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If the radii of the circular ends of a bucket $ 24 \mathrm{~cm} $ high are $ 5 \mathrm{~cm} $ and $ 15 \mathrm{~cm} $ respectively, find the surface area of the bucket.
Given:
The radii of the circular ends of a bucket \( 24 \mathrm{~cm} \) high are \( 5 \mathrm{~cm} \) and \( 15 \mathrm{~cm} \) respectively.
To do:
We have to find the surface area of the bucket.
Solution:
Height of the bucket (frustum) $h = 24\ cm$
Radius of the top of the bucket $r_1 = 15\ cm$
Radius of the bottom of the bucket $r_2 = 5\ cm$
Therefore,
Lateral height of the bucket $l=\sqrt{(h)^{2}+(r_{1}-r_{2})^{2}}$
$=\sqrt{(24)^{2}+(15-5)^{2}}$
$=\sqrt{24^{2}+10^{2}}$
$=\sqrt{575+100}$
$=\sqrt{676}$
$=26$
Total surface area of the bucket $=\pi(r_{1}+r_{2}) l+\pi r_{2}^{2}$
$=\pi(15+5) \times 26+\pi \times(5)^{2}$
$=20 \times 26 \pi+25 \pi$
$=520 \pi+25 \pi$
$=545 \pi \mathrm{cm}^{2}$
The surface area of the bucket is $545 \pi \mathrm{cm}^{2}$.
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