please solve it. volume of frustum of cone
Answers
Answer:
height of the frustum=16
upper and lower radii= 20cm ,8cm
volume of frustum= πh/3{R^2+r^2 +Rr}
= π16/3{20×20+8×8+20×8}
= 22÷7×16×1÷3{400+64+160}
= 3.14×16÷3{624}
= 3.14×16×208
= 10449.92cubic cm
volume of milk filled =10449.92 cubic cm
amount of milk filled =10.44992×15
= ₹ 156.74
cost of metal sheet required to make the container
= TSA of frustum
= πl(R+r)+πR^2+πr^2
but
l=√h^2+(R-r)^2
l=√400
l=20cm
20π(20+8)+π20×20+π8×8
= 56π+400π+64π
= 520π
= 1632.8sq cm
= 1632.8÷100×5
=. 16.328×5
=. ₹81.64