Question

In $△ABC$ with altitude $AD,\angle BAC=\frac{\pi }{4}$, $BD=\frac{3}{2}$ and $DC=1$, then the length of side $AB$ is

• A. $\frac{3\sqrt{5}}{2}$
• B. $\sqrt{3}$
• C. $3\sqrt{5}$
• D. $3$

Answer 1

The correct option is A $\frac{3\sqrt{5}}{2}$

Equating the area of triangle
$\frac{1}{2\sqrt{2}}\sqrt{h^2+\frac{9}{4}}\cdot \sqrt{h^2+1}=\frac{1}{2}\times h\times \frac{5}{2}$
$\Rightarrow (h^2+\frac{9}{4})(h^2+1)=\frac{25h^2}{2}$


$\Rightarrow (4h^2+9)(h^2+1)=50h^2$
$\Rightarrow 4h^4+13h^2+9=50h^2$
$\Rightarrow 4h^4-37h^2+9=0$
$\Rightarrow 4h^4-36h^2-h^2+9=0$
$\Rightarrow 4h^2(h^2-9)-1(h^2-9)=0$
$\Rightarrow h=\frac{1}{2}$ or $h=3$
$\Rightarrow AB=\sqrt{\frac{10}{4}}=\frac{\sqrt{10}}{2}=\frac{3\sqrt{5}}{2}$

Alternate:
Let $AD=h,\angle BAD=\alpha ,\angle CAD=\beta$
$\Rightarrow \tan \alpha =\frac{3}{2h}$ and $\tan \beta =\frac{1}{h}$
$\therefore \tan (\alpha +\beta )=\frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }$
$\Rightarrow \tan \frac{\pi }{4}=\frac{\frac{3}{2h}+\frac{1}{h}}{1-\frac{3}{2h}\times \frac{1}{h}}$
$\Rightarrow 2h^2-3=3h+2h\Rightarrow 2h^2-5h-3=0\Rightarrow h=-\frac{1}{2}\text{(rejected) or}h=3$
$\Rightarrow h=3$
$\therefore AB=\sqrt{{\left(\frac{3}{2}\right)}^2+3^2}=\frac{3\sqrt{5}}{2}$