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A bucket open at the top is in the form of a frustum of a cone with a capacity of $12308.8 cm^3$ . The radii of the top and bottom of a circular ends of the bucket are 20cm and 12cm respectively. Find the height of the bucket and also the area of the metal sheet used in making it. (Use $\pi=3.14$ ).

Answers (1)

We know that

$r_1=12cm$

$r_2=20cm$

Capacity of a bucket = Volume of the bucket

Capacity of bucket = $=\frac{\pi h}{3}\left [ r_1^2+r_2^2+r_1.r_2 \right ]$

Putting values in formula

$=\frac{1\times 3.14\times h}{3}\left [ 20^2+12^2+20.12 \right ]=12308.8$

$=\frac{1\times 3.14\times h}{3}\left [ 400+144+240 \right ]=12308.8$

$=\frac{1\times 3.14\times h}{3}\left [ 784 \right ]=12308.8$

$h=\frac{12308.8\times 100 \times 3}{314 \times 784}$

$h=15cm$

Metal required for bucket = Surface area of frustum

Metal required $=\pi r_2^2+\pi \left ( r_1+r_2 \right )l$ .....(1)

$l=\sqrt{h^2+(r_1-r_2)^2}$

Putting the values

$l=\sqrt{15^2+(20-12)^2}$

$l=\sqrt{225+64}$

$l=\sqrt{289}$

$l=17$

Putting values in eqn (1)

$=\pi \left (12 \right )^2+\pi \left (12+20 \right )17$

$=144 \pi +544 \pi$

$=688\pi$

$=688\times 3.14$

$=2160.32cm^2$

Posted by

Safeer PP

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