Mathematics Part ii Solutions Solutions for Class 10 Maths Chapter 7 Mensuration are provided here with simple step-by-step explanations. These solutions for Mensuration are extremely popular among class 10 students for Maths Mensuration Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part ii Solutions Book of class 10 Maths Chapter 7 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Mathematics Part ii Solutions Solutions. All Mathematics Part ii Solutions Solutions for class 10 Maths are prepared by experts and are 100% accurate.
Page No 145:
Question 1:
Answer:
Radius of the cone, r = 1.5 cm
Height of the cone, h = 5 cm
∴ Volume of the cone, V = 11.79 cm3
Thus, the volume of the cone is 11.79 cm3.
Page No 145:
Question 2:
Find the volume of a sphere of diameter 6 cm.
Answer:
Radius of the sphere, r = = 3 cm
∴ Volume of the sphere, V = = 113.04 cm3
Thus, the volume of sphere is 113.04 cm3.
Page No 145:
Question 3:
Answer:
Radius of the cylinder, r = 5 cm
Height of the cylinder, h = 40 cm
∴ Total surface area of cylinder, S = = 1413 cm2
Thus, the total surface area of cylinder is 1413 cm2.
Page No 145:
Question 4:
Find the surface area of a sphere of radius 7 cm.
Answer:
Radius of the sphere, r = 7 cm
∴ Surface area of the sphere, S = = 616 cm2
Thus, the surface area of sphere is 616 cm2.
Page No 145:
Question 5:
Answer:
The dimensions of the cuboid are 44 cm, 21 cm and 12 cm.
Let the radius of the cone be r cm.
Height of the cone, h = 24 cm
It is given that cuboid is melted to form a cone.
∴ Volume of metal in cone = Volume of metal in cuboid
Thus, the radius of the base of cone is 21 cm.
Page No 145:
Question 6:
Observe the measures of pots In the given figure. How many jugs of water can the cylindrical pot hold?
Answer:
Radius of conical water jug, r = 3.5 cm
Height of conical water jug, h = 10 cm
Radius of cylindrical pot, R = 7 cm
Height of cylindrical pot, H = 10 cm
Let n be the number of jugs of water that cylindrical pot can hold.
Thus, the cylindrical water pot can hold water of 12 conical water jugs.
Page No 145:
Question 7:
A cylinder and a cone have equal bases. The height of the cylinder is 3 cm and the area of its base is 100 cm2 .The cone is placed
upon the cylinder. Volume of the solid figure so formed is 500 cm3 . Find the total height of the figure.
Answer:
Height of the cylinder, h = 3 cm
Let the radius of the cylinder be r cm and the height of the cone be H cm.
Area of the base of cylinder = 100 cm2
.....(1)
Volume of the solid figure = 500 cm3
∴ Volume of the cylinder + Volume of the cone = 500 cm3
∴ Total height of the figure = h + H = 3 + 6 = 9 cm
Thus, the total height of the figure is 9 cm.
Page No 145:
Question 8:
In the given figure, a toy made from a hemisphere, a cylinder and a cone is shown. Find the total area of the toy.
Answer:
Radius of the sphere = Radius of the cylinder = Radius of cone = r = 3 cm
Height of the cone, h = 4 cm
Height of the cylinder, H = 40 cm
Let the slant height of the cone be l cm.
Total area of the toy
= Curved surface area of hemisphere + Curved surface area of cylinder + Curved surface area of cone
Thus, the total area of the toy is 273 cm2.
Page No 145:
Question 9:
In the given figure, a cylindrical wrapper of flat tablets is shown. The radius of a tablet is 7 mm and its thickness is 5 mm. How many such tablets are wrapped in the wrapper?
Answer:
Radius of a tablet, r = 7 mm
Thickness of a tablet, h = 5 mm
Radius of cylindrical wrapper, R = = 7 mm
Height of cylindrical wrapper, H = 10 cm = 100 mm (1 cm = 10 mm)
Let n be the number of tablets wrapped in the wrapper.
Thus, 20 tablets are wrapped in the wrapper.
Page No 145:
Question 10:
In the given figure shows a toy. Its lower part is a hemisphere and the upper part is a cone. Find the volume and surface area of the toy from the measures shown in the figure ()
Answer:
Radius of the hemisphere = Radius of the cone = r = 3 cm
Height of the cone, h = 4 cm
Let l be the slant height of the cone.
Volume of the toy = Volume of the hemisphere + Volume of the cone
Surface area of the toy = Curved surface area of cone + Curved surface area of hemisphere
Thus, the volume and surface area of the toy are 94.20 cm3 and 103.62 cm2, respectively.
