2022-07-22

In the following figure, $AD=AB$. Also $\mathrm{\angle }DAB=\mathrm{\angle }DCB=\mathrm{\angle }AEC={90}^{\circ }$ and $AE=5$. Find the area of quadrilateral ABCD.

$\sqrt{\left({x}^{2}-25\right)}$

Kali Galloway

Expert

Step 1
Join AC and hence see by cyclic quad.
$\mathrm{\angle }ACE=\mathrm{\angle }ACB={45}^{o}$
Hence $EC=5,AC=5\sqrt{2}$
Step 2
Thus, $2{x}^{2}={\left(\sqrt{{x}^{2}-25}+5\right)}^{2}+B{C}^{2}$
Now apply cosine rule in $\mathrm{\Delta }ABC$ to get
$\mathrm{cos}45=\frac{{x}^{2}-50-B{C}^{2}}{10\sqrt{2}BC}$

Francisco Proctor

Expert

Step 1
Take $\mathrm{\angle }BDC=\theta$, so $DE=xcos\left(\theta +45\right),xsin\left(\theta +45\right)=5,EC=x\sqrt{2}cos\theta -xcos\left(\theta +45\right),DC=x\sqrt{2}cos\theta ,\phantom{\rule{0ex}{0ex}}BC=x\sqrt{2}sin\theta$
$\mathrm{\Delta }ADB+\mathrm{\Delta }BDC=\mathrm{\Delta }ADE+◻AECB$
$\mathrm{\Delta }ADB+\mathrm{\Delta }BDC=\frac{{x}^{2}}{2}+{x}^{2}sin\theta cos\theta =\frac{{x}^{2}}{2}\left(sin\theta +cos\theta {\right)}^{2}$
Step 2
$\mathrm{\Delta }ADE+◻AECB={x}^{2}sin\theta cos\theta -\frac{{x}^{2}}{\sqrt{2}}sin\theta cos\left(\theta +45\right)+\frac{5}{\sqrt{2}}xcos\theta$
Equating both we get $x=\frac{5\sqrt{2}}{\left(sin\theta +cos\theta \right)}$.
Using this in $\frac{{x}^{2}}{2}\left(sin\theta +cos\theta {\right)}^{2}$ we get area $=25$.

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