## Maharashtra State Board Class 10 Maths Solutions Chapter 7 Mensuration Problem Set 7

Problem Set 7 Question 1. Choose the correct alternative answer for each of the following questions.

i. The ratio of circumference and area of a circle is 2 : 7. Find its circumference.

(A) 14 π

(B) \(\frac{7}{\pi}\)

(C) 7π

(D) \(\frac{14}{\pi}\)

Answer:

(A)

ii. If measure of an arc of a circle is 160° and its length is 44 cm, find the circumference of the circle.

(A) 66 cm

(B) 44 cm

(C) 160 cm

(D) 99 cm

Answer:

(D)

iii. Find the perimeter of a sector of a circle if its measure is 90° and radius is 7 cm.

(A) 44 cm

(B) 25 cm

(C) 36 cm

(D) 56 cm

Answer:

(B)

iv. Find the curved surface area of a cone of radius 7 cm and height 24 cm.

(A) 440 cm^{2}

(B) 550 cm^{2}

(C) 330 cm^{2}

(D) 110 cm^{2}

Answer:

(B)

v. The curved surface area of a cylinder is 440 cm^{2} and its radius is 5 cm. Find its height.

(A) \(\frac{44}{\pi}\) cm

(B) 22π cm

(C) 44π cm

(D) \(\frac{22}{\pi}\)

Answer:

(A)

vi. A cone was melted and cast into a cylinder of the same radius as that of the base of the cone. If the height of the cylinder is 5 cm, find the height of the cone.

(A) 15 cm

(B) 10 cm

(C) 18 cm

(D) 5 cm

Answer:

(A)

vii. Find the volume of a cube of side 0.01 cm.

(A) 1 cm

(B) 0.001 cm^{3}

(C) 0.0001 cm^{3}

(D) 0.000001 cm^{3}

Answer:

Volume of cube = (side)^{3}

= (0.01)^{3} = 0.000001 cm^{3}

(D)

viii. Find the side of a cube of volume 1 m^{3}

(A) 1 cm

(B) 10 cm

(C) 100 cm

(D) 1000 cm

Answer:

Volume of cube = (side)^{3}

∴ 1 = (side)^{3}

∴ Side = 1 m

= 100 cm

(C)

Problem Set 7 Geometry Class 10 Question 2. A washing tub in the shape of a frustum of a cone has height 21 cm. The radii of the circular top and bottom are 20 cm and 15 cm respectively. What is the capacity of the tub? = (π = \(\frac { 22 }{ 7 } \))

Given: For the frustum shaped tub,

height (h) = 21 cm,

radii (r_{1}) = 20 cm, and (r_{2}) = 15 cm

To find: Capacity (volume) of the tub.

Solution:

Volume of frustum = \(\frac { 1 }{ 3 } \) πh (r_{1}^{2} + r_{2}^{2} + r_{1} × r_{2})

∴ The capacity of the tub is 20.35 litres.

10th Geometry Problem Set 7 Question 3. Some plastic balls of radius 1 cm were melted and cast into a tube. The thickness, length and outer radius of the tube were 2 cm, 90 cm and 30 cm respectively. How many balls were melted to make the tube?

Given: For the cylindrical tube,

height (h) = 90 cm,

outer radius (R) = 30 cm,

thickness = 2 cm

For the plastic spherical ball,

radius (r_{1}) = 1 cm

To find: Number of balls melted.

Solution:

Inner radius of tube (r)

= outer radius – thickness of tube

= 30 – 2

= 28 cm

Volume of plastic required for the tube = Outer volume of tube – Inner volume of hollow tube

= πR^{2}h – πr^{2}h

= πh(R^{2} – r^{2})

= π × 90 (30^{2} – 28^{2})

= π × 90 (30 + 28) (30 – 28) …[∵ a^{2} – b^{2} = (a + b)(a – b)]

= 90 × 58 × 2π cm^{3
}

∴ 7830 plastic balls were melted to make the tube.

Problem Set 7 Geometry Question 4.

A metal parallelopiped of measures 16 cm × 11cm × 10cm was melted to make coins. How many coins were made if the thickness and diameter of each coin was 2 mm and 2 cm respectively?

Given: For the parallelopiped.,

length (l) = 16 cm, breadth (b) = 11 cm,

height (h) = 10 cm

For the cylindrical coin,

thickness (H) = 2 mm,

diameter (D) 2 cm

To find: Number of coins made.

Solution:

Volume of parallelopiped = l × b × h

= 16 × 11 × 10

= 1760 cm^{3}

Thickness of coin (H) = 2 mm

= 0.2 cm …[∵ 1 cm = 10 mm]

Diameter of coin (D) = 2 cm

∴ 2800 coins were made by melting the parallelopiped.

Mensuration Problem Question 5. The diameter and length of a roller is 120 cm and 84 cm respectively. To level the ground, 200 rotations of the roller are required. Find the expenditure to level the ground at the rate of ₹ 10 per sq.m.

Given: For the cylindrical roller,

diameter (d) =120 cm,

length = height (h) = 84 cm

To find: Expenditure of levelling the ground.

