Question

From a piece of cardbord, in the shape of a trapezium $ABCD$, and $AB\parallel CD$ and $\angle BCD={90}^{\circ }$, quarter circle is removed. Given $AB=BC=3.5cm$ and $DE=2cm$. Calculate the area of the remaining piece of the cardboard (Take $\pi$ to be $\frac{22}{7}$)

Answer 1

Given $AB=BC=3.5cmandDE=2cm$
$\Rightarrow CD=CE+DE=3.5+2=5.5cm$

Area of shaded region $=$Area of trapezium $ABCD-$ Area of a quarter circle

$=\frac{1}{2}(AB+CD)\times BC-\frac{\theta }{{360}^0}\times \pi r^2$

$=\frac{1}{2}(3.5+5.5)\times 3.5-\frac{{90}^0}{{360}^0}\times \frac{22}{7}\times 3.5\times 3.5$

$=(4.5\times 3.5)-(5.5\times 0.5\times 3.5)$

$=15.75-9.625$

$=6.125{cm}^2$