Question

2. The sides of some triangles are given below. Find out which ones are right-angled
triangles (i) 8, 15, 17 (ii) 11, 12, 15​

Answer 1

★ How to do :-

Here, we are given with the three sides of a triangles. There are two triangles each having three sides. We are also given with the measurement of all the three sides of the triangle. Ww are asked whether the given sides form a right-angled triangle or not. Here, we are going to use the concept of pythagoras theorem which is always used to right-angled triangle. This concept says that the sum of square of other two sides of a right-angled triangle always measures same as square of hypotenuse side. So, let's solve!!

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➤ Solution (1) :-

${\sf \leadsto \underline{\boxed{\sf {AB}^{2} + {BC}^{2} = {AC}^{2}}}}$

Substitute the given values.

${\sf \leadsto {8}^{2} + {15}^{2} = {17}^{2}}$

Find the square values of all the numbers.

${\sf \leadsto 64 + 225 = 289}$

Add the values on LHS.

${\sf \leadsto 289 = 289}$

So,

${\sf \leadsto LHS = RHS}$

By this, we can conclude that these three sides can form a right-angled triangle.

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➤ Solution (2) :-

${\sf \leadsto \underline{\boxed{\sf {AB}^{2} + {BC}^{2} = {AC}^{2}}}}$

Substitute the given values.

${\sf \leadsto {11}^{2} + {12}^{2} \neq {15}^{2}}$

Find the square values of all the numbers.

${\sf \leadsto 121 + 144 \neq 225}$

Add the values on LHS.

${\sf \leadsto 265 \neq 289}$

So,

${\sf \leadsto LHS \neq RHS}$

By this, we can conclude that these three sides cannot form a right-angled triangle.

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Hence solved !!