sheniblog-EM-SSLC Worksheets - Questions and Answers - John P A

Answers
√

a) AB = 52 + 122 = 13cm

b) Perimetre 13 + 13 + 10 = 36cm

c) See the diagarm

ABCE is a rhombus .Semicircle is completed into circle. Perimetre = 13 × 4 = 52cm ,

Area 60 × 2 = 120 sq.cm

Radius of the incircle r = A = 120 = 4.6cm
s 26

Another method

Let rbe the radius . It is the distance from center to the touching pointr.

Sum of the area of ABO and area of triangle ACO is equal to area of ABC

1 × 13 × r + 1 × 13 × r = 1 × 10 × 12
2 2 2
120
26r = 120, r = 26 = 4.6െസ.മീ ർ

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08

Solids

ഘനരൂപങ്ങൾ

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2020-21 Academic year Worksheets

Mathematics X
Solids

Concepts

⋆ There are 5 faces in a square pyramid. Four lateral faces and a base. Lateral faces are triangles
and base is a square.

⋆ There are 8 edges in a square pyramid. Four base edges and four lateral edges .Base edge is
denoted by a and lateral edges by e.

⋆ പാദ ിെ വികർ ംd, കളിെല ശീർഷ ിൽ നി ം പാദ ിേല ഉ തി hഎെ .

⋆ The lattitude of lateral face to the base edge is called slant height of the pyramid.

1) Manju has drawn an outline in a square card board for making a square pyramid as given below.

a) What is the total length of its edges?
b) What is the slant height of the square pyramid?
c) What is length of the side of the square paper in which outline is drawn.
Answers

a) Total length of the edges = 4a + 4e = 4 × 10 + 4 × 13 = 40 + 52 = 92cm
b) see figure

√
l = 132 − 52 = 12cm
c) One side of the square paper = a + 2l = 10 + 2 × 13 = 36cm

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2) A wire of length 96cm is cut into eight equal parts . The ends of the pieces are joined to make the pyramid.

a) What is the length of the edge of the pyramid?
b) What kind of triangle is its lateral edge?
c) What is its slant height?

Answers

a) Length of one edge = 96 = 12cm
8

b) Lateral faces are equilateral triangles

c) On dra√wing slant height on the lateral face we get 30◦ − 60◦ − 90◦ right triangle .
l = 6 3cm

3) The base perimetre of a square pyramid is 40 cm, height 12 cm.

a) What is the base edge of the pyramid?
b) what is the slant height of the pyramid?
c) What is the lateral edge of the pyramid?

Answers

a) Length of base edge = 40 = 10 cm
4
√
b) l = 122 + 52 = 13cm

c) latera√l edge e, slant√height l, half of base edge form a right triangle
e = 132 + 52 = 194cm

4) There is a square pyramid having its lateral faces equilateral triangles.Length of one leteral edge is 32cm
a) What is its base edge?
b) What is its slant height?

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c) What is the area of one lateral face?
d) Calculate the total area of its lateral faces

Answers

a) All edges are equal .base edge and lateral edge are 32cm

b) If l is th√e slant height and e is the lateral edge , l, e, a form a 30◦ − 60◦ − 90◦ triangle.
2

l = 16 3cm √√
× 32 × 16 3 = 256 3sq.cm
c) One lateral face area = 1
2
√√
d) Total lateral face area = 4 × 256 3 = 1024 3sq.cm

5) The base diagonal of a square pyramid is 12cm , height 8cm

a) What is its base edge?
b) What is its base area?
c) What is the length of its lateral edge?
d) Calculate the total length of its edges.

Answers

a) Two b√da2se=ed√1g22es=an6d√b2acsme diagonal form 45◦ − 45◦ − 90◦right triangle . d= √ 2a
a=

√ √
b) Base area = (6 2)2 = 72sq.cm

c) √Half of the base diagonal , height and lateral edge form a right triangle .e = 62 + 82 =
100 = 10cm
√√
d) Total length of the edges = 4a + 4e = 4 × 6 2 + 4 × 10 = 24 2 + 40cm

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2020-21 Academic year Worksheets

Mathematics X
Solids

Concepts

⋆ There are 5 faces in a square pyramid. Four lateral faces and a base. Lateral faces are triangles
and base is a square.

⋆ There are 8 edges in a square pyramid. Four base edges and four lateral edges .Base edge is
denoted by a and lateral edges by e.

1) The height of a square pyramid is 2 more than its base edge. Slant height is 13 cm

a) If the base edge is a what will be its height?
b) Write an equation connecting base edge , height and slant height
c) Calculate the length of base edge?
d) Calculate the total lateral surface area of the pyramid.

Answers

a) Height = a + 2

b) 132 = ( a )2 + (a + 2)2
2

c) a2 + a2 + 4a + 4 = 169, 5a2 + 4a = 165,5a2 + 16a − 660 = 0,a = 10.(solve second
4 4

degree equation)

Base edge a = 10cm

d) Lateral face area = 4× area of one lateral face = 4 × 1 × 10 × 13 = 260 sq.cm
2

2) Base area of a square pyramid is 400 sq.cm , lateral surface area 1040 sq.cm

a) What is the length of base edge?
b) What is the slant height?
c) Find the height of the pyramid.
d) Calculate the total surface area of the pyramid.

Answers
√

a) a = 400 = 20cm

b) 2al = 1040 , 2 × 20 × l = 1040, l = 1040 = 26 cm
40

c) l2 = h2 + ( a )2 → 262 = h2 + 102, h2 = 576, h = 24cm
2

d) Total suface area = 400 + 1040 = 1440 sq.cm

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3) A sectoral sheet of central angle 240◦ is taken from a circular sheet of radius 10cm. Four equal triangles
are made from the sector as in the figure.They are joined in such a way as to get a square pyramid.

a) What is the length of its edge?
b) What is the slant height of the pyramid ?
c) Find the height of the pyramid.

Answers

a) All edges are equal. It is equal to radius .Length of one end = 10cm
√
b) Lateral faces are equilateral triangles .Slant height = 5 3െസ ീമീ ർ

c) l2 = h2 + ( a )2, √ √√
2 (5 3)2 = h2 + 52, h2 = 75 − 25 = 50, h = 50 = 5 2 cm

4) A steel wire of length 120cm is cut into 8 equal parts, the ends are joined in such a way as to get a square
pyramid.

a) What is the length of its edge?
b) What is its slant height?
c) Calculate the area of paper used to cover the pyramid.

Answers

a) a= 120 = 15cm
8
√
b) l= 15 3cm
2
√
c) Total surface area =Base area +4×area of one lateral face = 225 + 225 3sq.cm.

5) Base edge of a square pyramid is a and slant height l.

a) Write a formula to find the lateral surface area of the pyramid.
b) Is it possible to make a square pyramid having base area and lateral surface area equal.
c) If the base edge is 10 and lateral surface area of a square pyramid is two times its base area. What

is its height?

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Answers

a) Lateral face area = 4× area of one lateral face

lareral face area = 4 × 1 × a × l = 2al
2
√
b) a2 = 2al → a = 2l, l = a , h = l2 − ( a )2 = l2 − l2 =0
2 2

Height becomes zero . Pyramid cannot be made .

c) 2al = 2a2, a =l √
,a= 5 3cm
l2 = h2 + a2 10 ,h =
4

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2020-21 Academic year Worksheets

Mathematics X
ഘന പ ൾ

Concepts

⋆ If base edge a,height h, and slant height lthen
l2 = h2 + ( a )2
2

If base edge a, slant height l, and lateral edge e then
e2 = l2 + ( a )2
2

Height h, base diagonal d, lateral edge e then

e2 = h2 + ( d )2
2

⋆ Lateral face area of the square pyramid = 2al

total surface area = a2 + 2al

volume = 1 × a2 ×h
3
√
⋆ If the lateral faces are equilateral triangles , total surface area = a2 + 3a2

1) There is a square pyramid of height 12cm , slant height 13cm.

a) What is the length of its base edge ?
b) What is the base area of the pyramid ?
c) Find the lateral face are of the pyramid.
d) Calculate the total surface area of the pyramid.

