Question

In the adjoining figure, $ PQ=24cm $, ${QR}=\text{26cm}$, $\angle PAR={{90}^{\circ }}$, PA=6cm and $ AR=8cm $. Find $\angle QPR$.

Answer 1

Hint: There are two triangles in the given figure $\Delta PQR$ and $\Delta PQR$. You can see one of them is right-angled. Use Pythagoras’ theorem in right angled triangles to find out the required angle.

Complete step by step answer:

 As given in the diagram there are two triangles $\Delta PQR$ and $\Delta PAR$. We can see that in $\Delta PAR$ the length of the sides as given in the question are $PA=6cm$ and $AR=8cm$ and the measurement of the angle $\angle PAR={{90}^{\circ }}$. We also see in $\Delta PQR$ , we are given the length of two sides $PQ=24cm$ and $QR=26cm$. We are asked find the measure of the angle $\angle QPR$\[\]
 

We know from the property of right angled triangles that in a right angled triangle there is a right angle otherwise known as an angle with measurement ${{90}^{\circ }}$. The side opposite to the right- angle is called hypotenuse and is denoted as $h$. Let us denote the other two sides as $p$ and $b.$ We see in the triangle we have drawn above that $\angle ABC={{90}^{\circ }},AC=h,AB=p,BC=b$\[\]
The Pythagoras theorem states that “ The triangle is right-angled if and only if the square of the hypotenuse is the sum of squares of the other two sides . ” We express it in symbols,
\[\Delta ABC\text{ is right angled }\Leftrightarrow A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}\]
In other words,
\[\Delta ABC\text{ is right angled }\Leftrightarrow {{h}^{2}}={{p}^{2}}+{{b}^{2}}\]
 Now as given in the figure $\Delta PAR$ is the right angle where $\angle PAR={{90}^{\circ }}$ is the right angle. Then its opposite side is hypotenuse $PR$ of $\Delta PAR$. Let us use Pythagoras theorem and put given values,
 \[\begin{align}
  & P{{R}^{2}}=P{{A}^{2}}+R{{A}^{2}}={{\left( 6cm \right)}^{2}}+{{\left( 8cm \right)}^{2}}=100c{{m}^{2}} \\
 & \Rightarrow PR=\sqrt{100c{{m}^{2}}}=10cm \\
\end{align}\]
 Now we shall observe $\Delta PQR$ we get, $QR=26cm$ and now $PR=10cm$. We calculate,
\[\begin{align}
  & PR=10cm\Rightarrow P{{R}^{2}}=100c{{m}^{2}} \\
 & PQ=24cm\Rightarrow P{{Q}^{2}}=576c{{m}^{2}} \\
 & QR=26cm\Rightarrow Q{{R}^{2}}=676c{{m}^{2}} \\
\end{align}\]
We observe that $Q{{R}^{2}}=676c{{m}^{2}}=576c{{m}^{2}}+100c{{m}^{2}}=P{{Q}^{2}}+P{{R}^{2}}$. So by Pythagoas’ theorem $QR$ is the hypotenuse of the triangle $\Delta PQR$and its opposite angle $\angle QPR$ have to be right angle in other words $\angle QPR={{90}^{\circ }}$.\[\]

So, the correct answer is “${{90}^{\circ }}$”.

Note: You need to be careful substitution and use of formula in this problem. The trick here is that the statement of Pythagoras theorem is treated as a bi-conditional statement. The question can also be framed to find other acute angles but for that you need trigonometric ratios.