Question

The base of an isosceles triangle is equal to $4$, the base angle is equal to ${45}^{\circ }$. A straight line cuts the extension of the base at a point $M$ at the angle $\theta$ and bisects the lateral side of the triangle which is nearest to $M$.

The area $A$ of the quadrilateral which the straight line cuts off from given triangles is

• A. $\frac{3+\tan \theta }{1+\tan \theta }$
• B. $\frac{3+2\tan \theta }{1+\tan \theta }$
• C. $\frac{3+\tan \theta }{1-\tan \theta }$
• D. $\frac{3+5\tan \theta }{1+\tan \theta }$

Answer 1

The correct option is D $\frac{3+5\tan \theta }{1+\tan \theta }$

Equation of line $PM$ is

$y-1=\tan \theta (x-1)$ ----- ( 1 )

For point $Q$, solving $MP$ and $AC,$

$4-x-1=\tan \theta (x-1)$

$\Rightarrow$ $3-x=\tan \theta x-\tan \theta$

$\Rightarrow$ $-x-\tan \theta x=-\tan \theta -3$

$\Rightarrow$ $x(1+\tan \theta )=3+\tan \theta$

$\Rightarrow$ $x=\frac{3+\tan \theta }{1+\tan \theta }$

Substituting value of $x$ in ( 1 ) we get,

$\Rightarrow$ $y-1=\tan \theta (\frac{3+\tan \theta }{1+\tan \theta }-1)$

$\Rightarrow$ $y=\tan \theta (\frac{3+\tan \theta }{1+\tan \theta }-1)+1$

$\Rightarrow$ $y=\frac{3\tan \theta +{\tan }^2\theta }{1+\tan \theta }-\tan \theta +1$

$\Rightarrow$ $y=\frac{3\tan \theta +{\tan }^2\theta -\tan \theta -{\tan }^2\theta +1+\tan \theta }{1+\tan \theta }$

$\Rightarrow$ $y=\frac{1+3\tan \theta }{1+\tan \theta }$

$\Rightarrow$ $Q=(x,y)=(\frac{3+\tan \theta }{1+\tan \theta },\frac{1+3\tan \theta }{1+\tan \theta })$

Now,

Area of $△APQ=\frac{1}{2}\Vert \begin{array}{ccc}1& 1& 1\\ 2& 2& 1\\ \frac{3+\tan \theta }{1+\tan \theta }& \frac{1+3\tan \theta }{1+\tan \theta }& 1\end{array}\Vert$

$=\frac{1}{2}[(2-\frac{1+3\tan \theta }{1+\tan \theta })-(2-\frac{3+\tan \theta }{1+\tan \theta })+(\frac{2+6\tan \theta }{1+\tan \theta }-\frac{6+2\tan \theta }{1+\tan \theta })]$

$=\frac{1}{2}\left[\frac{-1-3\tan \theta +3+\tan \theta +2+6\tan \theta -6-2\tan \theta }{1+\tan \theta }\right]$

$=\frac{1}{2}\left(\frac{2\tan \theta -2}{1+\tan \theta }\right)$

$=\frac{\tan \theta -1}{1+\tan \theta }$

$=\frac{1-\tan \theta }{1+\tan \theta }$ [ As $\theta <\frac{\pi }{4}$ ]

$\Rightarrow$ Area of quadrilateral $BPQC=$ Area of $△ABC-$ Area of $△APQ$

$=\frac{1}{2}\times 4\times 2-\left(\frac{1-\tan \theta }{1+\tan \theta }\right)$

$=4-\frac{1-\tan \theta }{1+\tan \theta }$

$=\frac{4+4\tan \theta -1+\tan \theta }{1+\tan \theta }$

$=\frac{3+5\tan \theta }{1+\tan \theta }$