Math, asked by reetagoyal540, 9 months ago

prove that in a right angle triangle the square of hypotenuse is equal to the sum of the square of the other two side

Answers

Answered by dsaluja1981
2

Step-by-step explanation:

The Pythagoras Theorem

Let’s have a look at what Mr Pythagoras stated when he came up with the Theorem,

Statement: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the remaining two sides.

To understand it better we break down the statement.

A right-angled triangle is a triangle with a 90-degree angle.

The hypotenuse is the longest side of the right-angled triangle.

The remaining sides of the triangle are called the base and the perpendicular.

Pythagoras Theorem

In the diagram above,

∠ABC is a right angle.

AC is the hypotenuse.

AB is known as the perpendicular.

BC is the base.

So according to the Pythagoras Theorem,

(AC)²=(AB)²+(BC)²

But then should we merely trust a single statement? I don’t think so. We need a proof!

Browse more Topics under Triangles

Properties of Triangles

Congruent Triangles

Similarity of Triangles

Inequalities of Triangles

Basic Proportionality Theorem and Equal Intercept Theorem

Proof of the Pythagoras Theorem using Similarity of Triangles:

Given: In ΔABC, m∠ABC=90°

Construction: BD is a perpendicular on side AC

To Prove: (AC)²=(AB)²+(BC)²

Proof:

In △ABC,

m∠ABC=90° (Given)

eg BD is perpendicular to hypotenuse AC (Construction)

Therefore, △ADB∼△ABC∼△BDC (Similarity of right-angled triangle)

△ABC∼△ADB

(AB/AC)=(AD/AB) (congruent sides of similar triangles)

AB2=AD×AC (1)

△BDC∼△ABC

CD/BC=BC/AC (congruent sides of similar triangles)

BC2=CD×AC (2)

Adding the equations (1) and (2),

AB2+BC2=AD×AC+CD×AC

AB2+BC2=AC(AD+CD)

Since, AD + CD = AC

Therefore, AC2=AB2+BC2

Hence Proved.

Answered by astha83
3

Given: In ΔABC, m∠ABC=90°

Construction: BD is a perpendicular on side AC

To Prove: (AC)²=(AB)²+(BC)²

Proof:

In △ABC,

∠ABC=90° (Given)

BD is perpendicular to hypotenuse AC (Construction)

Therefore, △ADB∼△ABC∼△BDC (Similarity of right-angled triangle)

△ABC∼△ADB

(AB/AC)=(AD/AB) (congruent sides of similar triangles)

AB2=AD×AC (1)

△BDC∼△ABC

CD/BC=BC/AC (congruent sides of similar triangles)

BC2=CD×AC (2)

Adding the equations (1) and (2),

AB2+BC2=AD×AC+CD×AC

AB2+BC2=AC(AD+CD)

Since, AD + CD = AC

Therefore, AC2=AB2+BC2

Hence Proved.

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