Question

A bucket is in the shape of a frustum with the top and bottom circles of radii $$15$$ cm and $$10$$ cm. its depth is $$12$$ cm. Find its curved surface area and total surface area. (Express the answer in terms of $$\pi$$)

Answer 1

Hint: Firstly we find the slant height. After that we will substitute that value in the CSA and TSA formula to find the $CSA$ & $TSA$.

Formula used: Using these formulas we will find the curved and total surface area of the given shape.
The curved surface area of the frustum is $CSA$ = $$\pi (R+r)s=\pi (R+r)\sqrt{{(R-r)}^2+h^2}$$
And the total surface area is $TSA$ = $$\pi (R+r)s+\pi r^2+\pi R^2=\pi (R+r)\sqrt{{(R-r)}^2+h^2}+\pi (r^2+R^2)$$

Complete step-by-step answer:
Let us consider, $R$, $r$, $s$ & $h$ be the radius of the lower base, radius of the upper base, the slant height and the perpendicular height of the frustum.
It is given that a bucket is in the shape of a frustum with the top and bottom circles of radii $$15$$ cm and $$10$$ cm and its depth is $$12$$ cm. We have to find its curved surface area and total surface area.
Now, substitute, $$R$$ = 15, $$r$$ = 10 and $$h$$ = 12 in lateral surface area we get,
 The lateral surface area $$CSA$$ = $$\pi (15+10)\sqrt{{(15-10)}^2+{12}^2}$$ $c{m}^2$
By simplifying the squares and square roots we get,
The lateral surface area $$CSA$$ = $$\pi \times 25\times 13$$ $c{m}^2$
And on further simplifications we get,
The lateral surface area of the bucket $$CSA$$ = $$325\pi$$ $c{m}^2$
Again, let us substitute, $$R$$ = 15, $$r$$ = 10 and $$h$$ = 12 in total surface area we get,
 The total surface area $TSA$ = $$\pi (15+10)\sqrt{{(15-10)}^2+{12}^2}+\pi ({10}^2+{15}^2)$$ $c{m}^2$
By simplifying the squares and square roots we get,
The total surface area $TSA$ = $$\pi \times 25\times 13+325\pi$$ $c{m}^2$
And on further simplification we get,
The total surface area of the bucket $TSA$ = $$650\pi$$ $c{m}^2$
Hence, the curved surface area is $$325\pi$$ $c{m}^2$ and total surface area is $$650\pi$$ $c{m}^2$

Note: The frustum is the sliced part of a right circular cone. If we eliminate the top corner part of the right circular cone we get a frustum.