Question

A pyramid with trapezium base with sides $AB=9cm,BC=12cm,CD=15cm$, and $DA=18cm$. The volume of pyramid $=1458{cm}^3$. Find the height, $H$, in cm?

Answer 1

In a trapezium $ABCD$,

$△ABC={90}^o$ and $ADC$ is a scalene triangle

In triangle $ABC$,

Area of $ABC=\frac{1}{2}\times AB\times BC$

$=\frac{1}{2}\times 12\times 9=54c{m}^2$

In triangle $ADC$,

Semi-perimeter $=S=(AC+AD+CD)/2=\frac{15+18+15}{2}=24cm$

Area of $△ADC=\sqrt{s(s-a)(s-b)(s-c)}$

$=\sqrt{24\left(9\right)\left(9\right)\left(6\right)}=6\times 2\times 9=108c{m}^2$

$\therefore$ Area of trapezium $ABCD=A\left(△ABC\right)+A\left(△ACD\right)=108+54=162c{m}^2$

Volume of pyramid $=\frac{1}{3}$Area of trapezium $\times$ height

$=\frac{1}{3}\times 162\times h$

$\therefore h=27cm$.

Hence, the answer is $27$.