edge AP along PB and edge AQ along

QC to form a rectangle FBCE. A
AA

h

h P 2 Q F 1E
Xh
22

B D CB D CB D C

-Thus, the area of triangle ABC = area of rectangle FBCE = length ×breadth

= BC× CE

= base× height

= b× h/2

= 1 b × h
2

OR, use GeoGebra tool to transform the triangle into rectangle.



(ii) Activity for finding the formula of the area of parallelogram D C
F
-Cut a parallelogram ABCD in a cardboard.

-Fold the base AB through D along DE such that DE ⊥ AB and h
mark DE as h which is the height of the parallelogram.

-Cut the paper along DE and place the edge AD along BC so that A Eb B
the rectangle DEFC is formed.

-Thus, the area of parallelogram ABCD = area of rectangle DEFC

= length ×breadth

= EF× CF

= AB× h

= b× h

OR, use GeoGebra tool to transform the parallelogram into rectangle.

(iii) Activity for finding the formula of the area of rhombus
-Cut a rhombus ABCD in a cardboard.
-Fold the diagonals AC and BD and mark them as d1 and d2. Also, mark point of intersection

of the diagonals AC and BD as O.

Vedanta Excel in Mathematics Teachers' Manual - 8 138

-Cut out the right angled triangles AOB, BOC, COD and AOD.

-Place the edge BC along AB and edge CD along AD so that the rectangle BEFG is formed.

-Thus, BD = BD = d2 and OC = BE = 1 AC = 1 d1
2 2
-Conclude the area of rhombus ABCD = area of rectangle DEFG

= length ×breadth

= BG× BE

= d2× 1 d1
2

= 1 d1× d2
2

OR, use GeoGebra tool to transform the rhombus into rectangle.

(iv) Activity for finding the formula of the area of kite
-Cut a kite ABCD in a cardboard.
- Fold the longer diagonal AC and perpendicular through BD to diagonal AC and mark AC

and BD as d1 and d2 respectively. Also, mark point of intersection of the diagonals AC and
BD as O.



-Cut out the right angled triangles AOB, BOC, COD and AOD.

-Place the edge AB along AD and edge BC along CD so that the rectangle ACEF is formed.

-Thus, AC = d1 and OD = CE = BD = d2
-Conclude the area of kite ABCD = area of rectangle ACEF

= length ×breadth

= AC× CE

= d1× 1 d2
2

= 1 d1× d2
2

OR, use GeoGebra tool to transform the kite into rectangle.

139 Vedanta Excel in Mathematics Teachers' Manual - 8

(v) Activity for finding the formula of the area of quadrilateral
- Cut a pair of congruent quadrilaterals ABCD and PQRS in a cardboard.
- Fold the longer diagonal AC in the quadrilateral ABCD and from B and D to diagonal AC

such that DE and BF perpendiculars to AC.
- Also, fold the longer diagonal PR in the quadrilateral PQRS and from S and Q to diagonal

PR such that SM and QN perpendiculars to PR.
-Name AC = PR = d1, DE = EM = p1 and BF = QN = p2


-Cut out the right angled triangles PMS, SMR, PQN and QNR.

-Place the edge SM of ∆ PMS along AD, edge QP of ∆ PQN along AB, edge RS of ∆ SMR
along DC and edge RQ of ∆ QNR along BC so that the rectangle WXYZ is formed.

-Thus, WX = AC = d1 and YX = DE + BF = p1 + p2
-The area of quadrilateral (ABCD + PQRS) = area of rectangle WXYZ

or, 2×area of quadrilateral ABCD = length ×breadth

or, 2×area of quadrilateral ABCD = WX× YX

or, 2×area of quadrilateral ABCD = d1× (p1 + p2)
1
The area of quadrilateral ABCD = 2 d1 (p1 + p2)

OR, use GeoGebra tool to transform the quadrilateral into rectangle.

-Similarly, discuss upon the area of trapezium and equilateral triangle.

4. Recall the formulae of area of triangles and quadrilaterals with the following table.
The plane surface enclosed by the boundary line of a plane closed figure is known as its

area. Area is measured in square units. For example, sq. centimeters (cm2), sq. metres (m2),

etc. Let’s review the formulae to calculate areas of the following plane geometrical figures.

