Question

In isosceles triangle $ABC$, base $BC=4cm$ and other sides are measure of $3cm$. Let $D$ and $E$ points on $BC$ and $AC$ respectively such that $CD+CE=$ semi-perimeter of triangle and area of quadrilateral $BDEA=\frac{1}{2}$ area of triangle $ABC$. If $DE=\sqrt{\lambda }$, then find the value of $\lambda$

Answer 1

Answer: 5

$\Delta =\text{ area of }△ABC=2\sqrt{5}$
$CD+CE=5$ .....(i)
$BDEA=\frac{1}{2}△ABC$
$2\sqrt{5}-\frac{1}{2}x\times (5-x)\sin C=\frac{1}{2}2\sqrt{5}$ .....(ii)
using $cosine\; law\cos C=\frac{2}{3}$
$\therefore \sin C=\frac{\sqrt{5}}{3}$
substituting in (ii) we get
$x=2$ or 3
$x=2\text{ is rejected }$
$\therefore x=3$
$\text{ now }\text{ using cosine law in }△ABC\text{ and EDC }$
$\cos C=\frac{9+16-9}{2\times 12}=\frac{4+9-\lambda }{12}$
$\lambda =5$