Answers

a) l2 = h2 + ( a )2
2
a2
132 = 122 + 4 √
100
a2 = 132 − 122 = 25, a2 = 100, a = = 10cm
4

b) Base area = 100sq.cm

c) Lateral surface area = 2al = 2 × 10 × 13 = 260 sq.cm

d) Total surface area = 100 + 260 = 360sq,cm

2) Height of the square pyramid is h , slant height l and lateral edge e .
a) If the base edge is a, write the relations between h, l and e
b) Prove that h2, l2, e2are in an arithmetic sequence
c) If the slant height is 13 , base edge 10 find height and lateral edge

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Answers

a) l2 = h2 + ((a2a2))22
e2 = l2 +

b) l2 = h2 + ( a )2 , e2 = l2 + ( a )2 → h2 +2× ( a )2 .
h2 2 in 2 2
, l2, e2 are a )2
arithmetic sequence.d = ( 2

c) a = 10, l = 13.

h2, l2, e2are in arithmetic sequence . √
144 = 12c√m
132 − h2 = 102 , h2 = 169 − 25 = 144, h =
4 169 + 25 = 194, e = 194
a2
e2 − l2 = 4 , e2 − 132 = 25, e2 = cm

3) Base perimetre of a square pyramid is 40cm, total length of the edges is 92cm

a) What is the length of base edge?
b) What is base diagonal?
c) Find the height of the pyramid
d) Calculate the total surface area.

Answers

a) Base edge a = 40 = 10cm
4
√√ √
b) d = 2a = 2 × 10 = 10 2cm

c) 4a + 4e = 92, 40 + 4e = 92, e = 13cm e2 = h2 + ( d )2 √
2 = 119, h = 119
√
132 h2 10 2 )2,169 h2 h2 50, h2
= + ( 2 = + 50, = 169 −

√√ √ √
d) l = e2 − a24 = 132 − 25 = 144 = 12. Total surface area = a2 + 2al =

100 + 2 × 10 × 12 = 340sq.cm

4) Base perimetre of a square pyramid is 40cm , height 12cm

a) What is the lenghth of base edge ?
b) Find the volume of the pyramid.
c) What is the volume of the square prism having same base area and height ?

Answers

a) a = 10cm

b) Volume of square pyramid = 1 × Base area × height
3
1 × 102 ×
Volume = 3 12 = 400cubic cm

c) . Volume = 3 × 400 = 1200cubic cm

5) Ratio of the base edges of two square pyramids is 1 : 2. Heights are in the ratio 2 : 3

a) If the base edge of the first pyramid is a then what is the ratio of their base area?
b) If the height of the first pyramid is h then what is the height of second pyramid?
c) What is the ratio of the volume ?
d) If the volume of the first pyramid is 10 cubic cm then what is the volume of the second pyramid?

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Answers

a) If base edges are a, 2athen base area are a2; (2a)2 .Ratio of base area → a2 : 4a2 = 1 : 4

b) If height of first h, height of second is 3h
2

c) Base edges a, 2a, heights 2h, 3h.

Ratio of volume → 1 × a2 × 2h : 1 × (2a)2 × 3h
3 3

V1 : V2 = 1 : 6

d) V2 = 6 × V1 = 60 cubic cm

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2020-21 Academic year Worksheets

Mathematics X
Solids

Concepts

⋆ If base edge a,height h, and slant height lthen
l2 = h2 + ( a )2
2

If base edge a, slant height l, and lateral edge e then
e2 = l2 + ( a )2
2

Height h, base diagonal d, lateral edge e then

e2 = h2 + ( d )2
2

⋆ Lateral face area of the square pyramid = 2al

total surface area = a2 + 2al

volume = 1 × a2 ×h
3
√
⋆ If the lateral faces are equilateral triangles , total surface area = a2 + 3a2

1) Angle between base edge and lateral edge of a square pyramid is 60◦, base edge is 12 cm

a) What is the length of its lateral edge?
b) What is its slant height?
c) Find the total surfsce area of the pyramid
d) Find the height of the pyramid
e) Calculate the volume of the pyramid

Answers

a) △ABD is a 30◦ − 60◦ − 90◦triangle .in the lateral face side opposite to 90◦ is e =

12െസ ീമീ ർ

√
b) l = 6 3cm
√
c) Total surface a√rea = a2 + 2al = 122 + 2 × 12 × 6 3
= 144 + 144 3 sq.cm
√√
√ √√
d) h = (6 3)2 − 62 = 108 − 36 = 72 = 6 2cm

e) Volume = 1 × a2 × h = 288√3cubic1 cm
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2) The base edge and height of a square pyramid are equal in length .Volume of the pyramid is 576 cubic cm
a) What is the base edge of the pyramid?
b) What is the slant height of the pyramid ?
c) calculate the total surface area of the pyramid.

Answers

a) 1 a2 ×h = 576 → 1 × a3 = 576, a3 = 576 × 3 = 1728, a = 12cm
3 3
√ √ √
b) l = h2 + ( a )2 , 122 + 62 = 180
2
√
c) Total surface area a2 + 2al = 122 + 2 × 12 × 180

3) Can isosceles right triangles be the lateral surfaces of a square pyramid. Justify your answer.

Answers

⋆ Base edge a, Laterasl edge e, Base diagonal d √

⋆ If the lateral faces are isosceles right triangles a = 2e.

e = √a 2

⋆ √ √d 2
Base diagonal d = 2a → a =

√ √ a2 a2
2 2
⋆ h, d , e make a right triangle h = e2 − ( d )2 = − =0
2 2

height becomes 0. Not possible to make pyrism

4) Base edge of a square pyramid is a and height h. The volume of the pyramid is 1 a2h.
3

a) What is the change of its volume if the base edge and height are doubled.

b) If the initial volume is 10 cubic cm then what will be the volume if height and base edge are doubled?

Answers

a) Base edge abecomes 2a, height hbecomes 2h

Volume 1 × (2a)2 × 2h = 8 × 1 a2h. Volume becomes 8times
3 3

b) തിയ പിക െട വ ാ ം= 8 × 10 = 80ഘനെസ ീമീ ർ

5) Base perimetre of a square pyramid is 64cm , volume is 1280 cubic cm

a) What is its base edge ?
b) What is the height of the pyramid?
c) What is the slant height of the pyramid ?
d) Calculate the total surface area of the pyramid.

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Answers

a) 4a = 64, a = 16cm

b) 1 × a2 × h = 1280
3 × 162 × h = 1280, h
1 = 1280×3 = 15 cm
3 256
√ √ √
c) l = h2 + ( a )2 = 152 + 82 = 289 = 17cm
2

d) Total surface area a2 + 2al = 162 + 2 × 16 × 17 = 800sq.cm

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2020-21 Academic year Worksheets

Mathematics X
Solids

Concepts

a) Cone can be made by rolling a sectorol sheet.While doing so the arc length of the sector becomes
the base perimetre of the cons.Area of the sector becomes lateral surface area of the cone. Lateral
surface area is also known as curved surface area .

b) The radius of the sector becomes slant height of the cone . It can be denoted by l for between
convenience.

c) Since arc length of the sector is equal to base perimetre of the cone we can make the relation given

below 2πl
x = 2πr

360

l radius of the sector , x central angle of the sector and r is the radius of the cone , lx = 360r.

d) Area of the sector becomes the lateral surface area of the cone.We can make a formula to calculate

lateral surface area of the cone

Curved surface area= Area of the sector
πr2
Curved surface area = 360 x = π×l×l×x
360

lx = 360r

Curved surface area = π×l×360r = πrl
360

1) A sectoral sheet of central angle 120◦is cut off from a circular sheet of radius 12cm . It is rolled in such a
way as to get a cone.

a) What is the slant height of the cone?
b) What is the radius of the cone ?
c) Find the curved surface area of the cone.

