# Right Angle Triangle Theorem

A **right-angle triangle theorem** is nothing but a Pythagoras theorem which states the relationship between hypotenuse, base and perpendicular of the triangle. According to this theorem, if the square of the hypotenuse of any right-angle triangle is equal to the sum of squares of base and perpendicular, then the triangle is a right triangle. It is expressed as:

If Hypotenuse^{2}=Perpendicular^{2}+Base^{2}

then, ∠θ = 90°

Where Hypotenuse is the side opposite to right angle (90 degrees), Perpendicular is the side adjacent to the right angle and connected to the hypotenuse, Base is the adjacent side to right angle, which is called legs of the triangle here. With the help of this formula, we can easily find the length of the side of a right angle triangle, if we know the length of the other two sides.

This triangle can also be mentioned as a right triangle. Here, only one angle is 90 degrees and the sum of other triangles is equal to 90 degrees, which are acute angles. The base and perpendicular of right triangle are interchangeable, depending on which acute angle we are considering. The three sides, i.e., base, perpendicular and hypotenuse are known as Pythagorean triples and if the all the three sides are integers then the triangle is called **Pythagorean triangle**.

## Proof of Right Angle Triangle Theorem

**Theorem**:In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

**To prove:** ∠B = 90°

**Proof:** We have a Δ ABC in which AC^{2} = AB^{2} + BC^{2}

We need to prove that ∠B = 90°

In order to prove the above, we construct a triangle PQR which is right-angled at Q such that:

PQ = AB and QR = BC

From triangle PQR, we have

PR^{2} = PQ^{2}+ QR^{2} (According to Pythagoras theorem,as ∠Q = 90°)

or, PR^{2} = AB^{2} + BC^{2 } (By construction) …… (1)

We know that;

AC^{2} = AB^{2}+BC (Which is given) …………(2)

So, AC = PR [From equation (1) and (2)]

Now, in Δ ABC and Δ PQR,

AB = PQ (By construction)

BC = QR (By construction)

AC = PR [Proved above]

So, Δ ABC ≅ Δ PQR (By SSS congruence)

Therefore, ∠B = ∠Q (CPCT)

But, ∠Q = 90° (By construction)

So, ∠B = 90°

Hence the theorem is proved.

**Also, read:**

### Right Angle Triangle Formula

The right-angled triangle formula is given by:

(Hypotenuse)^{2} = (Adjacent side)^{2} + (Opposite side)^{2}

If a, b and c are perpendicular, base and hypotenuse of a right triangle, then;

**c ^{2} = a^{2} + b^{2}**

Or we can also write it as;

**c = √(a ^{2} + b^{2})**

Which means hypotenuse is equal to the root of the sum of squares of base and perpendicular.

The area of a right triangle is equal to half of the product of base and altitude or perpendicular.

**A = 1/2 b x h**

where b is the base and h is the perpendicular length.

### Examples

Q.1: The angle ∠PRQ = 90∘ and RS is perpendicular to PQ. Prove that QR^{2}/PR^{2} = QS/PS.

Solution: We can see that;

△ PSR ~ △ PRQ

According to the property of similar triangles we have:

PR/PQ = PS/PR

or it can also be written as, PR^{2} = PQ.PS ……….(1)

Similarly, we have △ QSR ~ QRP

So we have, QS/QR = QR/QP

Or it can also be written as;

QR^{2} = QP.QS ……….. (2)

By dividing the equations (2) by (1) we can deduce that:

QR^{2}/PR^{2} = (QP.QS)/(PQ.PS) = QS/PS

Hence, proved.

**Q.2: If the measure of hypotenuse and base if 5cm and 3cm of a right triangle. Then find its height.**

Solution: By the formula of right angle triangle, we know;

(Hypotenuse)^{2} = (Adjacent side)^{2} + (Opposite side)^{2}

(5)^{2} = (3)^{2} + (Perpendicular)^{2}

25 = 9 + (P)^{2}

16 = (P)^{2}

P = √16 = 4cm

Hence, the measure of perpendicular is 4 cm.