Question

In the triangle ABC, MN is parallel to AB. Area of trapezium ABNM is twice the area of triangle CMN. What is ratio of CM : AM ?

• A. $\frac{1}{\sqrt{3}+1}$
• B. $\frac{\sqrt{3}-1}{2}$
• C. $\frac{\sqrt{3}+1}{2}$
• D. $Noneofthese$

Answer 1

The correct option is C $\frac{\sqrt{3}+1}{2}$

$\frac{\left(A\right)\Delta CMN}{\left(A\right)\square ABNM}=\frac{1}{2}$
$\therefore \frac{\left(A\right)\Delta CMN}{\left(A\right)\Delta CAB}=\frac{1}{3}$
$\Rightarrow \frac{MN}{AB}=\frac{CM}{CA}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{CM}{MA}=\frac{1}{\sqrt{3}-1}=\frac{\sqrt{3}+1}{2}$ MA = (CA - CM)