# Ex.6.5 Q1 The Triangle and Its Properties - NCERT Maths Class 7

## Question

\(PQR\) is a triangle right angled at \(P.\) If \(PQ\) \(=\rm{}\,10\, cm \) and \(PR\) \(= \rm{}\,24\, cm,\) find \(QR.\)

## Text Solution

**What is Known?**

\(PQR \) is a triangle right angled at \(P \) and the length of two sides \(PQ \) \(=\rm{}\,10\,cm \) and \(PR\) \(= \rm{}\,24\, cm.\)

**What is unknown?**

Length of one side of the triangle.

**Reasoning:**

This question is straight forward, as it is given in the question that \(PQR \) is a right- angled triangle and it is right angled at \(P.\) So, we can apply Pythagoras theorem here, if it is right angled at \(P\) then the side opposite to \(P\) will be the hypotenuse of the triangle i.e.\(QR\) and the other side is given \(PQ\) \(= \rm{}\,10\,cm\) and \(PR=24\). Now, by applying Pythagoras theorem i.e. in a right-angled triangle, the square of hypotenuse is equal to the sum of square of other two sides, we can find \(QR.\) For better visual understanding draw a right-angled triangle which is right angled at \(Q\) and consider the side opposite to it \(PR \) as hypotenuse

**Steps:**

** **Given, \(PQ\) \(= \rm{}\,10\,cm,\) \(PR\) \(= \rm{}\,24\,cm\) and \(QR =? \)

By applying Pythagoras theorem in triangle \(PQR,\) we get

\[\begin{align} \left( {{\rm{Hypotenuse}}} \right)^2 & \!=\! \left[ \begin{array}{l} \left( {{\rm{Perpendicular}}} \right)^2 \!+\! \\ \left( {{\rm{Base}}} \right)^2 \\ \end{array} \right] \\ QR^2 &\!=\! \left( {{\rm{PQ}}} \right)^2 \!+\! \left( {{\rm{PR}}} \right)^2 \\ QR^2 &\!=\! \left( {{\rm{10}}} \right)^2 \!+\! \left( {{\rm{24}}} \right)^2 \\ QR^2 &\!=\! 100 \!+\! 576 \\ QR^2 &\!=\! 676 \\ QR &\!=\! 26\;{\rm{cm}} \\ \end{align}\]

Thus, \(QR\) is equal to \(26\) cm

**Useful Tip:**

Whenever you encounter problems of this kind, it is best to think of the Pythagoras theorem for right angled triangle.

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