Page No 146:
Question 11:
Find the surface area and the volume of a beach ball shown in the figure
Answer:
Radius of the beach ball, r = = 21 cm
Surface area of the beach ball, S = = 5538.96 cm2
Volume of the beach ball, V = = 38772.72 cm3
Thus, the surface area and volume of the beach ball are 5538.96 cm2 and 38772.72 cm3, respectively.
Page No 146:
Question 12:
As shown in the figure, a cylindrical glass contains water. A metal sphere of diameter 2 cm is immersed in it. Find the volume of the water .
Answer:
Radius of the metallic sphere, r = = 1 cm
Radius of the cylindrical glass, R = = 7 cm
Height of water in cylindrical glass, H = 30 cm
∴ Volume of the water = Volume of water in the cylindrical glass − Volume of the metallic sphere
Thus, the volume of the water is 1468.67 cm3.
Page No 148:
Question 1:
Answer:
Radius of one circular end, r1 = 14 cm
Radius of other circular end, r2 = 7 cm
Height of the bucket, h = 30 cm
∴ Volume of water in the bucket = Volume of frustum of cone
Thus, the bucket can hold 10.780 litres of water.
Page No 148:
Question 2:
iii ) volume
Answer:
Here, r1 = 14 cm, r2 = 6 cm and h = 6 cm.
Slant height of the frustum, l = = 10 cm
i)
Curved surface area of frustrum
ii)
Total surface area of frustrum
iii)
Volume of the frustum
Page No 148:
Question 3:
The circumferences of circular faces of a frustum are 132 cm and 88 cm and its height is 24 cm. To find the curved surface area of the frustum complete the following activity.()
circumference1 = 2r1 = 132
r1 =
circumference2 =2r2= 88
r2=
slant height of frustum, l =
curved surface area of the frustum = (r1 + r2 ) l
=
Answer:
Circumference1 = 2r1 = 132
r1 =
Circumference2 = 2r2 = 88
r2 =
Slant height of frustum,
Curved surface area of the frustum = (r1 + r2 ) l
Page No 154:
Question 1:
Answer:
Radius of the circle, r = 10 cm
Measure of the arc, θ = 54º
∴ Area of the sector = = 47.1 cm2
Thus, the area of the sector is 47.1 cm2.
Page No 154:
Question 2:
Answer:
Radius of the arc, r = 18 cm
Measure of the arc, θ = 80º
∴ Length of the arc = = 25.12 cm
Thus, the length of the arc is 25.12 cm.
Page No 154:
Question 3:
Answer:
Radius of the sector, r = 3.5 cm
Length of the arc, l = 2.2 cm
∴ Area of the sector = = 3.85 cm2
Thus, the area of the sector is 3.85 cm2.
Page No 154:
Question 4:
Answer:
Radius of the circle, r = 10 cm
Area of the sector = 100 cm2
∴ Area of the corresponding major sector = Area of the circle − Area of the sector
Thus, the area of the corresponding major sector is 214 cm2.
Page No 154:
Question 5:
Answer:
Radius of the sector, r = 15 cm
Area of the sector = 30 cm2
∴ Length of the arc = = 4 cm (Area of the sector = × Length of the arc × Radius)
Thus, the length of the arc of the sector is 4 cm.
Page No 154:
Question 6:
find (1) Area of the circle .
Answer:
Radius of the circle, r = 7 cm
m(arc MBN) = ∠MON = θ = 60º
(1)
Area of the circle = = 154 cm2
(2)
A(O-MBN) = Area of the sector OMBN = = 25.7 cm2
(3)
A(O-MCN) = Area of the sector OMCN
= Area of the circle − Area of the sector OMBN
= 154 − 25.7
= 128.3 cm2
Page No 155:
Question 7:
In the given figure, radius of circle is 3.4 cm and perimeter of sector P-ABC is 12.8 cm . Find A(P-ABC).
Answer:
Radius of the circle, r = 3.4 cm
Perimeter of sector P-ABC = 12.8 cm
Let l be the length of the arc ABC.
∴ l + 2r = 12.8 cm
⇒ l + 2 × 3.4 = 12.8
⇒ l = 12.8 − 6.8 = 6 cm
∴ A(P-ABC) = Area of the sector PABC = = 10.2 cm2
Thus, A(P-ABC) is 10.2 cm2.
