ML Aggarwal Solutions Class 10 Mathematics Solutions for Mensuration Exercise 17.4 in Chapter 17 - Mensuration

Question 22 Mensuration Exercise 17.4

A toy is in the shape of a right circular cylinder with a hemisphere on one end and a cone on the other. The height and radius of the cylindrical part are 13 cm and 5 cm respectively. The radii of the hemispherical and conical parts are the same as that of the cylindrical part. Calculate the surface area of the toy if the height of the conical part is 12 cm.



Since the given solid is a combination of a cone, cylinder, and hemisphere, we must sum the curved surface areas of the three solids to determine its surface area.

M L Aggarwal - Understanding ICSE Mathematics - Class 10 chapter Mensuration Question 22 Solution image

Given height of the cylinder, H = 13 cm

Radius of the cylinder, r = 5 cm

Radius of the hemisphere, r = 5 cm

Height of the cone, h = 12 cm

Radius of the cone, r = 5 cm

Slant height of the cone, l = √(h2+r2)

= √(12)2+(5)2)

= √(144+25)

= √169

= 13 cm

Surface area of the toy = curved surface area of cylinder +curved surface area of hemisphere+ curved surface area of cone

= 2π rH+2π r2+π rl

= π r(2H+2r+l)

= (22/7)×5(2×13+2×5+13)

= (110/7)×(26+10+13)

= (110/7)×49

= 110×7

= 770 cm2

Hence the surface area of the toy is 770 cm2.

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