Question

Question 102

ABCD is a parallelogram in which AE is perpendicular to CD is shown in the given figure. Also, AC = 5 cm, DE = 4 cm and area of $\Delta AED=6c{m}^2$.
Find the perimeter and area of parallelogram ABCD.

Answer 1

Given, area $\Delta AED=6c{m}^2$ and AC = 5 cm and DE = 4 cm
$\therefore$ Area of $\Delta AED=\frac{1}{2}\times DE\times AE[\because$ area of triangle $=\frac{1}{2}\times base\times height]$
$\Rightarrow \frac{1}{2}\times 4\times AE=6\Rightarrow AE=\frac{6\times 2}{4}\Rightarrow AE=3cm$
Now, in right angled $\Delta AEC,$ AE = 3 cm and AC = 5 cm
So, $(EC{)}^2=(AC{)}^2-(AE{)}^2$ [by Pythagoras theorem]
$\Rightarrow (EC{)}^2=5^2-3^2=25-9\Rightarrow EC=\sqrt{16}\Rightarrow EC=4cm\because DE+EC=DC\Rightarrow DE+EC=DC\Rightarrow DC=4+4=8cm$
$\because ABCD$ is a parallelgoram.
So, AB = DC = 8 cm
Now, in right angled $\Delta ADE,A{D}^2=A{E}^2+E{D}^2$
$\Rightarrow A{D}^2=3^2+4^2=9+16\Rightarrow AD=\sqrt{25}\Rightarrow AD=5cm$
So, AD = BC = 5 cm $[\because$ ABCD is a parallelogram]
$\therefore$ Perimeter of paralelogram ABCD $=2(l+b)=2(DC+AD)=2(8+5)=2\times 13=26cm$
Area of parallelogram ABCD $=Base\times Height=DC\times AE=8\times 3=24c{m}^2$