Question

meter is placed over a
piece.
() A hemisphere of maximum possible diameter is placed over a
cuboidal block of side 7 cm. Find the surface area of the solid so formed.
(ii) A cubical block of side 10 cm is surmounted by a hemisphere.
What is the largest diameter that the hemisphere can have?Find
the cost of painting the total surface area of the solid so formed,at
the rate of 5 per 100 sq cm. (Use pie= 3.14.]
​

Answer 1

Step-by-step explanation:

1.Radius (r) of hemispherical part = 7/2cm

The total surface area of solid = Surface area of cubical part + CSA of hemispherical part - Area of the base of hemispherical part

TSA of solid= 6 (Edge)²+ 2πr² – πr²

= 6 (Edge)²+ πr²

TSA of solid = 6 (7)²+(22/7×7/2×7/2)

= 294 + 38.5

= 332.5 cm².........

2.The greatest diameter the hemisphere can have i 10cm

T.S.A of the solid = surface area of the cube +CSA of the hemisphere - area of the base of the hemisphere

=6×10²+2×22/7×5×5−22/7×5×5

=600+22/7×25

=600+78.57

=678.57cm2

∴ Cost of printing the block at the rate of Rs5/100 square is

100sq.cm=5

678.57=?

67.57100×5=33.93

Hence the amount is Rs.33.93.....

hope it help u.... xD