Question

$AD$ is an altitude of $cm$ equilateral $△ABC$. On a $AD$ base another equilateral triangle $ADE$ is construct then $ar\left(△ADE\right):ar\left(△ABC\right)=3:4$

• A. True
• B. False

Answer 1

The correct option is A True

We have an equilateral $△ABC$ in which $AD$ is altitude. An equilateral $△ADE$ is drawn using $AD$ as base.

Since, the two triangle are equilateral, the two triangles will be similar also.

$△ADE\sim △ABC$

We know that according to the theorem, the ratio of areas of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

$\Rightarrow$ $\frac{ar\left(△ADE\right)}{ar\left(△ABC\right)}={\left(\frac{AD}{AB}\right)}^2$ ----- ( 1 )

Now, $△ABC$ is an equilateral triangle.

$\therefore$ $\angle B={60}^o$

$\Rightarrow$ $\sin B=\frac{AD}{AB}$

$\Rightarrow$ $\sin {60}^o=\frac{AD}{AB}$

$\Rightarrow$ $\frac{\sqrt{3}}{2}=\frac{AD}{AB}$

$\Rightarrow$ ${\left(\frac{AD}{AB}\right)}^2=\frac{3}{4}$

Substituting above value in equation ( 1 ) we get,

$\Rightarrow$ $\frac{ar\left(△ADE\right)}{ar\left(△ABC\right)}=\frac{3}{4}$ ------ Hence proved.