Answers

a) Slant heightl = 12cm
b) lx = 360r

12 × 120 = 360 × r, r = 4 cm
c) Lateral surface area = πrl = π × 4 × 12 = 48πsq.cm

2) A cone is made by rolling a semicircular metal sheet of radius 10cm

a) What is the slant height of the cone.
b) What is the radius of the cone.
c) Find the curved surface area of the cone.
d) Base is made by a suitable circular sheet. What is its total surface area ?

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Answers

a) slant heightl = 10cm
b) lx = 360r

10 × 180 = 360 × r, r = 5cm
c) Lateral surface area = πrl = π × 5 × 10 = 50πsq.cm
d) Total surface area = Lateral surface area +base face = 50π + 25π = 75πsq.cm

3) A circular sheet of card board of radius 12cm .It is cut off into two sectors of central angle 120◦ and
240◦.Both of them are rolled into cones.

a) Name the measure coomon to both comes
b) What is the radius of small cone ?
c) What is the radius of the big cone.
d) How radii of the cones are related to the radius of the circular sheet.

Answers

a) slant height= 12 cm

b) lx = 360r1 → 12 × 120 = 360 × r1
12×120
r1 = 360 = 4cm

c) lx = 360r2 → 12 × 240 = 360 × r2
12×240
r2 = 360 = 8cm

d) r1 + r2 = 12.Sum of the base radii of cones is equal to the radius of the circular sheet

4) A sector of central angle 90◦ is cut off from a circular sheet of radius 16cm .It is rolled in such a way as to
get a cone.

a) What is the lateral surface are of the cone?
b) What is the radius of the cone?
c) The remaining part of the circular sheet is also rolled to get a cone . What is its base radius?
d) Which cone has more height ? Explain

Answers

a) Area of tsheectsoercitsor41=of14 the area of circular sheet.
Area of × π × 162 = 64πsq.cm

b) lx = 360r → 16 × 90 = 360 × r

r = 16×90 = 4cm
360

c) lx = 360r → 16 × 270 = 360 × r

r = 16×270 = 12cm
360

d) Radius , height and slant height form a right triangle.Slant height of both pyramids are equal.It
is equal to the hypotenuse of the triangle.Whenever the hypotenuse remains same , length
of one perpendicular side increases according to the decrease of other side. Cone made
from the sector of central angle has less height.

5) A cone is made by a sectoral sheet taken from a circular sheet.The slant height of the cone is two times its
radius.

a) What is the relation between lateral surface area and base area?

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b) If the base perimetre is 20πcm then what will be its lateral surface area ?
c) What is the central angle of this sector?
d) The remaining part is also rolled to get a cone. What is the ratio of the heights of cones so formed
Answers

a) l = 2r → lateral surface area = πrl = π × r × 2r = 2πr2 = 2×base area
b) 2πr = 20π → r = 10cm

l = 20cm .Lareral surface area π × 10 × 20 = 200πsq.cm
c) lx = 360r → 2r × x = 360 × r

x = 180◦
d) This is a semicircle. Remaining part is also a semicircle. Ratio of the height is 1 : 1

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2020-21 Academic year Worksheets

Mathematics X
Solids

Concepts

a) Cone can be made by rolling a sectorol sheet.While doing so the arc length of the sector becomes
the base perimetre of the cons.Area of the sector becomes lateral surface area of the cone. Lateral
surface area is also known as curved surface area .

b) The radius of the sector becomes slant height of the cone . It can be denoted by l for between
convenience.

c) Since arc length of the sector is equal to base perimetre of the cone we can make the relation given

below 2πl
x = 2πr

360

l radius of the sector , x central angle of the sector and r is the radius of the cone , lx = 360r.

d) Area of the sector becomes the lateral surface area of the cone.We can make a formula to calculate

lateral surface area of the cone

Curved surface area= Area of the sector
πr2
Curved surface area = 360 x = π×l×l×x
360

lx = 360r

Curved surface area = π×l×360r = πrl
360

1) A cone of radius r1 is made by using a sector of a circular sheet of radius R.The remaining part of the sheet
is rolled in such a way as to get another cone of radius r2

a) Which measure is common in both cones?
b) Write the relation between the radius , slant height and central angle of the sector in the case of first

cone.
c) Write the relation between the radius , slant height and central angle of the sector in the case of

second cone.
d) prove that R = r1 + r2

Answers

a) Common measure is slant height.Slant height of both cones is R

b) If the central angle of the sector for first cone is x then Rx = 360r1
c) The central angle of the sector for second cone is 360 − x and R(360 − x) = 360r2
d) Adding these equations, Rx + R(360 − x) = 360r1 + 360r2 → R = r1 + r2

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2) A cone is made by taking a sector from a circular sheet.The slant height of the cone is 25cm and its radius
110cm

a) What is the radius of the circular sheet?
b) What is the central angle of the sector?
c) What is the central angle of the remaining part?
d) What is the radius of the cone made by rolling the remaining part?

Answers

a) 25cm

b) lx = 360r → 25 × x = 360 × 10, x = 360×10 = 144◦
25

c) Central angle of sector = 360 − 144 = 216◦

d) Radius of the cone 25 − 10 = 15cm

3) The base perimetre of a cone is 20π cm, slant height 18cm . It is rolled to get a cone.

a) What is the radius of the sector?
b) What is the radius of the cone?
c) What is the central angle of the sector?
d) Find the lateral surface area of the cone?

Answers

a) 18cm

b) 2πr = 20π, r = 10cm

c) lx = 360r → 18 × x = 360 × 10, x = 360×10 = 200◦
18

d) പാർശ ഖപര ളവ് πrl = 180πsq.cm

4) A sector of central angle 288◦ and radius 25cm is taken from a circulat sheet .

a) What is the radius of the cone?
b) What is the height of the cone ?
c) Find the lateral surface area of the cone?
d) What is the radius of the cone made by rolling the remaining part?

Answers

a) lx = 360r → 288 × 25 = 360 × r, r = 288×25 = 20 cm
360
√
b) l2 = h2 + r2 , 252 = h2 + 202 → h2 = 625 − 400 = 225, h = 225 = 15cm

c) Lateral surface area πrl = π × 20 × 25 = 500π sq.cm

d) Radius of the remaining part25 − 20 = 5cm

5) A cone of maximum size is carved from a square prism of base edge 10cm and height 12cm.
a) What is the radius of the cone?
b) What is the slant height of the cone?
c) What is the lateral surface area of the cone?
d) Find the total surface area of the cone?

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Answers

a) 5cm
b) h =√12cm, r = 5cm

l = 52 + 122 = 13 cm
c) Lateral surface area = πrl = 65πsq.cm
d) Total surface area =Base area + Lateral surface area

= 25π + 65π = 90πsq.cm

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2020-21 Academic year Worksheets

Mathematics X
Solids

Concepts

a) There are three basic neasurements in a cone.Radius r, height h and slant height l. These
measurements form a right triangle. l2 = r2 + h2

b) Base perimetre of the cone= 2πr, Base area = πr2

c) Lateral surface area of the cone = πrl,Total surface area = πr2 + πrl

d) Volume of the cone= 1 πr2h
3

1) Radius of a cone is 5cm, height 12cm

a) what is the slant height of the cone?
b) What is the total surface area of the cone?
c) What is the volume of the cone ?
d) In a cone , radius and height are equal. If the volume and curved surface area are equal then what is

its radius ? What is its slant height?

Answers

√√ √
a) l2 = r2 + h2 → l = r2 + h2 = 52 + 122 = 169 = 13cm

b) Total surface area πr2 + πrl = π × 52 + π × 5 × 13 = 25π + 65π = 90π

c) Volume = 1 πr2 h = 1 π × 52 × 12 = 100πcubic cm
3 3
√
d) If r = hthen slant height l = 2r .Volume√= 1 × π × r2 ×r = 1 π × r3
3 3

Total surface ar√ea of the cone√= π × r × 2r.
1 πr3 = πr ×√ 2r →√r = 3 2cm
3

slant height= 2 × 3 2 = 6cm

2) The base perimetre of a cone is 30πcm, height 20 cm

a) What is the radius and slant height of the cone ?
b) What is the total surface area?
c) Find the volume of the cone?
d) What is the volume of a cyclidrical vessel of radius and height equal to that of the cone.