Solution:

Diameter of roller (d) = 120 cm

Now, area of ground levelled in one rotation = curved surface area of roller

= 3.168 m^{2}

∴ Area of ground levelled in 200 rotations

= 3.168 × 200 =

633.6 m^{2}

Rate of levelling = ₹ 10 per m^{2}

∴ Expenditure of levelling the ground

= 633.6 × 10 = ₹ 6336

∴ The expenditure of levelling the ground is ₹ 6336.

Question 6.

The diameter and thickness of a hollow metal sphere are 12 cm and 0.01 m respectively. The density of the metal is 8.88 gm per cm3. Find the outer surface area and mass of the sphere, [π = 3.14]

Given: For the hollow sphere,

diameter (D) =12 cm, thickness = 0.01 m

density of the metal = 8.88 gm per cm^{3}

To find: i. Outer surface area of the sphere

ii. Mass of the sphere.

Solution:

Diameter of the sphere (D)

= 12 cm

∴ Radius of sphere (R)

= \(\frac { d }{ 2 } \) = \(\frac { 12 }{ 2 } \) = 6 cm

∴ Surface area of sphere = 4πR^{2}

= 4 × 3.14 × 6^{2}

= 452.16 cm^{2}

Thickness of sphere = 0.01 m

= 0.01 × 100 cm …[∵ 1 m = 100 cm]

= 1 cm

∴ Inner radius of the sphere (r)

= Outer radius – thickness of sphere

= 6 – 1 = 5 cm

∴ Volume of hollow sphere

= Volume of outer sphere – Volume of inner sphere

∴ The outer surface area and the mass of the sphere are 452.16 cm^{2} and 3383.19 gm respectively.

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?

Given: For the cylindrical bucket,

diameter (d) = 28 cm, height (h) = 20 cm

For the conical heap of sand,

height (H) = 14 cm

To find: Base area of the cone (πR^{2}).

Solution:

Diameter of the bucket (d) = 28 cm

The base area of the cone is 2640 cm^{2}.

Question 8.

The radius of a metallic sphere is 9 cm. It was melted to make a wire of diameter 4 mm. Find the length of the wire.

Given: For metallic sphere,

radius (R) = 9 cm

For the cylindrical wire,

diameter (d) = 4 mm

To find: Length of wire (h).

Solution:

∴ The length of the wire is 243 m.

Question 9.

The area of a sector of a circle of 6 cm radius is 157t sq.cm. Find the measure of the arc and length of the arc corresponding to the sector.

Given: Radius (r) = 6 cm,

area of sector = 15 π cm^{2}

To find: i. Measure of the arc (θ),

ii. Length of the arc (l)

Solution:

∴ The measure of the arc and the length of the arc are 150° and 5π cm respectively.

Question 10.

In the adjoining figure, seg AB is a chord of a circle with centre P. If PA = 8 cm and distance of chord AB from the centre P is 4 cm, find the area of the shaded portion.

(π = 3.14, \(\sqrt { 3 }\) = 1.73)

Given: Radius (r) = PA = 8 cm,

PC = 4 cm

To find: Area of shaded region.

Solution:

Similarly, we can show that, ∠BPC = 60°

∠APB = ∠APC + ∠BPC …[Angle sum property]

∴ θ = 60° + 60° = 120°

Area of shaded region = A(P-ADB) – A(∆APB)

= 66.98 – 27.68

= 39.30 cm^{2}

∴ The area of the shaded region is 39.30 cm^{2}.

Question 11.

In the adjoining 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.

Solution:

Side of square ABCD

= radius of sector C-BXD = [20] cm

Area of square = (side)^{2} = 20^{2} = 400 cm^{2} ….(i)

Area of shaded region inside the square = Area of square ABCD – Area of sector C-BXD

Radius of bigger sector

= Length of diagonal of square ABCD

= \(\sqrt { 2 }\) × side

= 20 \(\sqrt { 2 }\) cm

Area of the shaded regions outside the square

= Area of sector A-PCQ – Area of square ABCD

= A(A – PCQ) – A(꠸ABCD)

Alternate method:

□ABCD is a square. … [Given]

Side of □ABCD = radius of sector (C-BXD)

= 20 cm

Radius of sector (A-PCQ) = Diagonal

= \(\sqrt { 2 }\) × side

= \(\sqrt { 2 }\) × 20

= 20 \(\sqrt { 2 }\) cm

Now, Area of shaded region

= A(A-PCQ) – A(C-BXD)

= 628 – 314

= 314 cm^{2}

∴ The area of the shaded region is 314 cm^{2}.

Question 12.

In the adjoining 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.

Solution:

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 + BQ = OB … [B – Q – O]

OQ = OB – BQ = R – 9

OE + DE = OD ….[D – E – O]

OE = OD – DE = [R – 5]

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 – 9) × R = (R – 5) × (R – 5) …[∵ OE = OF]

∴ R^{2} – 9R = R^{2} – 10R + 25

∴ -9R + 10R = 25

∴ R = [25units]

AQ = AB – BQ = 2r ….[B-Q-A]

∴ 2r = 50 – 9 = 41

∴ r = \(\frac { 41 }{ 2 } \) = 20.5 units