**Solving Word Problems Mensuration for Grade 10 :**

Here we are going to see an example problem on the topic mensuration.

**Question 1 :**

The slant height of a frustum of a cone is 4 m and the perimeter of circular ends are 18 m and 16 m. Find the cost of painting its curved surface area at ₹100 per sq. m.

**Solution :**

Perimeter of the top of the frustum cone = 2πR

2πR = 18

2^{ }⋅ (22/7)^{ }⋅ R = 18

R = 18 ⋅ (7/22) ⋅ (1/2)

R = 63/22

Perimeter of the bottom of the frustum cone = 2πr

2πr = 16

2^{ }⋅ (22/7)^{ }⋅ r = 16

r = 16 ⋅ (7/22) ⋅ (1/2)

r = 56/22

Slant height of frustum cone (l) = 4 cm

Curved surface area = π(R + r)l

= (22/7) [(63/22) + (56/22)] 4

= (22/7) [(63 + 56)/22] 4

= 119(4)/7

= 68 cm^{2}

cost of painting its curved surface area at ₹100 per sq. m

Required cost = 68 (100)

= ₹ 6800

**Question 2 :**

A hemi-spherical hollow bowl has material of volume 436π/3 cubic cm. Its external diameter is 14 cm. Find its thickness.

**Solution :**

External radius (R) = 14/2 = 7 cm

Volume of hemispherical bowl = (2/3)π(R^{3} - r^{3})

(2/3) π(R^{3} - r^{3}) = 436π/3

(2/3) (7^{3} - r^{3}) = 436/3

(343 - r^{3}) = (436/3)^{ }⋅ (3/2)

343 - r^{3} = 218

r^{3} = 125

r = 5

Thickness = R - r

= 7 - 5

= 2 cm

**Question 3 :**

The volume of a cone is 1005 5/7 cu. cm. The area of its base is 201 1/7 sq. cm. Find the slant height of the cone.

**Solution :**

Area of base of cone = 201 1/7

π r^{2} = 1408/7

(22/7) r^{2} = 1408/7

r^{2} = (1408/7) (7/22)

r^{2} = 64

r = 8

Volume of a cone = 1005 5/7

(1/3) π r^{2 }h = 7040/7

(1/3) (22/7) 8^{2 }h = 7040/7

h = (7040/7) (7/22) ⋅ 3 (1/64)

h = 15

Slant height of the cone = √8^{2} + 15^{2}

= √(64 + 225)

= √289

= 17 cm

**Question 4 :**

A metallic sheet in the form of a sector of a circle of radius 21 cm has central angle of 216°. The sector is made into a cone by bringing the bounding radii together. Find the volume of the cone formed.

**Solution :**

Since the sector is made into a cone by bringing the bounding radii together,

length of arc = Perimeter of the base of the cone

length of arc = (θ/360) ⋅ 2 π r

2π r = (216/360) ⋅ 2 ⋅ π ⋅ 21

r = 12.6

Radius of sector = Slant height of cone

l = 21 cm

h = √l^{2} - r^{2}

h = √21^{2} - (12.6)^{2}

h = √441 - 158.76

h = √282.24

h = 16.8

Volume of cone = (1/3) π r^{2} h

= (1/3) (22/7) 12.6 (12.6) (16.8)

= 2794.18 cm^{3}

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