Page No 155:
Question 8:
In the given figure, O is the centre of the sector. ROQ = MON = 60° . OR = 7 cm, and OM = 21 cm. Find the lengths of arc RXQ and arc MYN. ( )
Answer:
In the given figure, ∠ROQ = ∠MON = θ = 60º
Radius of the sector ORXQ = OR = 7 cm
∴ Length of the arc RXQ = = 7.3 cm
Radius of the sector OMYN = OM = 21 cm
∴ Length of the arc MYN = = 22 cm
Thus, the lengths of the arc RXQ and arc MYN are 7.3 cm and 22 cm, respectively.
Page No 155:
Question 9:
In the given figure, if A(PABC) = 154 cm2 radius of the circle is 14 cm,
(1) APC.
Answer:
Radius of the circle, r = 14 cm
(1)
A(PABC) = Area of the sector PABC = 154 cm2
Thus, the measure of APC is 90º.
(2)
l(arc ABC) = Length of the arc ABC = 22 cm
Thus, the length of arc ABC is 22 cm.
Page No 155:
Question 10:
Answer:
Radius of the sector of the circle, r = 7 cm
(1)
Measure of arc of the sector = θ = 30º
∴ Area of the sector = = 12.83 cm2
(2)
Measure of arc of the sector = θ = 210º
∴ Area of the sector = = 89.83 cm2
(3)
Measure of arc of the sector = θ = 3 × 90º = 270º
∴ Area of the sector = = 115.5 cm2
Page No 155:
Question 11:
Answer:
Area of minor sector of the circle = 3.85 cm2
Measure of central angle, θ = 36º
Let the radius of the circle be r cm.
Now,
Area of minor sector = 3.85 cm2
Thus, the radius of the circle is 3.5 cm.
Page No 155:
Question 12:
In the given figure, PQRS is a rectangle. If PQ = 14 cm, QR = 21 cm, find the areas of the parts x, y and z .
Answer:
PQRS is a rectangle.
∴ ∠Q = ∠R = 90º
Radius of sector PTQ = PQ = 14 cm
∴ Area of part x = Area of the sector PTQ = = 154 cm2
Radius of sector TUR = TR = QR − QT = QR − PQ = 21 − 14 = 7 cm (QT = PQ)
∴ Area of part y = Area of the sector TUR = = 38.5 cm2
Now,
Area of rectangle PQRS = QR × PQ = 21 × 14 = 294 cm2
∴ Area of part z = Area of rectangle PQRS − Area of part x − Area of part y = 294 − 154 − 38.5 = 101.5 cm2
Page No 155:
Question 13:
Answer:
âLMN is an equilateral triangle.
∴ LM = MN = LN = 14 cm
∠L = ∠M = ∠N = 90º
(1)
Area of âLMN = = 84.87 cm2
(2)
Radius of the each sector, r = 7 cm
Area of any one of the sectors = = 25.67 cm2
(3)
Total area of all the three sectors = 3 × Area of any one of the sectors = 3 × 25.67 = 77.01 cm2
(4)
Area of the shaded region = Area of âLMN − Total area of all the three sectors = 84.87 − 77.01 = 7.86 cm2
Page No 159:
Question 1:
In the given figure, A is the centre of the circle. ABC = 45° and AC = 7cm. Find the area of segment BXC.
Answer:
In âABC,
AB = AC (Radii of the circle)
∴ ∠ACB = ∠ABC = 45º (Equal sides have equal angles opposite to them)
Using angle sum property, we have
∠ACB + ∠ABC + ∠BAC = 180º
∴ 45º + 45º + ∠BAC = 180º
⇒ ∠BAC = 180º − 90º = 90º
Here,
Radius of the circle, r = cm
Measure of arc BXC, θ = 90º
∴ Area of segment BXC
Thus, the area of the segment BXC is 28 cm2.
Page No 159:
Question 2:
Answer:
Radius of the circle, r = 10 cm
m(arc PQR) = ∠POR = θ = 60º
∴ Area of the shaded region = Area of segment PQR
Thus, the area of the shaded region is 9.08 cm2.
Page No 160:
Question 3:
In the given figure, if A is the centre of the circle. PAR = 30°, AP = 7.5, find the area of the segment PQR ( = 3.14)
Answer:
Radius of the circle, r = 7.5 cm
∠âPAR = θ = 30º
∴ Area of segment PQR
Thus, the area of the segment PQR is 0.65625 cm2.
Page No 160:
Question 4:
In the given figure, if O is the centre of the circle, PQ is a chord. POQ = 90°, area of shaded region is 114 cm2 , find the radius of the circle.
( = 3.14)
Answer:
∠POQ = θ = 90º
Let the radius of the circle be r cm.
