Telanagana SCERT Solution Class VIII (8) Math Chapter 14 Surface Areas And Volume Exercise 14.1
(1)
For fig A
The total surface area = 2 ( 50 x 60 + 50 x 40 + 40 x 60)
= 2 (3000 + 2000 + 2400)
= 14800 cm2
For fig B
The total surface area = 2(50 x 50 + 50 x 50 + 50 x 50)
= 2 x 3 x 502
= 6 x 502
= 15000 cm2
So, fig A require less amount of material.
(2) We know,
Surface area = 6 x (Side)2
=> 6(Side)2 = 600
=> Side = √100
=> Side = 10 cm
Side of cube = 10 cm
(3)
Total surface area of the cuboid = 2(1×2 + 2×1.5 + 1×1.5)
= 13 m2
And the area of bottom part of the cabinet = 1 x 2 = 2 m2
The area she covered =(13 – 2) = 11 m2
(4) The surface area of the cuboid = 2( 20 x 15 + 15 x 12 + 12 x 20)
= 2 (300 + 180 + 240)
= 1440 cm2
Total cost of pointing at the of 5 paisa = 1440 x 0.05 = Rs. 72