Question

The line $x+y=a$ meets the axes of $x$ and $y$ at $A$ and $B$ respectively. A triangle $AMN$ is inscribed in the triangle $OAB$, $O$ being the origin, with right angle at $N,M$ and $N$ lie respectively on $OB$ and $AB$. If the area of the triangle $AMN$ is $3/8$ of the area of the triangle $OAB$, then $AN/BN=$

• A. $2/3$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $2/3$

Area ${\Delta }^{\prime \prime }OAB=\frac{1}{2}\times a\times a=\frac{a^2}{2}$

area of ${\Delta }^{\prime \prime }AMN=\frac{1}{2}\times MN\times AN$

$=\frac{3}{8}{\Delta }^{\prime \prime }OAB=\frac{{3a}^2}{16}$

$\Rightarrow MN.AN=\frac{{3a}^2}{8}$

$\therefore$ $\frac{AN}{BN}=\frac{2}{3}$