Answers

a) 2πr = 30π, r =√15cm √ √
slant heightl = r2 + h2 = 152 + 202 = 625 = 25cm

b) Total surface area πr2 + πrl = 225π + 375π = 600πsq.cm

c) Volume = 1 πr2 × h = 1 × π × 152 × 20 = 1500πcubic cm
3 3

d) Volume becomes 3 times .Volume = 4500πcubic cm
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3) Diametre and height of a cone are equal.

a) What is the relation between radius and slant height?
b) What is the ratio of radius , height and slant height?
c) If the radius is 6 cm then what is its volume?
d) If the radius is 6 cm then what is the total surface area ?

Answers

√ √√
a) h = 2r, l = r2 + (2r)2 = 5r2 = 5r

√
b) r : h : l = r : 2r :√ 5r

r:h:l=1:2: 5

c) If radius is 6cm then height 10cm .Volume = 1 π × 62 × 10 = 360πcubic cm
3
√√
d) If radius is 6cm then l = 5 × 6 = 6 5cm√.
√
Total surfac√e area = π × 62 + π × 6 × 6 5 = 36 + 36 6
= 36(1 + 6)sq.cm

4) Radius of a cone is r and height h.

a) What will be the change in volume if radius and height are doubled?
b) What will be the change in volume if radius is doubled and height is halved?
c) How many solid cones can be made by melting a solid cone of radius 10cm and height 6cm with half

the radius and height of the melted cone?

Answers

a) V = 1 πr2h.
3
1 π(2r)2 × × 1 πr2h
radius 2r, height 2h Volume = 3 (2h) = 8 3 = 8V

Volume becomes 8times

b) If radius is 2r and height h then
2
1 (2r)2 h 1 πr2h
volume 3 π × × 2 = 2 × 3 = 2V

Volume becomes 2 times .

c) When radius and height become halved then volume become 1 part . 8 cones can be made.
8

5) A conical measuring vessel is made by rolling a sectoral sheet of central angle 288◦ and radius 10cm.

a) What is the radius of the vessel?
b) What is the height of the vessel?
c) What is the capacity of the vessel in litres ?

Answers

a) lx = 360r → 10 × 288 = 360 × r

r = 10×288 = 8 cm
360
√√
b) h = l2 − r2 = 102 − 82 = 6cm

c) Volume = 1 π × r2 × h = 1 × π × 82 × 6 = 128πcubic cm
3 3

1000cubic cm =1litre

Volume = 128×3.14 = 0.4litre
1000

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2020-21 Academic year Worksheets

Mathematics X
Solids

Concepts

a) There are three basic neasurements in a cone.Radius r, height h and slant height l. These
measurements form a right triangle. l2 = r2 + h2

b) Base perimetre of the cone= 2πr, Base area = πr2

c) Lateral surface area of the cone = πrl,Total surface area = πr2 + πrl

d) Volume of the cone= 1 πr2h
3

1) Radius of a cone is 21cm , height 28cm.

a) Calculate slant height.
b) Find the leteral surface area .
c) Calculate the total surface area
d) Calcualte the volume of the cone..

Answers
√

a) l2 = r2 + h2, l2 = 212 + 282 = 1225, l = 1225 = 35 cm

b) Lateral surface area = πrl = π × 21 × 35 = 735πsq.cm

c) Total surface area = πr2 + πrl = π × 212 + π × 21 × 35 = 441π + 735π = 1176πsq.cm

d) Volume = 1 πr2 h = 1 × π × 212 × 28 = 4116π cubic cm
3 3

2) Ratio of radius and height of a cone are 3 : 4.Volume of the cone is 301.44 cubic cm
a) Find the radius of the cone.
b) Find the height of the cone.
c) Calculate the slant height of the cone.
d) Calculate the lateral surface area of the cone.

Answers

a) r : h = 3 : 4, r = 3x, h = 4x
x313×=πr30221h2.4×=49××334×017.4=4,813 × 3.14 × (3x)2 × (4x) = 301.44
x = 2, r = 3x = 6cm

b) Height h = 4x = 8cm
√√

c) l = r2 + h2 = 100 = 10cm
d) Lateral surface area = π × r × l = 60πsq.cm

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3) Lateral surface area of a cone is 4070 sq.cm , diametre 70 cm

a) Find the slant height of the cone .
b) Find the height of the cone?
c) Calcualte the volume of the cone.

Answers

a) πrl = 4070, 22 × 35 × l = 4070, l = 4070×35 = 37 cm
7 22×5
√√ √
b) h = l2 − r2 = 372 − 352 = 144, h = 12cm

c) Volume= 1 × π × r2 × h = 1 × π × 352 × 12 = 4900πcubic cm
3 3

4) The height of a cone is 24cm , its lateral surface area 550 sq.cm

a) What is the radius of the cone?
b) Find the height of the cone?
c) Calcualte the volume of the cone?

Answers

a) πrl = 550, 22 × rl = 550, rl = 175.
7
r2l2 = 1752, r2(r2 + h2) = 1752

If r2 = x then, x(x + 242) = 1752

x2 + 576x = 30625, x2 + 576 − 30625 = 0

x = 49 → r2 = 49, r = 7

b) rl = 175, l = 175 = 25cm
7

c) Volume = 1 πr2 h = 1 × π × 72 × 24 = 49 × 8 × π = 392πcubic cm
3 3

5) A semicircular sheet of radius 28cm is rolled in such a way as to get a cone.

a) What is the slant height of the cone?
b) Find the radius of the cone?
c) Find the height of the cone?
d) Calcualte the volume of the cone

Answers

a) Slant height = 14cm

b) lx = 360r → 14 × 180 = 360 × r, r = 14×180 = 7cm
360
√√ √
c) h = l2 − r2 = 142 − 72 = 147. h = 147 = 12.12 cm

d) Volume = 1 πr2 h = 1 × π × 72 × 12.12 = 621.6cubic cm
3 3

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2020-21 Academic year Worksheets

Mathematics X
Solids

Concepts

a) The basic measurement of a sphere is its radius. Surface area and volume can be calculated using
radius of the sphere.

b) If r is the radius of the sphere , its surface area is 4πr2

c) Volume of the sphere with radius r is 4 πr3
3

d) Volume of the hemisphere is 2 πr3
3

e) Curved surface area of the hemisphere = 2πr2. Total surface area of the hemisphere = 3πr2

1) Calculate the following measures of sphere with radius 3cm

a) Find the total surface area of the sphere.
b) Calculate the volume of the sphere
c) Calculate the curved surface area of the hemisphere taken from this sphere .
d) If it is solid sphere find the total surface area of the hemisphere.

Answers

a) Total surface area 4πr2 = 4π × 32 = 36πsq.cm

b) Volume= 4 πr3 = 4 × π × 33 = 36πcm
3 3

c) Curved surface area of the hemisphere = 18πsq.cm

d) 4πr2 = 36π, πr2 = 9π
Total surface area of the hemisphere 3πr2 = 27πsq.cm

2) Volume and surface area of a sphere are equal in number.

a) What is its radius
b) Calculate the volume or total surface area
c) How many spheres of radius 1cm can be made by melting this solid sphere ?

Answers

a) 4πr2 = 4 πr3 → 3r2 = r3, r = 3
3

b) Volume= 4 π33 = 36π, Total surface area = 36πsq.cm
3

c) Volume of melted sphere = 4 π × 33
3
=sph43eπre×=3343÷π × 13
Volume of the small
Number of spheres 4 π × 13 = 27
3

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3) Find the volume of the sphere accoring the changes of radius as given below

a) What will be the change in volume if radius is doubled?
b) What will be the change in volume if radius is halved?
c) The volume of a sphere is 10 cubic cm. What will be the volume of the sphere of diametre two times

the first one.

