Question

Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the
other two sides.
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Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

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Given:

A right angled ∆ABC, right angled at B

To Prove- AC²=AB²+BC²

Construction: draw perpendicular BD onto the side AC .

Proof:

We know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, than triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

We have

△ADB∼△ABC. (by AA similarity)

Therefore, AD/ AB=AB/AC

(In similar Triangles corresponding sides are proportional)

AB²=AD×AC……..(1)

Also, △BDC∼△ABC

Therefore, CD/BC=BC/AC

(in similar Triangles corresponding sides are proportional)

Or, BC²=CD×AC……..(2)

Adding the equations (1) and (2) we get,

AB²+BC²=AD×AC+CD×AC

AB²+BC²=AC(AD+CD)

( From the figure AD + CD = AC)

AB²+BC²=AC . AC

Therefore, AC²=AB²+BC²

This theroem is known as Pythagoras theroem.

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#Maks#