Question

37. Prove that "In a right triangle, the square of the liypotemuse is equal to the sum of the squares of
the other two sides.​

Answer 1

Answer:

Step-by-step explanation:

AC² = AB² + BC² .

Step-by-step explanation:

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)

BC²=CD×AC……..(2)

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

AB²+BC²=AD×AC+CD×AC

AB²+BC²=AC(AD+CD)

( AD + CD = AC)

AB²+BC²=AC . AC

AC²=AB²+BC² .

Answer 2

Step-by-step explanation:

Given : A ∆ABC right angle at B.

To prove : AC² = AB² + BC².

Construction : Draw BD parallel AC.

Proof : In Δ ADB and Δ ABC,

Δ ADB and Δ ABC,∠ ADB = ∠ ABC (each 90°)

Δ ADB and Δ ABC,∠ ADB = ∠ ABC (each 90°)∠ BAD = ∠ CAB (common)

Δ ADB and Δ ABC,∠ ADB = ∠ ABC (each 90°)∠ BAD = ∠ CAB (common)

Δ ADB and Δ ABC,∠ ADB = ∠ ABC (each 90°)∠ BAD = ∠ CAB (common) Δ ADB ~ Δ ABC (By AA similarity criterion)

Δ ADB and Δ ABC,∠ ADB = ∠ ABC (each 90°)∠ BAD = ∠ CAB (common) Δ ADB ~ Δ ABC (By AA similarity criterion)Now, AD/AB = AB/AC (corresponding sides are proportional)

AB² = AD × AC … (i)

= AD × AC … (i)Similarly, Δ BDC ~ Δ ABC

BC²= CD × AC … (ii)

= CD × AC … (ii)Adding (i) and (ii)

AB² + BC² = (AD × AC) + (CD × AC)

AB² + BC² = AC × (AD + CD)

AB² + BC² = AC²

Hence Proved.