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

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## 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( \text{Hypotenuse} \right)}^2} &= \rm{{\left(\text {Perpendicular} \right)}^2} + {\rm{ }}{{\left(\text {Base} \right)}^2}\\&= \rm{{{\left( {QR} \right)}^2} = {\rm{ }}{{\left( {PQ} \right)}^2} + {\rm{ }}{{\left( {PR} \right)}^2}}\\&= \rm{{{ }}{{\left( {QR} \right)}^2} = {\rm{ }}{{\left( {10} \right)}^2} + {\rm{ }}{{\left( {24} \right)}^2}}\\&= \rm{{{ }}{{\left( {QR} \right)}^2}= {\rm{ }}100{\rm{ }} + 576}\\&= \rm{{{ }}{{\left( {QR} \right)}^2}= {\rm{ }}676}\\QR{\rm{ }} &= \rm{{ }}26\,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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