Question

Two isosceles triangles have equal vertical angles and their areas are in the ratio $36:25$. What is the ratio of their corresponding heights?

• A. $\frac{1}{4}$
• B. $\frac{3}{4}$
• C. $\frac{6}{5}$
• D. $\frac{7}{4}$

Answer 1

The correct option is C

$\frac{6}{5}$

Similar Triangles:

If two triangles are said to be similar, then

(i) their corresponding angles are equal

(ii) their corresponding sides are in the same proportion (or) ratio.

Proving the similarity of the two triangles:

Consider two isosceles triangles ABC and XYZ.

It is given that, two vertical angles are equal in the triangles ABC and XYZ.

That is, $\angle BAC=\angle YXZ$

Since, $ABC$ and $XYZ$ are isosceles triangle, $AB=AC$ and $XY=YZ$.

By side - side - angle similarity criterion, the two triangles $ABC$ and $XYZ$ are similar.

Area of similar triangle theorem:

The ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides.

$\frac{ar\left(ABC\right)}{ar\left(XYZ\right)}={\left(\frac{AB}{XY}\right)}^2={\left(\frac{BC}{YZ}\right)}^2={\left(\frac{CA}{ZX}\right)}^2$

Calculating the ratio of the corresponding heights:

It is given that, two isosceles triangles have equal vertical angles and their areas are in the ratio $36:25$.

$\frac{ar\left(ABC\right)}{ar\left(XYZ\right)}={\left(\frac{BC}{YZ}\right)}^2=\frac{36}{25}$

$$\begin{gathered} \Rightarrow \left(\frac{BC}{YZ}\right)=\sqrt{\frac{36}{25}} \\ \Rightarrow \left(\frac{BC}{YZ}\right)=\frac{6}{5} \end{gathered}$$

Then,

$$\begin{gathered} \frac{ar\left(ABC\right)}{ar\left(XYZ\right)}=\frac{{\displaystyle \frac{1}{2}}\times AD\times BC}{{\displaystyle \frac{1}{2}}\times XM\times YZ} \\ \Rightarrow \frac{{\displaystyle \frac{1}{2}\times AD\times BC}}{{\displaystyle \frac{1}{2}\times XM\times YZ}}=\frac{36}{25} \\ \Rightarrow \frac{AD}{XM}\times \frac{BC}{YZ}=\frac{36}{25} \\ \Rightarrow \frac{AD}{XM}\times \frac{6}{5}=\frac{36}{25}[\mathrm{Since},\frac{BC}{YZ}=\frac{6}{5}] \\ \Rightarrow \frac{AD}{XM}=\frac{36}{25}\times \frac{5}{6} \\ \Rightarrow \frac{AD}{XM}=\frac{6}{5} \end{gathered}$$

Hence, the ratio of corresponding heights of the two triangles is $\frac{6}{5}$.