Question

find the sum of the areas of the two end surfaces of a right circular of which the height is 20 m and curved surface area is 8800 sq.m.

Answer 1

$h = 20cm \\ csa = 2\pi \: rh = 8800sq.cm \\ r = \frac{8800 \times 7}{2 \times 22 \times20 } = 70cm\\ sum \: of \: areas \: of \: two \: end \: surfaces = 2 \times \pi \: {r}^{2} \\ \: \: \: \: \: \: \: = 2 \times \frac{22}{7} \times 70 \times 70 = 44 \times 700 =30800 {cm}^{3}$

Answer 2

Step-by-step explanation:

h= 20c

curved surface area=8800

8800*7/2*22*20=70

area of surface end=2*π r*r

2×22/7×70×70

44×700=30800

=solve by Bhanu pratap pal.