Question

How much paper of each shade is needed to make a kite given in the figure, in which $ABCD$ is a square with diagonal $40cm?$

Answer 1

We know that all the sides of a square are equal.

Therefore, $AB=BC=CD=AD$

In $△ACD$, we have

$AC=40cm$ and $\angle D={90}^{\circ }$

By Pythagoras theorem, we have

${AC}^2={AD}^2+C{D}^2⟹{40}^2={AD}^2+C{D}^2⟹2{AD}^2=1600⟹{AD}^2=800⟹AD=20\sqrt{2}cm\therefore AB=BC=CD=AD=20\sqrt{2}cm$

Area of square $=sid{e}^2=20\sqrt{2}cm\times 20\sqrt{2}cm=800c{m}^2$

Area of two red portions $=2\times \frac{areaofsquare}{4}$ ...[As area of square is divided into four equal quadrants]

$=2\times \frac{800}{4}=400c{m}^2$

Similarly, area of two green triangular portion $=400c{m}^2$

Consider the $△CEF$, we have

$a=13cm,b=10cm$ and $c=13cm$

Semi-perimeter of the triangle, $s=\frac{a+b+c}{2}=\frac{13cm+10cm+13cm}{2}=18cm$

Area of the $△CEF$$=\sqrt{18\times (18-13\times (18-10)\times (18-13)}c{m}^2$

$=\sqrt{18\times 5\times 8\times 5}$

$=60{cm}^2$

Hence, area of black portion = $60{cm}^2$