Name of plane figure Shape of plane figure Formula of area

Area = 1 base × height
2
h
Triangle b A = 1 bh
2

Rectangle b Area = length × breadth
A=l×b

l

Vedanta Excel in Mathematics Teachers' Manual - 8 140

d Area = (side)2 or, Area = 1 (diagonal)2
A = l2 2
1
Square or, A = 2 d2

l

h Area = base × height
A = b × h
Parallelogram
b

d1 Area = 1 d1 × d2
d2 2
Rhombus

d1 Area = 1 (Product of diagonals)
d2 2

Kite A = 1 d1 × d2
2

l2 Area = 1 (Sum of parallel sides) ×h
h 2

Trapezium l1 A = 1 h (l1 + l2)
2

p1 d Area = 1 diagonal (Sum of
p2 2 perpendiculars)

Quadrilateral A = 1 d (p1 + p2)
2

5. Solve the problems related to Pythagoras theorem in group.

Solution of selected questions from Excel in Mathematics -Book 8

1. Find the area of the given figure. A

Solution: 1 1 10 cm
Area of ∆ABC = 2 b × p = 2 ×12 cm × 10 cm = 60 cm2 10 cm
12 cm
Area of ∆ACD = 1 b × p = 1 ×8 cm × 10 cm = 40 cm2 B 12 cm D
2 2 8 cm
Area of the figure = 60 cm2 + 40 cm2 = 100 cm2 P

QT C

2. Find the area of the given figure.

Solution:

Area of square QRST = l2 = (15 cm) 2 = 225 cm2

Height of ∆PQT = 18 cm – 15 cm = 3 cm R 15 cm S

141 Vedanta Excel in Mathematics Teachers' Manual - 8

Area of ∆PQT = 1 b × h = 1 ×15 cm × 3 cm = 22.5 cm2
2 2
∴Area of the figure = 225 cm2 + 22.5 cm2 = 247.5 cm2
5cm D C 5cm

3. Find the area of the given figure. E 15 cm F
Solution:

Area of rectangle ABCD = l × b = 20 cm × 15 cm = 300 cm2 A 30 cm B

Area of ∆ADE and ∆BCF = 2 × 1 b × h = 15 cm × 5 cm = 75 cm2
2

∴Area of the figure = 300 cm2 + 75 cm2 = 375 cm2

4. Find the area of the given figure. HG

Solution: 7 cm

Area of otrwfeceotaascnehgmlseie-mEciFir-GcclHiersc=laetl=e×ac21bh×=en71dc2m=cm=2××3.2157×ccmmπr=2 =8427c2m×2 E 12 cm F
Radius cm2
(3.5 cm) 2 = 38.5
Area of

∴Area of the figure = 84 cm2 + 38.5 cm2 = 122.5 cm2

5. Find the area of shaded region in the given figure. 9 cm
4cm
Solution: = 1 × h (a + b) = 1 × 10 cm (9 cm + 15 cm) = 120 cm2 3cm
Area of trapezium 2 2 15 cm
10 cm 5cm
Area of parallelogram = b × h = 4cm × 3 cm = 12 cm2 5cm
15 cm
∴Area of the shaded region = 120 cm2 – 12cm2 = 108 cm2
14 cm
6. Find the area of shaded region in the given figure. 12 cm 4cm 4cm
Solution:
Area of bigger rectangle = l × b = 15cm × 12 cm = 180 cm2
Area of two smaller rectangles = 2× (l × b) = 2 × 5cm × 4 cm = 40 cm2
∴Area of the shaded region = 180 cm2 – 40 cm2 = 140 cm2

7. Find the area of shaded region in the given figure. 7 cm 5 cm

Solution:

Area of urepcptaenr gwlehi=telt×riabng=le1=4c12m××b1×2 chm==2116×81c4mc2m× 5 cm = 35 cm2
Area of

Area of lower white triangle = 1 × b × h = 1 × 14 cm× 7 cm = 49 cm2
2 2
∴Area of the shaded region = 168 cm2 – (35cm2 + 49cm2) = 84 cm2

8. Find the area of shaded region in the given figure. 5 cm 5 cm4 cm 4 cm
4 cm 12cm
AASorreeluaatooioffnttwr:aoperzigiuhmt an=gl12ed×trhia(nag+lebs )==221××121×0 cm (8 cm + 16cm) = 120 cm2
4 cm × 5 cm = 20 cm2

∴Area of the shaded region = 120 cm2 – 20cm2 = 100 cm2

Vedanta Excel in Mathematics Teachers' Manual - 8 142

9. The perimeter of a square ground is 100 m. Find the cost of plastering ground at Rs 80

per. Sq. meter.

