Question

In the following, figure, seg DE || side BC in $△ABC$. If $3A\left(△ADE\right)=A\left(DECB\right),$ then find the ratio $BC:DE$ ?

• A. 1 : 2
• B. 16 : 1
• C. 1 : 16
• D. 2 : 1

Answer 1

Answer: (D)

2 : 1

$3A\left(△ADE\right)=A\left(DECB\right)⟹A\left(△ABC\right)=4A\left(△ADE\right)$

$⟹\frac{A\left(△ABC\right)}{A\left(△ADE\right)}=4$

In $△ADE$ and $△ABC$, $\angle DAE=\angle BAC$

$DE\parallel BC$, $\angle ADE=\angle ABC$ and $\angle AED=\angle ACB$

$⟹△ABC$ and $△ADE$ are similar.

$⟹\frac{A\left(△ABC\right)}{A\left(△ADE\right)}={\left(\frac{BC}{DE}\right)}^2$

$⟹4={\left(\frac{BC}{DE}\right)}^2$

$⟹\frac{BC}{DE}=\frac{2}{1}$

Hence, option D is correct.