Answers

a) Initial radius r, Initial volumeV

V = 4 πr3.
3
4 π(2r)3 23 × 4 πr3
When radius is doubled volume = 3 = 3 = 8V

Volume becomes 8 times

b) When radius becomes r , volume = 4 × π × ( r )3 = 1 ×V.
2 3 2 8
1
becomes 8 part

c) When diametre doubles , radius becomes two times . Volume = 80cubic cm

4) A sphere of maximum side is carved from a wooden cube of side 6cm
a) What is the radius of the sphere?
b) Calculate the surface area of the sphere.
c) calculate the volume of the sphere .

Answers
a) Side of the cube and diameter of the sphere are equal . r = 3
b) Total surface area 4πr2 = 4 × π × 32 = 36πsq.cm
c) Volume = 36πcubic cm

5) A sphere is fixed inside a conical vessel of diametre 10cm and height 12cm. It touches the curved surface
of the cone and its base.

a) What is the slant height of the cone ?
b) Find the radius of the sphere?
c) Calculate the volume of the sphere ?
d) What fraction of the inner volume of the cone is occupied by the sphere?

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Answers
√√

a) l = h2 + r2 = 122 + 52 = 13cm

b) When the solid is divided by two equal parts along the diameter we can see a triangle and

its incircle.

Perimeter of the triangle 13 + 13 + 10 = 36cm

s = 18. Area of triangle = 1 × 10 × 12 = 60sq.cm
2
A 60 10
r = s = 18 = 3 cm

c) Volume of sphere = 4 × π × ( 10 )3 = 4000π cubic cm
3 3 81

d) Volume of cone 1 π × 52 × 12 = 100π.
3
4000π ÷ 40
81 100π = 81 part .

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2020-21 Academic year Worksheets

Mathematics X
Solids

Concepts

a) The basic measurement of a sphere is its radius. Surface area and volume can be calculated using
radius of the sphere.

b) If r is the radius of the sphere , its surface area is 4πr2

c) Volume of the sphere with radius r is 4 πr3
3

d) Volume of the hemisphere is 2 πr3
3

e) Curved surface area of the hemisphere = 2πr2. Total surface area of the hemisphere = 3πr2

1) There is a sphere of radius 1cm.It is melted an recast into small spheres of radius 1 cm .
2

a) What part of the volume of the melted sphere is the volume of the small sphere ?

b) What part of the surface area of the melted sphere is the surface area of the small sphere ?

c) How many spheres of radius 1 can be made ?
2

d) What is the difference between the surface area of big sphere and sum of the surface area of small

spheres.

Answers

a) VV18ooplluuammrteeooof ffthtthheeevssoppluhhmeeerreeooof ffbrrigaaddsiiuupsshe121recismis43isth×e43 π × 13 = =34 π18 ×
)3 a small
π × ( 1 4 π
2 of 3

volume sphere.

b) Surface area of the sphere of radius 1 cm is 4π.

Surface area of tahesmsaplhl esrpeheorferaisdi41usof21 is 4π × ( 1 )2 = 1 × 4π
Surface area of 2 area big sphere.
4
the surface
of

c) 8 spheres

d) total surface area of 8 small spheres is 8 × 1 × 4π = 2 × 4π.
4
The difference of the surace area of big sphere and all small spheres is 2 × 4π − 4π = 4π

2) Total surface area of a solid sphere is 64π sq.cm

a) What is the radius of the sphere?
b) What is the volume of the sphere ?
c) The sphere is split up into two hemispheres . What is the total surface area of a hemisphere?

Answers

a) 4πr2 = 64π ,r2 = 16, r = 4cm

b) Volume of the sphere = 4 πr3 = 4 π × 43 = 256π cubic cm
3 3 3

c) Total surface area of the hemisphere = 3πr2 = 3π × 42 = 48πsq.cm

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3) Total surface area of a solid hemisphere is 27π sq.cm
a) What is the radius of the hemisphere ?
b) What is the curved surface area of this hemisphere?
c) What is the total surface area of the sphere made by joining two such hemispheres?

Answers
a) 3πr2 = 27π → r2 = 9, r = 3cm
b) Curved surface area of the hemisphere = 2πr2 = 18πsq.cm
c) When two hemispheres are joined in their circular faces it make a sphere of surface area
36π sq.cm

4) A cone of maximum size is carved from a hemisphere of radius 10cm

a) What is the height of the cone?
b) What is the slant height of the cone?
c) Find the total surface area of the cone?
d) Find the volume of the cone?
e) What part of the volume of the hemisphere is the volume of the cone?

Answers

a) h = 10cm
√√

b) l = 2 × r = 10 2cm
√√

c) Total surface area = πr2 + πrl = π × 102 + π × 10 × 10 2 = 100π(1 + 2)sq.cm

d) Volume of the cone = 1 π × 102 × 10 = 1000 πcubic cm
3 3

e) Volume of the hemisphere is 2 πr3 . Volume of the cone is 1 πr3.Since h = r
The volume of the cone is 3 sphere. 3
1
the volume of the
2

5) A hemisphere is carved from a solid cone of radius 15cm and height 20 cm.

a) What is the slant height of the cone?
b) What is the radius of the hemisphere ?

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c) Calculate the surface area of the hemisphere
d) Calculate the volume of the hemisphere

Answers

√√
a) AB = 202 + 152 = 625 = 25cm

b) Let rbe the radius of the semicircle.

△BP D and △BDAare similar.

BD = AD
AB PD
15 r
25 = 20 , r = 12cm

c) Surface area of the hemisphere = 3πr2 = 3π × 122 = 432πsq.cm

d) Volume = 2 π × 123 = 1152πcubic cm
3

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2020-21 Academic year Worksheets

Mathematics X
ഘന പ ൾ

Concepts

a) The basic measurement of a sphere is its radius. Surface area and volume can be calculated using
radius of the sphere.

b) If r is the radius of the sphere , its surface area is 4πr2

c) Volume of the sphere with radius r is 4 πr3
3

d) Volume of the hemisphere is 2 πr3
3

e) Curved surface area of the hemisphere = 2πr2. Total surface area of the hemisphere = 3πr2

1) Total surface area of a solid sphere is 100 sq.cm . It splits up into two hemispheres .
a) What is the curved surface area of the hemisphere ?
b) What is the total surface area of the hemisphere?

Answers

a) 4πr2 = 100 → πr2 = 25
Curved surface area = 2πr2 = 2 × 25 = 50sq.cm

b) Total surface area of the hemisphere = 3πr2 = 3 × 25 = 75sq.cm

2) A cylindrical vessel of radius 4cm contains water to the height 10cm. A small solid sphere of radius 2 cm is
immersed in it.

a) Find the volume of the sphere.
b) Calculate the level to which water rises.
c) A child said: If a sphere of double the radius is immersed the level also rises twice. Is this statement

true? Justify

Answers

a) Volume of the sphere = 4 πr3 = 4 π × 23 = 32π cubic cm
3 3 3

b) Volume of sphere is equal to the volume of water rised .Let h be the level of water rises .

π × 42 × h= 32π
h =
32 = 2 3
3×16 3
= 0.66sq.cm

c) Ir radius is rthen volume V = 4 × π × r3 . If radius is 2rthen volume 4 ×π × (2r)3 = 8×V
3 3
When radius doubles valume become 8 times . Water level rises 8 × 0.66 = 5.28cm

3) A cone of height 16cm and maximum size is carved from a solid sphere of radius 10cm.

a) calculate the radius of the cone? 1
b) Find the slant height of the cone?
c) Calculate the volume of the cone?

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Answers

a) If radius of the cone is rcm
102 = r2 + 62 → r2 = 64, r = 8 cm
√√

b) Slant height l, l2 = 82 + 162 = 64 + 256 = 320, l = 320 = 8 5cm

c) Volume of cone = 1 π × r2h = 1 × π × 82 × 16 = 1024 πcubic cm
3 3 3

4) Ratio of the volume of two cones is 64 : 27. The sum of the radii is 21cm

a) What is the ratio of their radii?
b) Calculate the radius of each cone?
c) What is the ratio of the surface area of the cones?

