Chapter 11.7, Problem 8PSA

To determine

To calculate: The ratio of the areas of two right-angled triangles.

Answer to Problem 8PSA

The ratio of areas of the given right-angled triangles is $49:16.$

Explanation of Solution

Given Information:

Each of the two triangles are right-angled triangle.

In both the triangles, one side and one angle are given.

Formula used:

Sum of all three internal angles of a triangle is ${180}^{\circ }$

Ratio of areas of two similar triangles is equal to the ratio of squares of the corresponding sides.

Calculation:

  

In $\Delta ABC,\angle ABC={90}^{\circ }$ and $\angle ACB={30}^{\circ }$

  $\angle CAB+\angle ABC+\angle ACB={180}^{\circ }$ … (Sum of interior angles of a triangle is ${180}^{\circ }$ )

  $\therefore \angle CAB+{90}^{\circ }+{30}^{\circ }={180}^{\circ }$

  $\therefore \angle CAB={60}^{\circ }$ … (i)

In $\Delta PQR,\angle PQR={90}^{\circ }$ and $\angle QPR={60}^{\circ }$

  $\angle QPR+\angle PQR+\angle PRQ={180}^{\circ }$ … (Sum of interior angles of a triangle is ${180}^{\circ }$ )

  $\therefore {60}^{\circ }+{90}^{\circ }+\angle PRQ={180}^{\circ }$

  $\therefore \angle PRQ={30}^{\circ }$ … (ii)

In $\Delta ABC$ and $\Delta PQR,$

  $\angle ABC=\angle PQR$ … (Given)

  $\angle CAB=\angle QPR$ … (Each ${60}^{\circ })$

  $\angle ACB=\angle PRQ$ … (Each ${30}^{\circ })$

  $\therefore \Delta ABC\sim \Delta PQR$ … (By Angle-Angle-Angle similarity test)

In $\Delta ABC,$

  $\sin \theta =\frac{Oppositeside}{Hypotenuse}$

  $\begin{array}{l}\sin {30}^{\circ }=\frac{AB}{AC}\\ \therefore \frac{1}{2}=\frac{AB}{14}\\ \therefore AB=7\end{array}$

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

  $\begin{array}{l}\frac{ar\left(\Delta ABC\right)}{ar\left(\Delta PQR\right)}=\frac{A{B}^2}{P{Q}^2}=\frac{B{C}^2}{Q{R}^2}=\frac{C{A}^2}{R{P}^2}\\ \frac{ar\left(\Delta ABC\right)}{ar\left(\Delta PQR\right)}=\frac{A{B}^2}{P{Q}^2}=\frac{7^2}{4^2}=\frac{49}{16}\end{array}$

Hence, ratio of areas of two right-angled triangles is $49:16.$