Question

prove that in a right angled triangle the square of hypotenuse is equal to the sum of square of other two sides​

Answer 1

Answer:

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Step-by-step explanation:

Given:- A right triangle ABC right angled at B.

To prove:- AC

2

=AB

2

+BC

2

Construction:- Draw BD⊥AC

Proof:-

In △ABC and △ABD

∠ABC=∠ADB(Each 90°)

∠A=∠A(Common)

∴△ABC∼△ABD(By AA)

AC

AB

=

AB

AD

(∵Sides of similar triangles are proportional)

⇒AB

2

=AD⋅AC.....(1)

Similarly, in △ABC and △BCD

∠ABC=∠BDC(Each 90°)

∠C=∠C(Common)

∴△ABC∼△BCD(By AA)

∴

BC

DC

=

AC

BC

⇒BC

2

=DC⋅AC.....(2)

Adding equation (1)&(2), we have

AB

2

+BC

2

=AD⋅AC+DC.AC

⇒AB

2

+BC

2

=AC(AD+DC)

⇒AB

2

+BC

2

=AC

2

Hence proved.

solution

Answer 2

Answer:

this is proved so simply by pythagoras theorem.

pythagoras theorem says that

(hypotenuse)² = (perpendicular)²+(base)²