Answers

a) 4 πr13 : 4 πr23 = 64 : 27
3 3
r13
r23 = 64 , r1 : r2 =4:3
27

b) r1 + r2 = 21 → 4r + 3r = 21, 7r = 21, r = 3
Radius :r1 = 12 cm , r2 = 9cm

c) Ratio of the surface area is equal to ratio of the radii
Ratio of total surface area = 122 : 92 = 144 : 81 = 16 : 9

5) Radius of a cone and s hemisphre are equal. Volume of cone and hemisphere are equal.

a) What is the ratio of the heights?
b) If the radius of the sphere is 10cm then what is its height?
c) If the radius of the sphere is 2cm then what is its curved surface area ?
d) Find the ratio of the surface area of the cone and hemisphere.

Answers

a) Radius of the semicircle and height are equal.

Let r be the radius r, Height of cone h

1 πr2h : 2 πr3 = 1 : 1
3 3

h:r=2:1

b) Height = 20cm
c) If radius is 2 , the curved surface area of the hemisphe√re is 2πr2 = 8√π

Radius of the cone 2cm , height 4cm . s√lant heig√ht l = 42 + 22 = 2 5cm
Curved surface area πrl = π × 2 × 2 5 = 4 5πsq.cm

√√
d) Ratio of the curved surface area of hemisphere and cone is = 8 : 4 5 = 2 : 5

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2020-21 Academic year Worksheets

Mathematics X
Solids

1) A solid is made by fixing a hemisphere in the base of a cone . The height of the cone is 12 cm. Cone and
hemisphere have smae base area.Total height of the solid is 17cm.

a) What is the radius of the hemisphe?
b) What is the slant height of the cone?
c) Calculate the surface area of the solid.
d) Calculate the volume of the solid.

Answers

a) r = 17 − 12 = 5cm
√√

b) l = 122 + 52 = 169 = 13cm

c) Curved surface area of the solid = 2πr2 + πrl = 2π × 52 + π × 5 × 13 = 50π + 65π =
115πsq.cm

d) V31oπlu×me52=×131π2r2+h + 2 πr3
3
2 × 53 250π 550π
3 π = 100π + 3 = 3 cubic cm

2) Hemispheres are fixed at the ends of a cylinder.Radius of the hemispheres and cylinder are equal.Total
height of the solid is 19 cm.If the common radius is 7 cm then

a) What is the height of the cylinder?
b) Calculate the total surface area of the solid.
c) Calculate the volume of the solid.

Answers

1

a) h = 19 − 14 = 5cm

b) Totavl sisuriftacwe awrewa =.s2h×e2πnri2s+c2hπorho=l.i4nπ×o7r2+w2hπa×t7s×a5p=p19760π1+2704π9=826606π6sq.cm

c) Two hemispheres make a sphere .

Answers

a) Height of the cone = 23.2 − (4.2 + 14) = 5cm

b) Volume of the cone = 1 π × r2h = 1 π × 4.22 × 5 = 29.4πcubic cm
3 3

c) Volume of the hemisphere = 2 × π × 4.23 = 49.39π cubic cm
3

d) Volume of cylinder= πr2h = π × 4.22 × 14 = 246.96πcubic cm

e) Volume of solid= 29.4π + 49.39π + 246.96π = 1022.85cubic cm

4) A cone of radius 10cm and 10cm height 8 cm is divided equally by a plane parallel to the base half of its
height.

a) What is the radius of the small cone taken from the big cone?
b) Calculate the volume of the small cone .
c) What is the volume of the remaining part ?ബാ ിവ ഭാഗ ിെ വ ാ ം കണ ാ ക
d) What is the ratio of the volume of two parts .

Answers

a) △ABC, △ADEare similar .

4 = r → r = 5 cm
8 10

b) Volume of small cone = 1 πr2h = 1 × π × 52 × 4 = 100π cubic cm
3 3 3

c) Volume of rbeigmcaoinnineg31pπarrt28h030=π 1 × π × 102 × 8 = 800π
Volume of 3 3
100π 700π
− 3 = 3 cubic cm

d) അംശബ ം 100π : 700π = 1 : 7
3 3

5) A sphere is fixed inside a cone in such a way that the sphere touches the lateral face and base of the
cone.The base radius of the cone is 6 cm and height 8 cm

a) What is the slant height of the cone?
b) What is the radius of the sphere ?
c) What is the volume of the sphere?
d) Calculate the volume of the cone.

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Answers

Divide the solid along its diametre of the base. Now we can see a triangle and incircle. Using the
relation A = rs we can find r

√√
a) l = 62 + 82 = 100 = 10cm

b) s= 10+10+12 = 16, A = 1 × 12 × 8 = 48
2 2
A 48
r = s = 16 = 3cm

c) Volume of sphere = 4 πr3 = 36πcubic cm
3

d) Volume of cone = 1 πr2h = 1 × π × 62 × 8 = 96πcubic cm
3 3

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09

Geometry and
Algebra

ജ്യാമിതിയും
ബീജഗണിതവും

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2020-21 Academic year Worksheets

Mathematics X
Geometry and Algebra

Concepts
⋆ In this unit we discuss the relation between the coordinates of points in a geometric figure

algebraically.
⋆ The relation between the coordinates of points on a line is the equation of the line. A line will have

an equation.
⋆ The coordinates of two points on a line and the coordinates of any point on that line in between the

given points are related to eachother by a ratio.
⋆ The slope of a line is the measure of its inclination.
1) ABCD is a parallelogram . If A(1, 1), B(4, 2), C(6, 7)then

a) Write the difference between x coordinates of A and B
b) Write the difference between y coordinates of A and B
c) Write the coordinates of D.
Answers

a) The difference between x coordinates of A and B is = 4 − 1 = 3
b) The difference between y coordinates of A and C is = 2 − 1 = 1
c) D(6 − 3, 2 − 1) = D(3, 1)
2) In the figure P QRS is a parallelogram .If P (0, 3), P S = 4, Q(5.4)then

1

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a) Write the coordinates of S
b) Write the coordinates of R
c) Find the length of the sides.
Answers

a) S(0, 3 + 4) = S(0, 7)
b) R(5, 4 + 4) = R(5, 8)

√ √√
c) P Q = (5 −√0)2 + (4 − 3)2 = 25 + 1 = 26

P Q = RS = 26, P S = QR = 4

3) P (1, 4)the mid point of the side AB ,Q(2, 3) is the mid point of side BC , R(5, 6)is the mid point of side
AC
a) Draw a suitable diagram representing the position of points
b) Write the coordinates of B
c) Write the coordinates of C
d) Write the coordinates of A
Answers
a) Look at the picture

b) BP RQis a parallelogram .The difference between x coordinates of P, R is 4
The difference between ycoordiantes is 2
B(2 − 4, 3 − 2) = B(−2, 1)

c) P RCQ is a parallelogram .The difference between x coordinates of P, R is 4
The difference of y coordinates of P, R is 2.
C(2 + 4, 3 + 2) = C(6, 5)

d) P QRA is a parallelogram. The difference between x coordinates of Q, R is 3
The difference between ycoordinates is 3
A(1 + 3, 4 + 3) = B(4, 7)

4) (1, 5), (4, 1), (4, 5) are the mid points of the sides of a triangle.
a) Draw a diagram and mark the mid points
b) Find the coordinates of the vertices of the triangle.
c) What kind of triangle is this ?
d) Calculate the area of this triangle.

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Answers
a) see picture

b) In P QRB ,the difference between x coordinates of P, Q is 3.
The difference between ycoordinates is 0.
Therefore B(4 − 3, 1) = B(1, 1)
In P QCR, the difference between x coordinates of P, Q is 3.
The difference between ycoordinates is 0.
Therefore C(4 + 3, 1) = B(7, 1)
In AP RQ,the difference between x coordinates of Q, R is 0 .
The difference of y coordinates is 4.
Therefore A(1, 5 + 4) = A(1, 9)

c) AB = 8, BC = 6, AC = 10. AB2 + BC2 = AC2. ABC is a right triangle

d) Area = 1 × BC × AB = 1 × 6 × 8 = 24 sq.unit
2 2

5) In the parallelogram ABCD the line AB and line CD are parallel to xaxis. A(1, 5) .Length of these
parallel sides is 6cm and the area of the parallelogram is 48.

a) What is the distance between the parallel sides AB and CD
b) Write the coordinates of B
c) If x coordinates of C is 9, then what is the y coordinates of C. Write the point C .
d) Write the coordinates of D
e) Find the perimetre of the parallelogram.

