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# Ex.6.5 Q2 The Triangle and Its Properties - NCERT Maths Class 7

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## Question

$$ABC$$ is a triangle right angled at $$C.$$ If $$AB$$ $$= \rm{}25\,cm$$ and $$AC$$ $$=\rm{}7\, cm,$$ find $$BC.$$

Video Solution
Triangle & Its Properties
Ex 6.5 | Question 2

## Text Solution

What is  known?

$$ABC$$ is a triangle right angled at $$C$$ and the length of two sides $$AB$$ $$=\rm{}25\, cm$$ and $$AC$$ $$=\rm{}7\, cm.$$

What is unknown?

Length of $$BC.$$

Reasoning:

This question is straight forward, as it is given in the question that $$ABC$$ is a triangle, right-angled at $$C.$$ So, we can apply Pythagoras theorem here, if it is right angled at $$C$$ then the side opposite to $$C$$ will be the hypotenuse of the triangle i.e. $$AB$$ $$= \rm{}25\, cm$$ and the other two side is $$AC$$ $$= \rm{}7\, cm$$ and $$BC.$$ 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 $$BC.$$

For better visual understanding draw a right-angled triangle which is right angled at $$C$$ and consider the side opposite to it $$AB$$ as hypotenuse.

Steps:

Given, $$ABC$$ is a triangle which is right-angled at $$C.$$

$$AB$$ $$= \rm{}25\,cm,$$ $$AC = \rm{}7\, cm$$

and $$BC =$$$$?$$

In triangle $$ACB,$$ By Pythagoras theorem,

\begin{align}&\left(\text{Hypotenuse} \right)^2\\ &= \left(\text {Perpendicular} \right)^2 + \left(\text {Base} \right)^2\end{align}

\begin{align}&= {{\left( {AB} \right)}^2} = {{\left( {AC} \right)}^2} + {{\left( {BC} \right)}^2}\\&= {{\left( {25} \right)}^2} = {{\left( 7 \right)}^2} +{{\left( {BC} \right)}^2}\\ 625 &= 49 + {{\left( {BC} \right)}^2}\\{{\left( {BC} \right)}^2} &= 625-49\\{{\left( {BC} \right)}^2} &= 576\\BC\rm{ } &= \rm{ }24\rm{ }cm\end{align}

Thus, $$BC$$ is equal to $$\rm{}24\,cm$$

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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