# SOLVING WORD PROBLEMS MENSURATION FOR GRADE 10

Solving Word Problems Mensuration for Grade 10 :

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

## Solving Word Problems Mensuration for Grade 10 - Questions

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 cm2

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/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(R3 - r3)

(2/3) π(R3 - r3)  =  436π/

(2/3) (73 - r3)  =  436/3

(343 - r3)  =  (436/3) ⋅ (3/2)

343 - r3  =  218

r3  =  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 r2  =  1408/7

(22/7) r2  =  1408/7

r2  =  (1408/7) (7/22)

r2  =  64

r  =  8

Volume of a cone  =  1005  5/7

(1/3) π r2  =  7040/7

(1/3) (22/7) 8 =  7040/7

h  =  (7040/7) (7/22) ⋅ 3 (1/64)

h  =  15

Slant height of the cone  =  √82 + 152

=  √(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  =  √l2 - r2

h  =  √212 - (12.6)2

h  =  √441 - 158.76

h  =  √282.24

h  =  16.8

Volume of cone  =  (1/3) π r2 h

=  (1/3)  (22/7) 12.6 (12.6) (16.8)

=  2794.18 cm3 After having gone through the stuff given above, we hope that the students would have understood, "Solving Word Problems Mensuration for Grade 10".

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