Answers

a) Draw figure .ah = 48 → 6 × h = 48, h = 8
Distance between parallel sides 8.

b) B(1 + 6, 5) = B(7, 5)

c) y coordinates of C is 5 + 8 = 13
C(9, 13)

d) D(3, 13)
√ √√

e) BC = (9 − 7√)2 + (13 − 5)2 = 22 + 82 = 68
Perimetre= 2 × 68 + 12

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2020-21 Academic year Worksheets

Mathematics X
Geometry and Algebra

Concepts

In the parallelogram ABCD, A(x1, y1), B(x2, y2), C(x3, y3), D(x4, y4) then x3 = x2 + x4 −
x1, y3 = y2 + y4 − y1.
Using this the vertices of a triangle can be determined by using the mid points of the sides.

1) The mid point of the side AB of triangle ABC is P (1, 1).The mid point of the side BC is (5, 4), mid point
of the side AC is (7, 4)
a) Find the coordinate of the vertex A.
b) Find the coordinate of the vetex B
c) Find the coordiante of the vertex C.

Answers

a) A(1 + 7 − 5, 1 + 4 − 4) → A(3, 1)
b) B(1 + 5 − 7, 1 + 4 − 4)→ B(−1, 1)
c) C(7 + 5 − 1, 4 + 4 − 1)→C(11, 7)

2) There is a line joining the points A(x1, y1), B(x2, y2). The point P (x, y) divides AB in the ratio m : n.

a) Establish the relation x = x1 + m (x2 − x1)
m+n

b) Establish the relation y = y1 + m (y2 − y1)
m+n

Answers

a) In the figure AS, P Q, BRare the perpendiculars from A, B, P to x axis .Line AN intersect

P Q at M . This line is parallel to x axis .

M (x, y1), N (x2, y1) AM = x − x1, AN = x2 − x1, P M = y − y1, BN = y2 − y1

△AM P , △AN Bare similar traingles . Sides opposite to eaual angles are propotional.

AP = AM
AB AN
AM m
AN = m+n

x−x1 = m (x2 − x1), x = x1 + m (x2 − x1)
x2 −x1 = m+n m+n

x − x1 m
m+n

b) Similarly y = y1 + m (y2 − y1)
m+n

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3) P (x, y) is a point on the line AB in between A and B which divides AB in the ratio 2 : 3.If A(1, 3) and
B(3, 4) then

a) Find the x coordinates of P
b) Find the y coordinates of P

Answers

a) x = x1 + m (x2 − x1)
m+n

x = 1 + 3 (3 − 1) = 1 + 3 × 2
3+4 7

x= 1+ 6 = 13
7 7

b) y = y1 + m (y2 − y1)
m+n

y = 3 + 3 (4 − 3) = 3 + 3 × 1
3+4 7

y = 3+ 3 = 24
7 7

4) There is a line jpining the points A(x1, y1) and B(x2, y2). P (x, y) is the mid point of AB.

Prove that the coordinates of P are (x1 +x2 ) , (y1 +y2 )
2 2
Find the coordinates of the mid point of the line joining A(−3, 5) and B(7, 7)

Answers

a) P (x, y) is the mid point of of AB.

m:n=1:1

x = x1 + m (x2 − x1) → x1 + 1 (x2 − x1)
m+n 1+1
x1 +x2
x = 2

b) y = y1 + m (y2 − y1) → y1 + 1 (y2 − y1)
m+n 1+1
y( x=+2xy21,+2yy+22y2 )

Coordinates of mid point = ( −3+7 . 5+7 ) = (2, 6)
2 2

5) In triangle ABC, A(−3, 2), B(1, 5), C(3, −4)then

a) Find the coordinates of the mid point of AB
b) Find the coordiantes of the mid point of BC
c) Find the coordinates of the mid point of AC

Answers

a) mid point of AB is ( −3+1 , 2+1 ) = (−1, 3 )
2 2 2

b) Mid point of BC is ( 1+3 , 5+−4 ) = (2, 1 )
2 2 2

c) Mid point of AC is( −3+3 , 2+−4 ) = (0, −1)
2 2

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2020-21 Academic year Worksheets

Mathematics X
Geometry and Algebra

Concepts

⋆ The line joining vertex to the mid point of the opposite side is called median.There are three medians
in a triangle.

⋆ The point of intersection of the medians is called centroid of the triangle.It is denoted by G.
⋆ The centroid of a triangle divides the median in the ratio 2 : 1 from the vertex.

1) In triangle ABC, A(0, 0), B(4, 0), C(2, 10)
AB, BC, AC has the mid points P, Q, Rrespectively

a) Write the coordinates of P, Q, R
b) Find the coordinates of the point which divides CP in the ratio 2 : 1.
c) Find the coordinates of the point which divides AQ in the ratio2 : 1
d) Find the coordinates of the point which divides BR in the ratio 2 : 1

Answers

a) P (2, 0) , Q( 2+4 , 10+0 ) → Q(3, 5) , R( 0+2 , 10+0 ) → R(1, 5)
2 2 2 2

b) x coordinate of the point which divides CP in the ratio 2 : 1 x = x1 + m (x2 − x1) =
m+n

2 + 2 (2 − 2) = 2
2+1

ycoordinate = y1 + m (y2 − y1) = 10 + 2 (0 − 10) = 10 − 20 = 10
m+n 2+1 3 3

Point is (2, 10 )
3

c) The point which divides AQ is (0 + 2 (3 − 0), 0 + 2 (5 − 0))→ Q(2, 10 )
3 3 3

d) Point which divides BR is (4 + 2 (1 − 4), 0 + 2 (5 − 0))→ (2, 10 )
3 3 3

2) The line joining A(−1, 3), B(4, 1) intersect y axis at P

a) What is the x coordinate of P ?
b) Find AP : BP
c) Find the coordinates of P .

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Answers

a) Since P is on y axis x coordiante is 0

b) x = x1 + m (x2 − x1)
m+n
m −− m
0 = −1 + m+n (4 1), 1 = m+n × 5

m +n = 5m, 1 = m , m : n = 1 : 4
4 n

c) y = y1 + m (y2 − y1) y = 3 + 1 (1 − 3) = 13
m+n 1+4 5

P (0, 13 )
5

3) Triangle ABC is an equilateral triangle.A(1, 0), B(5, 0)

a) Find side of the triangle
b) What is the coordinates of the mid point of AB
c) Find the coordinates of C
d) Find the coordinates of the mid point of the centroid .

Answers

a) AB =| 5 − 1 |= 4

b) Mid point of AB is M .

M ( 1+5 , 0) → M (3, 0)
2
√
c) C(3, 2 3)

d) C M is the median . Coodinate of the 2c√en3tr+oid32G(0(x−, y)√, √
x = 2 3) 23
3, y = y1 + m (y2 − y1) = =
m+n 3
√
G(3, 2 3 3 )

4) In triangle ABC , CA = CB, A(1, 4), B(9, 4), Altitude from C is 6

a) Find the coordinates of the mid point of AB
b) Find the coordinates of C
c) Find the coordinates of the centriod of the triangle.

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Answers

a) M (5, 4)is the mid point of AB

b) C(5, 10)

c) Since G(x, y)then x = 5, y = y1 + m (y2 − y1)
m+n

y = 10 + 2 (4 − 10) = 10 − 4 = 6
2+1

G(5, 6)

5) There is a line AB joining (1, 6) and B(5, 2).Find the coordinates of the points which divides AB into three
equal parts.

