Question

A bucket open at the top is of the form of frustum of cone. the diameter of it's upper and lower ends are 40cm and 20cm respectively. if total 17600cm^3 of water can be filled in the bucket, find it's total surface area.​

Answer 1

Answer:

GIVEN VOLUME. BH THAT U CAN FIND HEIGHT AND BY THAT HEIGHT U CAN FIND SLANT HEIGHT i.e l

AND PUT THE TSA FORMULA UR ANSWER WILL COME

Answer 2

Total surface area of bucket=4022.9 square cm

Step-by-step explanation:

Diameter of upper ends=$d_1=40 cm$

$r_1=\frac{40}{2}=20 cm$

$d_2=20 cm$

$r_2=\frac{20}{2}=10 cm$

Volume of water can be filled in the bucket=$V=17600 cm^3$

Volume of water in bucket=$\frac{1}{3}\pi h(r^2_1+r^2_2+r_1r_2)$

Using $\pi=\frac{22}{7}$

Substitute the values

$17600=\frac{1}{3}\times \frac{22}{7}h((20)^2+(10)^2+20\times 10)$

$h=\frac{17600\times 3\times 7}{22\times ((20)^2+(10)^2+20\times 10)}$

$h=24 cm$

Slant height of bucket, l=$\sqrt{(r_1-r_2)^2+h^2}$

Slant height of bucket, l=$\sqrt{(20-10)^2+(24)^2}=26 cm$

Total surface area of bucket=$\pi (r_1+r_2)l+\pi r^2_2+\pi r^2_1$

Total surface area of bucket=$\frac{22}{7}\times 26(20+10)+\frac{22}{7}(20)^2+\frac{22}{7}(10)^2$

Total surface area of bucket=4022.9 square cm

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