Question

Show that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Answer 1

Answer:

Given: A right triangle ABC right angled at B.

To Prove: AC2 = AB2 + BC2

Construction: Draw BD ⊥ AC

Proof:

In 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)

Similarly, Δ BDC ~ Δ ABC

BC2 = CD × AC … (ii)

Adding (i) and (ii)

AB2 + BC2 = (AD × AC) + (CD × AC)

AB2 + BC2 = AC × (AD + CD)

AB2 + BC2 = AC2

Hence Proved.

Answer 2

Answer:

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)

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...