2015)
pherical part of the toy at the rate of 10 per cm. (Take it =
33. In Fig. 14.58, from a cuboidal solid metalic block, of dimensions 15 cm x 10 cm
x 5 cm, a cylindrical hole of diameter 7 cm is drilled out. Find the surface area of the
remaining block. (Take a = 22/7).
[CBSE 2015]
7 cm
5 cm
10 cm
15 cm
Answers
Answer:
Given: Length of cuboidal block =15 CM
Breadth of cuboidal block =10 CM
Height of cuboidal block =5 CM
Also, Diameter of Cylindrical hole =7 CM
To find : Surface Area of remaining block
Solution :
As Length, Breadth and height is given then,
TSA of Cuboidal block = 2( lb+bh+hl)
=2 ( 15*10+ 10*5+5*15)
= 550 CM^2
And, CSA of Cylindrical hole drilled out from cuboidal block = 2πrh
=2* 22/7 *7/2 * 5
=110 Cm^2
So, surface area of the remaining block =
TSA of Cuboidal block+ CSA of Cylindrical hole - 2* ar( base)
= 2( lb+bh+hl) + 2πrh- 2*πr^2
= 550 cm^2 + 110 cm^2- 2*22/7*(7/2) ^2
= 660 cm^2 - 77 cm^
= 583 cm^2
Hope it helped you...