Area of the shaded region = Area of the segment PRQ = 114 cm2
Thus, the radius of the circle is 20 cm.
Page No 160:
Question 5:
Answer:
Radius of the circle, r = 15 cm
Let O be the centre and PQ be the chord of the circle.
∠POQ = θ = 60º
Area of the minor segment = Area of the shaded region
Now,
Area of the circle = = 706.5 cm2
∴ Area of the major segment = Area of the circle − Area of the minor segment = 706.5 − 20.44 = 686.06 cm2
Thus, the areas of the minor segment and major segment are 20.44 cm2 and 686.06 cm2, respectively.
Page No 160:
Question 1:
(1) The ratio of circumference and area of a circle is 2:7. Find its circumference.
(A) (B) (C) 7 (D)
(3) Find the perimeter of a sector of a circle if its measure is 90° and radius is 7 cm .
(4) Find the curved surface area of a cone of radius 7 cm and height 24 cm.
(5) The curved surface area of a cylinder is 440 cm 2 and its radius is 5 cm. Find its height.
(A) cm (B) 22cm (C) 44cm (D) cm
(8) Find the side of a cube of volume 1 m3 .
Answer:
(1)
Let r be the radius of the circle.
Circumference of the circle : Area of the circle = 2 : 7
∴ Circumference of the circle =
Hence, the correct answer is option (A).
(2)
Measure of arc of circle, θ = 160º
Let r be the radius of the circle.
Length of the arc = 44 cm
Thus, the circumference of the circle is 99 cm.
Hence, the correct answer is option (D).
(3)
Here, r = 7 cm and θ = 90º.
∴ Perimeter of the sector = = 11 + 14 = 25 cm
Hence, the correct answer is option (B).
(4)
Radius of the cone, r = 7 cm
Height of the cone, h = 24 cm
Let l be the slant height of the cone.
= 25 cm
∴ Curved surface area of the cone = = 550 cm2
Hence, the correct answer is option (B).
(5)
Radius of the cylinder, r = 5 cm
Let the height of the cylinder be h cm.
Curved surface area of the cylinder = 440 cm2
Thus, the height of the cylinder is cm.
Hence, the correct answer is option (A).
(6)
Radius of the cone = Radius of the cylinder = r cm (Say)
Height of the cylinder, h = 5 cm
Let the height of the cone be H cm.
It is given that the cone melted and recasted into a cylinder.
∴ Volume of the cone = Volume of the cylinder
Thus, the height of the cone is 15 cm.
Hence, the correct answer is option (A).
(7)
Side of the cube = 0.01 cm
∴ Volume of the cube = (Side)3 = (0.01 cm)3 = 0.000001 cm3
Hence, the correct answer is option (D).
(8)
Volume of the cube = 1 m3
∴ (Side)3 = 1 m3 = (1 m)3
⇒ Side = 1 m = 100 cm
Thus, the side of the cube is 100 cm.
Hence, the correct answer is option (C).
Page No 161:
Question 2:
Answer:
Radius of the circular top of washing tub, r1 = 20 cm
Radius of the circular bottom of washing tub, r2 = 15 cm
Height of the washing tub, h = 21 cm
∴ Capacity of the washing tub = Volume of frustum of cone
Thus, the capacity of the tub is 20.35 litres.
Page No 161:
Question 3:
Answer:
Radius of each plastic ball, R = 1 cm
Outer radius of the tube, r2 = 30 cm
Thickness of the tube = 2 cm
∴ Inner radius of the tube, r1 = Outer radius of the tube − Thickness of the tube = 30 − 2 = 28 cm
Length of the tube, h = 90 cm
Let the number of plastic balls melted to make the tube be n.
It given that plastic balls are melted to form the tube.
∴ n × Volume of each plastic ball = Volume of the tube
Thus, the number of plastic balls melted to make the tube are 7830.
Page No 161:
Question 4:
Answer:
Radius of each coin, r = = 1 cm
Thickness of each coin, h = 2 mm = = 0.2 cm (1 cm = 10 mm)
Let the number of coins made be n.
It is given that a metal parallelopiped is melted to make the coins.
∴ n × Volume of metal in each coin = Volume of the metal parallelopiped
Thus, the number of coins made are 2800.
