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

Question 7 Mensuration Exercise 17.4

A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter that the hemisphere can have? Also, find the surface area of the solid.

Solution:

Here, to get the surface area of the solid , we need to add the surface area of cube and hemisphere but surface area of base of hemisphere is subtracted as it is counted twice.

Given edge of the cube, a = 7 cm

Diameter of the hemisphere, d = 7 cm

Radius, r = d/2 = 7/2 = 3.5 cm

Surface area of the hemisphere = 2π r2

= 2×(22/7)×3.52

= 44×12.25/7

= 539/7

= 77 cm2

Surface area of the cube = 6a2

= 6×72

= 6×49

= 294 cm2

Surface area of base of hemisphere = π r2

= (22/7)×3.52

= 22×12.25/7

= 38.5 cm2

Surface area of the solid = surface area of the cube + surface area of hemisphere – surface area of the base of hemisphere

= 294+77-38.5

= 332.5 cm2

Hence the surface area of the solid is 332.5 cm2.

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