Question

prove that the square of the hypotenuse in a right angle triangle the equal of the sum of the square of the two sides​

Answer 1

Given:

ABC is a triangle in which ∠ABC=90 ∘

Construction:

Draw BD⊥AC.

Proof:

In △ADB and △ABC

∠A=∠A [Common angle]

∠ADB=∠ABC [Each 90∘]

△ADB∼△ABC [A−A Criteria]

So, AD/AB=AB/AC

Now, AB² =AD×AC ..........(1)

Similarly,

BC² =CD×AC ..........(2)

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

AB² + BC² = AD×AC+CD×AC

=AC(AD+CD)

=AC×AC

∴AB² + BC² = AC²

[hence proved]

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Answer 2

Step-by-step explanation:

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