Math, asked by BrainlyHelper, 1 year ago

A decorative block which is made of two solids , a cube and a hemisphere. The base of the block is a cube with edge 5 cm , and the hemisphere fixed on the top has a diameter 4.2 cm.Find the total surface area of the block .
(class 10 CBSE SAMPLE PAPER 2017-18 MATHS)

Answers

Answered by nikitasingh79
165
SOLUTION:
The decorative block is a combination of a cube and the hemisphere.

For cubical portion:
Each edge= 5 cm


For hemispherical portion:
Diameter= 4.2 cm
Radius(r)= 4.2/2= 2.1 cm

Total surface area of the cube= 6 × (edge)²

= 6 (5)²= 6 × 25= 150 cm²

Here the part of the cube where the hemisphere is attached is not included in the surface area.

So the total surface area of the decorative block= total surface area of the cube+ area of base of hemisphere + curved surface area of hemisphere

total surface area of the decorative block= 150 - πr² + 2πr²
= 150 +πr²
= 150 + (22/7) × 2.1× 2.1
= 150 + 13.86 = 163.86cm²

Hence,total surface area of the decorative block=163.86 cm²

HOPE THIS WILL HELP YOU....
Answered by FuturePoet
93

Hi!

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Total Surface area of the Cube

6 * (edge)^2

6 * (5)^2

6 * 5 * 5 cm^2

150 cm^2

The Surface area of the block

TSA of Cube - base area of hemisphere + CSA of hemisphere

150- \pi r^2+ 2\pi r^2 = (150 + \pi r^2) cm^2

150 \ cm^2 +( \frac{22}{7} * \frac{4.2}{2} * \frac{4.2}{2} ) cm^2

(150 + 13.86) cm^2

= 163.86 cm^2

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Thanks !!

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