Page No 161:
Question 5:
Answer:
Radius of the roller, r = = 60 cm
Length of the roller, h = 84 cm
Area of the ground levelled in one rotation of the roller = Curved surface area of the roller = = 31680 cm2
∴ Area of the ground levelled in 200 rotations of the roller
= 200 × Area of the ground levelled in one rotation of the roller
= 200 × 31680
= 6336000 cm2
= (1 m2 = 10000 cm2)
= 633.6 m2
Rate to level the ground = Rs 10/m2
∴ Expenditure (or total cost) to level the ground
= Total area of the ground × Rate to level the ground
= Area of the ground levelled in 200 rotations of the roller × Rate to level the ground
= 633.6 × 10
= Rs 6,336
Thus, the expenditure to level the ground is Rs 6,336.
Page No 161:
Question 6:
Answer:
Outer radius of the sphere, R = = 6 cm
Thickness of the sphere = 0.01 m = 0.01 × 100 cm = 1 cm (1 m = 100 cm)
∴ Inner radius of the sphere, r = R − 1 = 6 − 1 = 5 cm
Outer surface area of the sphere = = 452.16 cm2
Volume of metal in the sphere = = 380.97 cm3
Density of the metal = 8.88 g/cm3
∴ Mass of the sphere = Volume of metal in the sphere × Density of the metal = 380.97 × 8.88 = 3383.01 g
Thus, the outer surface area and mass of the sphere are 452.16 cm2 and 3383.01 g, respectively.
Page No 161:
Question 7:
A cylindrical bucket of diameter 28 cm and height 20 cm was full of sand. When the sand in the bucket was poured on the ground, the sand got converted into a shape of a cone. If the height of the cone was 14 cm, what was the base area of the cone ?
Answer:
Radius of the bucket, r = = 14 cm
Height of the bucket, h = 20 cm
Height of the cone, H = 14 cm
Let the radius of the base of the cone be R cm.
∴ Area of the base of the cone = R2
Now,
Volume of sand in the cone = Volume of sand in the cylindrical bucket
Thus, the base area of the cone is 2640 cm2.
Page No 161:
Question 8:
Answer:
Radius of the sphere, R = 9 cm
Radius of the wire, r = = 2 mm = = 0.2 cm (1 cm = 10 mm)
Let the length of the wire be l cm.
It is given that the metallic sphere is melted to make the wire.
∴ Volume of metal in the wire = Volume of the metallic sphere
Thus, the length of the wire is 243 m.
Page No 161:
Question 9:
Answer:
Radius of the sector, r = 6 cm
Let the measure of the arc be θ and the length of the arc corresponding to the sector be l cm.
Area of the sector = 15 cm2 (Given)
Length of the arc =
Thus, the measure of the arc and length of the arc corresponding to the sector are 150º and 5 cm, respectively.
Page No 161:
Question 10:
Answer:
Draw PQ ⊥ AB.
∴ AQ = QB (Perpendicular from the centre of the circle to the chord bisects the chord)
In right âAPQ,
∴ AB = 2AQ =
Also,
Similarly,
∴ ∠APB = ∠APQ + ∠BPQ = 60º + 60º = 120º
Radius of the circle, r = 8 cm
Measure of arc AB, θ = 120º
∴ Area of the shaded portion = Area of the sector ABP − Area of âAPB
Thus, the area of the shaded portion is 39.29 cm2.
Page No 162:
Question 11:
In the given figure, square ABCD is inscribed in the sector A - PCQ. The radius of sector C - BXD is 20 cm. Complete the following activity to find the area of shaded region
Answer:
Side of square ABCD = radius of sector C-BXD = cm
Area of square = (side)2 = = cm2
Area of shaded region inside the square
= Area of square ABCD − Area of sector C-BXD
=
=
= − 314
= cm2
Radius of bigger sector = Length of diagonal of square ABCD = cm
Area of the shaded regions outside the square
= Area of sector A-PCQ − Area of square ABCD
= A(A-PCQ) − A( ABCD)
= −
= −
= −
= cm2
∴ Total surface area of the shaded region = 86 + 228 = 314 cm2
Page No 163:
Question 12:
In the given figure , two circles with centres O and P are touching internally at point A. If BQ = 9, DE = 5, complete the following activity to find the radii of the circles.
Answer:
Let the radius of the bigger circle be R and that of smaller circle be r.
OA, OB, OC and OD are the radii of the bigger circle.
∴ OA = OB = OC = OD = R
PQ = PA = r
OQ = OB − BQ =
OE = OD − DE =
As the chords QA and EF of the circle with centre P intersect in the interior of the circle, so by the property of internal division of two chords of a circle,
OQ × OA = OE × OF
× R = × .....(âµ OE = OF)
R2 − 9R = R2 − 10R + 25
R =
AQ = 2r = AB − BQ (âµAB = 2R)
2r = 50 − 9 = 41
r = =
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