Solution:

Here,

The perimeter of a square ground =100 m

or, 4l = 100 m ∴l = 25 m

Now, area of the garden = l2 = (25m) 2 = 225 m2

Hence, the cost of plastering the ground = Area × Rate of cost

= 625 × Rs 80

= Rs 50,000

10. The length of the floor of a room is two times its breath and the perimeter of the floor is

48 m. Find the cost of paving marbles on the floor at Rs 125 per sq. metre.

Solution:

Let the breath of the room (b) = x m

Then the length of the room (l) = 2x m

Now,

The perimeter of the room = 2(l + b)

or, 48 m = 2 (2x + x) or, 48 m = 6x ∴ x = 8 m

Breath of the room (b) = x = 8 m and the length (l) = 2x m = 2×8 m = 16 m

Also, area of the floor = l × b = 16 m × 8 m = 128 m2

Hence, the cost of paving marbles on the floor = Area × Rate of cost

= 128 × Rs 125

= Rs 16,000

11. The length of the rectangular ground is three times its breath. If the perimeter of the

ground is 96 m, find the cost of growing grass on it at Rs 65 per sq. metre.

Solution:

Let the breath of the room (b) = x m

Then the length of the room (l) = 3x m

Now,

The perimeter of the room = 2(l + b)

or, 96 m = 2 (3x + x) or, 96 m = 8x ∴ x = 12 m

∴Breath of the room (b) = x = 12 m and the length (l) = 3x m = 3×12 m = 36 m

Also, area of the floor = l × b = 36 m × 12 m = 432 m2

Hence, the cost of paving marbles on the floor = Area × Rate of cost

= 432 × Rs 65

= Rs 28,080

12. A rectangular park of length 60 m and breadth 50 m encloses a volleyball court of

length 18 m and breadth 10 m. Find the area of the park excluding the court. Also, find

the cost of paving stone in the park excluding the court at the rate of Rs 110 per square

metre.

143 Vedanta Excel in Mathematics Teachers' Manual - 8

Solution:

Area of the park = l × b = 60 m × 50 m = 3000 m2

Area of the volleyball court = l × b = 18 m × 10 m = 180 m2

Area of the park excluding the court = 3000 m2 – 180 m2 = 2820 m2

Hence, the cost of paving stone in the park excluding the court = Area × Rate of cost

= 2820 × Rs 110

= Rs 3, 10, 200

13. Mr. Thapa build a circular pond of diameter 28 ft. for fish farming in his rectangular

field of length 70 ft. and breadth 40 ft. Find the area of the field excluding the pond.

Also, estimate the cost of planting vegetable at Rs 11 per square feet.

Solution:

Here,

Area of the rectangular filed = l × b = 70 m × 40 m = 2800 ft2

Radius of the pond = 14 ft.

Area of the pond = π2r22 × (14 ft)2 = 616 ft2
= 7

∴Area of the filed excluding the pond = 2800 ft2 - 616 ft2 = 2184 ft2

Again,

The cost of planting vegetable in the filed excluding the pond = Area × Rate of cost

= 2184 × Rs 11 = Rs 24,024

Extra Questions

1. Find the area of shaded portion in the given figure.

2. A rectangular ground of length 50 m and breadth 40 m encloses a rectangular basketball

court of length 30 m and breadth 20 m. Find the area of the park excluding the court.

Also, find the cost of paving stone in the park excluding the court at the rate of Rs 160 per

square metre. [Ans: 1400m2, Rs 2,24,000]

3. Ram had a piece of land in the shape of kite of diagonals 120 ft. and 60 ft. He built a house

on the rectangular part of the land along an edge with length 50 ft. and breath 45 ft. Now,

he has grown vegetable in his remaining part of the land at the rate of Rs 20 per sq. ft, find

the cost of growing vegetable. [Ans: Rs 27,000]

4. A rectangular park is 200 m long and 150 m broad. A path of uniform wide 2m is
constructed inside the park. Find the cost of paving the path with stones at Rs 250 per sq.

metre. [Ans: Rs 346,000]

