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The right-angle triangle theorem is nothing but the Pythagoras theorem. It deals with the relationship between the adjacent side, hypotenuse and base of a right-angled triangle.
According to the right-angle triangle theorem: The square of the hypotenuse in a right-angled triangle is equal to the sum of squares of its adjacent side and base.
Mathematically, we can denote the theorem as:
Hypotenuse² = Adjacent Side² + Base²
Hypotenuse is the largest side in a right-angled triangle and is opposite to the 90 degree angle. The base is the shortest side in the triangle.
Pythagoras theorem is one of the most widely used theorems in mathematics. There are ample ways to prove the Pythagoras theorem. The two most common methods used to prove Pythagoras Theorem are algebraic methods and using similar triangles. We will focus on learning how to prove the Pythagoras theorem using similar triangles.
First, let us see what similar triangles are. The triangles whose corresponding angles or sides are equal or proportional to each other are similar. Moreover, we can use sine or cosine law to prove the similarity of the triangles as well.
Consider a triangle ABC with another triangle inside, ABD. ∠B and ∠D are right angles, and ∠A is common in both triangles.
In triangles ABD and ABC,
1. ∠A is the common angle. Therefore, ∠A = ∠A.
2. ∠B = ∠D because both are right angles.
'Therefore, triangles ABC and ABD are similar to each other from AA (angle-angle) criterion. Also, AB/CA = DA/BA from the similarity criterion. Therefore, CA x DA = BA2.
Now, from triangles ABC and BDC,
1. ∠C = ∠C, due to common angle in both triangles.
2. ∠B = ∠D, both are right angles.
Therefore, both the triangles ABC and BDC are similar to each other. Also, DC/BC = CB/CA from the similarity criterion. Therefore, DC x CA = CB2.
Using both the similarity conditions, we get,
AC2 = AB2 + BC2, which is the required Pythagoras theorem.
Prove that QR2/PR2 = QS/PS if the angle ∠PRQ = 90∘ and RS is perpendicular to PQ in a right angled triangle.
Solution:
We can observe,
△PSR ~ △PRQ
According to the property of similar triangles we have:
PR/PQ = PS/PR
It can be rewritten as: PR2 = PQ . PS (1)
Similarly, we have △ QSR ~ QRP
QS/QR = QR/QP QR2 = QP.QS (2)
Dividing equation (2) by (1), we get,
QR2/PR2 = (QP . QS) / (PQ . PS) = QS/PS
Hence, proved.
We know, the right angle triangle formula is denoted by:
(Hypotenuse)² = (Adjacent side)² + (Base)²
If a, b and c are perpendicular, base and hypotenuse of a right triangle, respectively, then;
c² = a² + b²
The above equation can be rewritten as:
c = √(a² + b²)
It 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.