Answers

⋆ AP = P Q = QB, AP : P Q : QB = 1 : 1 : 1

AP : PB = 1 : 2, P (x, y),x = x1 + m (x2 − x1)
m+n

x = 1+ 1 (5 − 1), x = 1+ 4 = 7
1+2 3 3

y = 6 + 1 (2 − 6) = 6 + 1 × −4 = 14
1+2 3 3

P ( 7 , 14 )
3 3

⋆ AQ : QB = 2 : 1, Q(x, y), , x = 11 , y = 10
3 3
11 10
Q( 3 , 3 )

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2020-21 Academic year Worksheets

Mathematics X
Geometry and Algebra

Concepts

⋆ The line joining vertex to the mid point of the opposite side is called median.There are three medians
in a triangle.

⋆ The point of intersection of the medians is called centroid of the triangle.It is denoted by G.
⋆ The centroid of a triangle divides the median in the ratio 2 : 1 from the vertex.

1) ABCD is a parallelogram.If A(1, 2), B(4, y), C(x, 6), D(3, 5)then

a) What is the x coordinate of the mid point of the diagonal BD.
b) Write the coordinates of C
c) Write the y coordinate of the diagonal AC
d) Write the coordiantes of B.

Answers

a) Diagonals bisect eachother. x coordinate of the mid point of BD is 4+3 = 7
2 2

b) 1+x = 7 , 1 + x = 7, x = 6 C(6, 6)
2 2

c) y coordinate of the mid point of AC is 2+6 =4
2

d) 5+y = 4, 5 + y = 8, y = 3 B(4, 3)
2

2) The centre of a circle is (2, −3),AB is the diametre of the circle.If B(4, −3) then

a) What is the radius of the circle?
b) Write the coordinates of the end A of the diametre
c) CD is another diametre perpendicular to AB. Write the coordiantes of C and D
d) What is the area of ACBD?

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Answers

a) AB is parallel to x axis. Radius OB =| 4 − 2 |= 2

b) A(2 − 2, −3) → A(0, −3)

c) CD is parallel to y axis . C(2, −3 + 2) → C(2, −1)
D(2, −3 − 2) → D(2, −5)

d) Area of ABCD = d2 = 42 = 8 sq.unit
2 2

3) In the line AB, A(−2, −2), B(2, −4).P is a point in between A and B.

Length of AP is 3 of the length of AB
7

a) What is the ratio at which P divides AB

b) Write the coordinates of P

c) Find the coordinates of the mid point of AB
begintcolorbox[colback=blue!5!white,colframe=blue!75!black,title=Answers]

a) AP : BP = 3; 4

b) P (x, y) is the point. x = x1 + m (x2 − x1)
m+n
−14+12 −2
x = −2 + 3 (2 −− 2), x = −2 + 3 ×4 = −2 + 12 = 7 = 7
3+4 7 7

y = y1 + m (y2 − y1), y = −2 + 3 (−4 −− 2) = − 20
m+n 7 7
−2 −20
P ( 7 , 7 )

c) Coordinates of the mid point = ( −2+2 , −2+−4 ) = (0, −3)
2 2

T

1. bo vertices of △ABCare on x axis .If A(1, 3)then

a) Write the coordiantes of B and C from the figure
b) Find the point at which the median from A intersect BC
c) Find the coordinates of the centroid of the triangle.

Answers

a) B(−1, 0), C(5, 0)

b) Mid point of BC is the point where the median intersect the side BC .M ( −1+5 , 0) =
2

M (2, 0)

c) Centroid G(x, y), x = x1 + m (x2 − x1)
m+n

x = 1 + 2 (2 − 1), x = 5
3 3
m − 2 × −3
y = y1 + m+n (y2 y1), y = 3 + 3 = 1

G( 5 , 1)
3

5) A(4, 2), B(6, 5), C(1, 4)are the vertices of a triangle. P is a point on AB, Q is a point on BC

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a) If AP : BP = 1 : 2then write the coordinates of P
b) If Q is the mid point of BC write the coordinates of Q
c) Join these points suitably to get three equal area triangles by dividing the area of triangle ABC.

Answers

a) P (x, y) , x = x1 + m (x2 − x1)
m+n

x = y41++13m×m+n(6(y−2 4) = 14
y = 3
− y1) , y 1 × − 14
= 2 + 3 (5 2), y = 3 , P ( 3 , 3)

b) Q( 6+1 , 5+4 ), Q( 7 , 9 )
2 2 2 2

c) Draw CP , Area of triangle CPA is 1 of area △AB C .
Area of △CP B is 3
2 area of △ABC.
3 of the

Draw P Q. Area of △CP Q and area of △BP Q are aequal .

Area of △CP Q is equal to area of △BP Q = 1 × area of △ABC .
2
1 × 2 1 × △ABC
= 2 3 area ABC = 3 area of .

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2020-21 Academic year Worksheets

Mathematics X
Geometry and Algebra

Concepts

⋆ The inclination of a line with x axis is very important for higher level learning coordinate geometry.
There is a slope to a line. It is the measure of inclination. But mere inclination cannot determine a
perticular line. There are more more one line with same slope.

⋆ Let A(x1, y1), B(x2, y2)be two lines. Slope of this line is .y2 −y1

x2 −x1

⋆ The tan measure of the angle made by a line with the positive direction of x axis is also the slope

of this line

1) In the figure you can see a horizontal plane and an inclined plane.A is walking from A to B along the inclined
plane .

a) What is the horizontal distance and vertical distance from A to B

b) If BC = 1 then what is DE
AC 2 AE

c) If AG = 20then what is F G?

d) Write your inference on this concept.

Answers

a) Horizotal diatance = AC, Vertical distance = BC

b) BC = 1 . Since DE = 1
AC 2 AE 2
(△ACB, △AED are similar )

c) FG = 1 , FG = 1 , FG = 10
AG 2 20 2

d) The vertical distance is proportional to the horizontal distance

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2) Three lines passes through a point on x axis .

a) What is the slope of the line AD?
b) What is the slope of the line AC?
c) What is the slope of the line AB?
d) Write the coordinates of two points on AC

Answers

a) Angle made by AD with x axis is 30◦. Slope of AD is = tan 30◦ = √1
3

b) Angle made by ACwith x axis is 30 + 15 = 45◦
Slop of AC is = tan 45◦ = 1

c) Angle omfaAdeBbyisA=Btwanith6x0◦a=xis√is360◦ .
Slope

d) A(1, 0) is a point on AC , slope = 1 . Joining these two cocepts we can write the points
.Make same change for x and y coordinates . (1, 0), (1 + 1, 0 + 1) → (2, 1) . We can
write many points like this .

3) Draw x axis and y axis (rough diagram), mark the points A(4, 3) and B(12, 7)

a) What is the slope of this line?
b) Write the coordinates of another point on this line?
c) How many lines are there having the same slop?

4) Consider the points A(2, 3), B(3, 4), C(4, 5)

a) Find the slope of the line passing through A(2, 3) and B(3, 4)
b) Find the slope of the line passing through B(3, 4) and C(4, 5)
c) Are these points on a line? How can we realize it.
d) Write the coordinates of one more point on the line?

Answers

a) Slope of AB is = 4−3 =1
3−2

b) Slope of BC is = 5−4 =1
4−3

c) Slope of AB and slope of BC are equal. Bis the common point . So A, B, C are on a line.

d) Find the linear relation between the coordinates of points. Using this relation we can write
points .

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5) Consider the points A(2, 0), B(−6, −2), C(−4, −4), D(4, −2)

a) Find the slope pf the lines AB and CD
b) Find the slope of the line AD and BC
c) Is ABCD a parallelogram ?Explain

Answers

a) Slope of AB is −2−0 = −2 = 1
−6−2 −8 4

Slope of CD is = −2−− 4 = 2 = 1
4−− 4 8 4

AB is parallel to CD ് സമാ രമാണ്

b) Slope of AD is = −2−0 = −2 = −1
4−2 2

Slope of BC is −4−− 2 = −2 = −1
−4−− 6 2

AD is parallel to BC

c) Since opposite sides are parallel ABCD is a parallelogram.

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