Question

From a tin sheet, a sector of radius 20 centimetres and central angle ${240}^o$ is divided into four equal parts as shown in the figure. Then the shaded portion is cut off. Using this, a vessel in the shape of a square pyramid is made.
What is the capacity of this vessel?

Answer 1

From a tin sheet, a sector of radius 20 cm and central angle ${240}^o$ is divided in to four equal parts.
So each triangle has central angle of ${60}^o$ and sides adjacent to the angle are of length $20cm$.
Therefore each triangle is isosceles triangle and angles opposite to them are congruent.
By using angle sum property of triangle we get,
${60}^o+x+x={180}^o$
$x={60}^o$
Hence each triangle is equilateral triangle and length of base is also $20cm$.
Now the base edge and lateral edge of pyramid are $20cm$.
Slant height of the pyramid is
$\sqrt{{20}^2-{10}^2}=\sqrt{4000-1000}=\sqrt{3000}=10\sqrt{3}cm$
Height of the pyramid is
$\sqrt{(10\sqrt{3}{)}^2-{10}^2}=\sqrt{300-100}=\sqrt{200}=10\sqrt{2}cm$
Capacity of vessel
= Volume of pyramid
$=\frac{1}{3}\times (baseedge{)}^2\times height$
$=\frac{1}{3}\times (20{)}^2\times 10\sqrt{2}$
$=\frac{4000\sqrt{2}}{3}c{m}^3$
Capacity of the vessel is $\frac{4000\sqrt{2}}{3}c{m}^3$.