A B C D is a parallelogram in which A E C D as shown in figure, a… - Vidyadip

Question

$A B C D$ is a parallelogram in which $A E \perp C D$ as shown in figure, also $A C=A D$ and $A C=5 cm, D E=4 cm$ and area of $\triangle A E D=6 cm^2$. Then, find
(i) area of $\triangle A D C$
(ii) area of parallelograr $A B C D$.
(iii) Perimeter of parallelogram $A B C D$

✓

Answer

Given, area of $\triangle A E D=6 cm^2$,
$A C=5 \text { and } D E=4 cm$$\therefore$ Area of $\triangle A E D=\frac{1}{2} \times D E \times A E$ $\qquad\left[\because\right.$ Area of triangle $=\frac{1}{2} \times$ Base $\times$ Height $]$
$\begin{aligned} \Rightarrow & \frac{1}{2} \times 4 \times A E=6 \\ \Rightarrow & A E=\frac{6 \times 2}{4} \Rightarrow A E=3 cm\end{aligned}$
Now, in right angled $\triangle A E C, A E=3 cm$ and $A C=5 cm$
So, $\quad(E C)^2=(A C)^2-(A E)^2\quad$[by Pythagoras theorem]
$\begin{array}{lr}\Rightarrow & (E C)^2=5^2-3^2=25-9 \\ \Rightarrow & E C=\sqrt{16} \\ \Rightarrow & E C=4 cm \\ \because & D E+E C=D C \\ \Rightarrow & D C=4+4=8 cm\end{array}$
$\because A B C D$ is a parallelogram.
$\begin{array}{l}\text { So, } \quad A B=D C=8 cm \\\because \quad A C=A D \text { and } A C=5 cm \\\therefore \quad A D=B C=5 cm\quad[\because A B C D \text { is a parallelogram }] \\\text { (i) Area of } \triangle A D C=\frac{1}{2} \times C D \times A E\end{array}$
$=\frac{1}{2} \times 8 \times 3=\frac{24}{2}=12 cm^2$
(ii) Area of parallelogram $A B C D$
$\begin{array}{l}=\text { Base } \times \text { Height }=D C \times A E \\ =8 \times 3=24 cm^2\end{array}$
(iii) Perimeter of parallelogram $A B C D$
$\begin{array}{c}=2(l+b)=2(D C+A D)=2(8+5) cm \\=2 \times 13 cm=26 cm\end{array}$

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