5. A circular dining table of diameter 1.4 m is covered with a decorated squared plastic with
each side 1.5 m. Find the area of the cover hanging around the table.[Ans: 0.71 m2]

Vedanta Excel in Mathematics Teachers' Manual - 8 144

Teaching Activities for Solids and their Nets

1. Take some solids like cube, cuboid, cylinder, triangular pyramids made of cardboard
paper.

(i) Discuss about their faces, edges and vertices.
(ii) Open out them into flat faces and discuss about the nets.
2. If possible, show the visualization the nets of the polyhedrons by using GeoGebra tool.
3. Give the group work to trace out the nets of cube, cuboid, cylinder, triangular prism, and

tetrahedron in cardboard paper. And tell to make the solids by pasting their nets with

glue and present in the classroom.

4. Make a discussion on solving the problems given in the exercise.

Teaching Activities for Area of Solids

1. With a cuboidal object or rectangular classroom, ask the following questions.
(i) How many faces does it have?
(ii) Are all of its faces rectangular?
(iii) What is the area of its base?
2. Discuss about the area of bases, lateral surface area (L.S.A.) and total surface area (T.S.A)

of cube and cuboid and list the following formulae.

(a) In any Cube,
(i) Area of its each face = l2
(ii) Area of its 4 lateral faces (lateral surface area) = 4l2
(iii) Total surface area = 6l2
(b) In any Cuboid,
(i) Area of its base = l ×b
(ii) Lateral surface area = 2l h + 2 bh = 2h (l + b)
(iii) Total surface area = 2( lb + bh + lh)

3. Show different cylindrical objects like can, bottle, pipe etc. and discuss about its bases
and curved surface area.

4. Take a rectangular sheet of paper and roll it to form a cylinder. Then ask the following
questions.

(i) What is the formula to find area of rectangular sheet of paper?
(ii) What is the formula to find the circumference of the base of cylindrical paper?
(iii) Is the length of rectangular sheet equal to base of cylindrical shape?
(iv) Is the area of cylindrical shape equal to the area of rectangular sheet?
(v) What is the formula of curved surface area of the cylindrical shape?

Area of circular base
= pr2

Area = l × b b Area = l × b b = h Area of curved surface
l = 2pr ×h = 2prh

l = 2pr Area of circular base
=pr2

145 Vedanta Excel in Mathematics Teachers' Manual - 8

Under discussion, list out the following formulae

(i) Circumference of the base = 2πr

(ii) Curved surface area (C.S.A.) of cylinder = 2πrh

(iii) Total surface area (T.S.A.) of cylinder = C.S.A. + 2×Area of base

= 2πrh + 2πr2 = 2πr (r + h)

5. Take a triangular prism and discuss about its properties.

(i) It has two triangular bases and three rectangular faces.

(ii) Its opposite ends are congruent and parallel.

(iii) Its three rectangular faces are called lateral surfaces

6. Take out its net and discuss about the formulae of lateral surface area and total surface

area. Then list out the following formulae.

(i) Lateral surface area (L.S.A.) = area of 3 rectangular surfaces

= l.a + l.b + l. c

= l (a + b + c)

= Perimeter of triangular base ×length

= P.l

(ii) Total surface area = L.S.A. + 2 × area of triangular base

= P.l + 2× 1 b × h
2
= P.l + b.h

7. Discuss about the problems given in the exercise.

Solution of selected questions from Excel in Mathematics -Book 8

1. Find the area of base, curved surface area and total surface area of

the given cylinder. 20 cm

Solution:

In the given cylinder, radius of base (r) = 7 cm and height (h) = 20 cm

(i) Area of base = πarr2e=a =2722π×rh(7=cm2×) 227=2 154 cm2 r = 7 cm
(ii) Curved surface 1188 cm2
(iii) × 7 cm × 20 cm = 880 cm2
20 cm) =
Total surface area = 2πr (r + h) = 2× 22 × 7 cm (7 cm +
7

2. Find the lateral surface area and total surface area of the 8 cm 5

given triangular prism. cm 15 cm

4 cm

Solution: 6 cm

(i) Perimeter of the triangular base (P) = 8 cm + 6 cm + 5 cm = 19 cm

(ii) Area of base (A) = 1 × b× h = 1 × 6 cm × 4 cm = 12 cm2
2 2
(iii) Lateral surface area = P.l = 19 cm × 15 cm = 285 cm2

