**Chapter 7 Mensuration Set 7**

## Chapter 7 Mensuration 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.**

Answer:

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 cmAnswer:**

(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 cmAnswer:**

(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.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 cmAnswer:**

(A)

**vii. Find the volume of a cube of side 0.01 cm.(A) 1 cm(B) 0.001 cm**
Volume of cube = (side)

^{3}(C) 0.0001 cm

^{3}(D) 0.000001 cm

^{3}Answer:

^{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 cmAnswer:**

Volume of cube = (side)

^{3}

∴ 1 = (side)

^{3}

∴ Side = 1 m

= 100 cm

(C)

**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 = 22/7)Given: For the frustum shaped tub,height (h) = 21 cm,radii (r**

_{1}) = 20 cm, and (r

_{2}) = 15 cmTo find: Capacity (volume) of the tub.Solution:

Volume of frustum = 1/3 πh (r

_{1}

^{2}+ r

_{2}

^{2}+ r

_{1}× r

_{2})

∴ The capacity of the tub is 20.35 litres.

**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 cmFor the plastic spherical ball,radius (r**

_{1}) = 1 cmTo 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.

**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 cmFor the cylindrical coin,thickness (H) = 2 mm,diameter (D) 2 cmTo 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.

**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 cmTo 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 mdensity of the metal = 8.88 gm per cm**

^{3}To find: i. Outer surface area of the sphereii. Mass of the sphere.Solution:

Diameter of the sphere (D)

= 12 cm

∴ Radius of sphere (R)

∴ 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 cmFor the conical heap of sand,height (H) = 14 cmTo 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 cmFor the cylindrical wire,diameter (d) = 4 mmTo 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, √3 = 1.73)Given: Radius (r) = PA = 8 cm,PC = 4 cmTo 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

= √2 × side

= 20 √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

= √2 × side

= √2 × 20

= 20 √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 = 44/2 = 20.5 units