In an obtuse angled triangle, the sum of the squares of the sides making the obtuse angle is:

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less than the square of the third side

greater than the square of the third side

equal to the square of the third side

none of these

answer is A.

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Detailed Solution

In an obtuse-angled triangle, one of the angles is greater than 90 degrees and the other two are less than 90 degrees.
Let ABC is an obtuse triangle, where
∠B = 120°, AB = 5 unit, BC = 7 unit, AC = 10 unit
Question Image

\Rightarrow

{(AC)}^{2} = {(10)}^{2} = 100

Sides making the obtuse angle are AB and BC

\Rightarrow

{(AB)}^{2} + {(BC)}^{2} = 5^{2} + 7^{2} = 25 + 49 = 74

Therefore,

\Rightarrow

{(AB)}^{2} + {(BC)}^{2} < {(AC)}^{2}

\Rightarrow

74 < 100

Hence, in an obtuse-angled triangle, the sum of the squares of the sides making the obtuse angle is less than the square of the third side.

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