(iv) Total surface area = L.S.A + 2A = 285 cm2 + 2×12 cm2 = 309cm2

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3. Find the lateral surface area and total surface area of the given 8 cm

triangular prism. 14 cm

Solution: 6 cm

(i) Using the Pythagoras theorem to find the side AC in right angles ∆ABC,

AC = AB2 + BC2 = 82 + 62

(ii) Perimeter of the triangular base (P) = 8 cm + 6 cm + 10 cm = 24 cm

(iii) Area of base (A) = 1 × b× p = 1 × 6 cm × 8 cm = 24 cm2
2 2
(iv) Lateral surface area = P.l = 24 cm × 14 cm = 336 cm2

(v) Total surface area = L.S.A + 2A = 336 cm2 + 2×24 cm2 = 384cm2

4. A cubical block 10 cm long is placed on the table. How much area 35 cm
does it cover on the surface of the table? r = 14 cm

Solution:
The area covered by the cubical block on the surface of the table = Area
of the base of the block =l2 = (10 cm) 2 = 100 cm2

Teaching Activities for Area of Solids

1. Discuss about the meaning of volume of solid objects.
2. Recall the formula to calculate the volume of cube and cuboid.
3. Make a discussion about the prism with proper examples and make clear that the volume

of prism is obtained by multiplying the base area and the height.

4. Under discussion, list out the following formulae
(i) Volume of cube = Area of base × height = l2 × l = l3
(ii) Volume of cuboid = Area of base × height = l×b×h
(iii) Volume of cylinder = Area of base × height = πr2×h = πr2h
(iv) Volume of triangular prism = Area of triangular base height (or length)
5. Give proper guidance to solve the problems as given in the exercise.

Solution of selected questions from Excel in Mathematics -Book 8

1. Find the volume of the given solid. d=28 cm

Solution:

In the given cylinder, radius of base (r) = 14 cm and height (h) = 40 cm 40 cm

Now, volume of the cylinder = πr2h = 22 × (14 cm)2 × 40 cm = 24,640 cm3
7

2. Find the volume of the given solid.

Solution: 10 cm

In the given t(rAia)n=gu12la×r pbri×smh, = 1 × 8 cm × 10 cm = 40 cm2 8 cm 15 cm
Area of base 2

Now, volume of the prism = Area of base (A) × h = 40 cm2 × 15 cm = 600cm3

147 Vedanta Excel in Mathematics Teachers' Manual - 8

3. Find the volume of the given solid.
Solution:

In the given triangular prism, 6 cm

Area of base (A) = 1 × b × p = 1 × 8 cm × 6 cm = 24 cm2 14 cm 8 cm
2 2

Now, volume of the prism = Area of base (A) × h = 24 cm2 × 14 cm = 336cm3

4. A rectangular tank is 1.5 m × 1.4 m × 2 m. How many litres of water does it hold when
it is completely filled?

Solution:
Here,
Volume of tank = 1.5 m × 1.4 m × 2 m = 4.2 m3
We know, 1 m3 = 1000 litre
∴The capacity of the tank = 4.2 ×1000 litre = 4200 litre
Hence, the tank can holds 4200 litre of water when it is completely filled.

5. A rectangular box is 1 m long and 75 cm high. If the volume of the box is 375000 cm3,
find the width of the box.

Solution:
Let the width of the box be x cm.
Now,
Volume of the box = 1 m × 75 cm × x cm
or, 375000 cm3 = 100 cm × 75 cm × x cm
∴x = 50
Hence, the width of the box is 50 cm.

6. A rectangular vessel can hold 40 litres of milk when it is full. If the area of the base is
of the vessel is 2000 cm2, find the height of the vessel.

Solution:
The volume of the vessel = 40 litres = 40×1000 cm3 = 40000 cm3
Now,
Volume of the vessel = Area of base × height (h)
or, 40000 = 2000h
∴h = 20
Hence, the height of the vessel is 20 cm.

7. A cubical box is 75 cm high. How many rectangular packets each of 25 cm × 15 cm ×
5 cm are required to fill the box?

Solution:
Volume of cubical box (V) = l3 = (75 cm) 3 = 5625 cm3
Volume of each box (v) = 25 cm × 15 cm × 5 cm = 1875 cm3
Now,

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Required no. of boxes (N) = V = 5625 cm3 = 3
v 1875 cm3

8. A wall is 15 m long, 0.2 m wide and 2m high. How many bricks of size 15 cm × 8 cm ×

5 cm each are required to construct the wall?

