Question

If two isosceles triangles have equal vertical angles and their areas are in the ratio of 9:16, then the ratio of their corresponding heights (altitudes) is
A. 3:4
B. 4:3
C. 81:256
D. 9:16

Answer 1

We have to find the ratio of altitudes of the two isosceles triangles. We have been given two isosceles triangles whose vertical angles are equal. The vertical angle is the angle enclosed by the two equal sides of the isosceles triangles. When two isosceles triangles have the same vertical angles, then they are similar. We have also been given the ratio of the areas of those triangles. The ratio of the area of two similar triangles is equal to the square of the ratio of their sides. Also, the ratio of the altitudes of the triangle is equal to the ratio of the sides of the triangles. Using this, we will find our answer.

Complete step-by-step solution:
We have been given two isosceles triangles $\Delta ABC$ with $AB=AC$ and $\Delta PQR$ with $PQ=PR$.
AM is the altitude of $\Delta ABC$ and PN is the altitude of $\Delta PQR$.
The vertical angles in $\Delta ABC$ and $\Delta PQR$ are $\angle A$ and $\angle P$.

It is given that the vertical angles in these triangles are equal, thus we can say:
$\angle A=\angle P$
Now, in $\Delta ABC$:
$\begin{align}
  & AB=AC \\
 & \Rightarrow \angle B=\angle C \\
\end{align}$
Using angle sum property in $\Delta ABC$ we get:
$\angle A+\angle B+\angle C={{180}^{\circ }}$
Since, $\angle B=\angle C$ we can write:
$\begin{align}
  & \angle A+2\angle B={{180}^{\circ }} \\
 & \Rightarrow \angle A={{180}^{\circ }}-2\angle B \\
\end{align}$
Now, in $\Delta PQR$:
$\begin{align}
  & PQ=PR \\
 & \Rightarrow \angle Q=\angle R \\
\end{align}$
Now, using angle sum property in $\Delta PQR$ we get:
$\angle P+\angle Q+\angle R={{180}^{\circ }}$
Since, $\angle Q=\angle R$ we can write:
$\begin{align}
  & \angle P+2\angle Q={{180}^{\circ }} \\
 & \Rightarrow \angle P={{180}^{\circ }}-2\angle Q \\
\end{align}$
Since, $\angle A=\angle P$we can say that:
\[{{180}^{\circ }}-2\angle B={{180}^{\circ }}-2\angle Q\]
On solving this, we will get:
\[\begin{align}
  & {{180}^{\circ }}-2\angle B={{180}^{\circ }}-2\angle Q \\
 & \Rightarrow 2\angle B=2\angle Q \\
 & \Rightarrow \angle B=\angle Q \\
\end{align}\]
Thus, \[\Rightarrow \angle Q=\angle R=\angle B=\angle C\]
Now, in $\Delta ABC$ and $\Delta PQR$
$\angle A=\angle P$
$\angle B=\angle Q$
Therefore, $\Delta ABC$ and $\Delta PQR$ are similar triangles by AA (two angles are equal) property.
Since $\Delta ABC$ and $ \Delta PQR$ are similar, their sides are proportional. Thus, we can say that:
$\dfrac{AB}{PQ}=\dfrac{BC}{QR}$ …………........(1)
Now, in $\Delta ABM$ and $\Delta PQN$
$\angle B=\angle Q$
$\angle BMA=\angle QNP={{90}^{\circ }}$ (AM and PN are altitudes)
Thus, $\Delta ABM$ and $\Delta PQN$ are similar triangles by AA (two angles are equal) property.
Since $\Delta ABM$ and $\Delta PQN$ are similar, their sides are proportional. Thus, we can say that:
$\dfrac{AB}{PQ}=\dfrac{AM}{PN}$ ......(2)
From equations (1) and (2), we get that:
\[\dfrac{BC}{QR}=\dfrac{AM}{PN}\]
Now, areas of triangles $\Delta ABC$ and $\Delta PQR$ are in the ratio 9:16
Now areas of $\Delta ABC$ and $\Delta PQR$ are given as:
$\begin{align}
  & ar\left( ~\Delta ABC \right)=\dfrac{1}{2}\times BC\times AM \\
 & ar\left( ~\Delta PQR \right)=\dfrac{1}{2}\times QR\times PN \\
\end{align}$
Thus, we can say that:
$\begin{align}
  & \dfrac{ar\left( \Delta ABC \right)}{ar\left( \Delta PQR \right)}=\dfrac{9}{16} \\
 & \Rightarrow \dfrac{\dfrac{1}{2}\times BC\times AM}{\dfrac{1}{2}\times QR\times PN}=\dfrac{9}{16} \\
 & \Rightarrow \dfrac{BC\times AM}{QR\times PN}=\dfrac{9}{16} \\
\end{align}$
Now, we have already established that \[\dfrac{BC}{QR}=\dfrac{AM}{PN}\]
Thus, we can say that:
$\begin{align}
  & \Rightarrow \dfrac{AM\times AM}{PN\times PN}=\dfrac{9}{16} \\
 & \Rightarrow {{\left( \dfrac{AM}{PN} \right)}^{2}}=\dfrac{9}{16} \\
 & \Rightarrow \dfrac{AM}{PN}=\dfrac{3}{4} \\
\end{align}$
Thus, the ratio of the altitudes of the triangles is 3:4.
Thus, option (A) is the correct option.

Note: Two triangles are said to be similar if they have proportional sides or have all three angles equal. Here, we have used AA property to prove them similar not AAA because once two angles are equal, the other angle is automatically equal through angle sum property. So having a AA property is enough to prove the triangles similar.