Solution:

Volume of wall (V) = 15 m × 0.2 m × 2 m = 1500 cm × 20 cm × 200 cm

= 6000000 cm3

Volume of each brick (v) = 15 cm × 8 cm × 5 cm = 600 cm3

Now, V 6000000 cm3
v 600 cm3
Required no. of boxes (N) = = = 10000

9. Find the volume of the given solid.

Solution:

In the bigger cuboid; length (l) = 8 cm, breadth (b) = 7 cm

and height (h) = 2 cm

∴Volume (V1) = l × b × h = 8 cm× 7 cm × 2 cm
= 112 cm3

In the smaller cuboid;

length (l) = 12 cm – 8 cm = 4 cm, breadth (b) = 3 cm and height (h) = 2 cm

∴Volume (V2) = l × b × h = 4 cm× 3 cm × 2 cm = 24 cm3
Hence, the volume of the solid (V) = V1 + V2 = 112 cm3 + 24 cm3 = 136 cm3

10. Find the volume of the given solid.
Solution:
In the bigger cuboid;
length (l) = 8 cm, breadth (b) = 7 cm and height (h) = 2 cm
∴Volume (V1) = l × b × h = 8 cm× 7 cm × 2 cm = 112 cm3
In the smaller cuboid;
length (l) = 12 cm – 8 cm = 4 cm, breadth (b) = 3 cm and height (h) = 2 cm
∴Volume (V2) = l × b × h = 4 cm× 3 cm × 2 cm = 24 cm3
Hence, the volume of the solid (V) = V1 + V2 = 112 cm3 + 24 cm3 = 136 cm3

11. Find the volume of the given solid.
Solution:
In the uppermost cube; each side (l) = 3 cm
∴Volume (V1) = l3 = (3 cm) 3 = 27 cm3
In the lowermost cuboid;
Length (l) = 6 cm, breadth (b) = 3 cm and height (h) = 6 cm – 3 cm = 3 cm
∴Volume (V2) = l × b × h = 6 cm× 3 cm × 3 cm = 54 cm3
Hence, the volume of the solid (V) = V1 + V2 = 27 cm3 + 54 cm3 = 81 cm3

149 Vedanta Excel in Mathematics Teachers' Manual - 8

12. Find the volume of the given solid.
Solution:
In each of two sidewise cuboids;
Length (l) =4 cm, breadth (b) = 6 cm and height (h) = 12 cm
Volume (V1) = 2 (l × b × h)
= 2 × 4 cm× 6 cm × 12 cm = 576 cm3
In the middle cuboid;
length (l) = 10 cm, breadth (b) = 6 cm and height (h) = 3 cm
Volume (V2) = l × b × h = 10 cm× 6 cm × 3 cm = 180 cm3
Hence, the volume of the solid (V) = V1 + V2 = 576 cm3 + 180 cm3 = 756 cm3

13. The length, breadth and height of a rectangular room are in the ratio 4:3:2. If the room
contains 192 m3 of air, find the area of the floor of the room.

Solution:
Let the length of the room = 4x m, breadth (b) = 3x m and height (h) = 2x m
Now,
Volume (V) = l × b × h
or, 192 = 4x × 3x× 2x
or, 192 = 24x3
or, 8 = x3
∴x = 2
Thus, length (l) = 4x = 8 m, breadth (b) = 3x = 6 m and height (h) = 2x = 4 m
Again,
Area of the floor of the room = l × b = 8 m× 6 m = 24 cm2

14. The length of a room is three times the height and breadth is twice of its height. If the
volume of the room is 384 m3, find its length, breadth and height.

Solution:
Let the height of the room = x m. Then, length (l) = 3x m and breadth (b) = 2x
Now,
Volume (V) = l × b × h
or, 384 = 3x × 2x× x
or, 384 = 6x3
or, 64 = x3
∴x = 4
Thus, length (l) = 3x = 3 × 4 m = 12 m, breadth (b) = 2x = 2 × 4 m = 8 m and height (h) =

x=4m

15. The length of a rectangular cartoon is twice the breadth and thrice the height. If the
volume of the cartoon is 4500 cm3, find its length, breadth and height.

Solution:

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