Question

1. Prove by theorem the relation between three sides of the triangle, if one of the angle of triangle is 90 °

Answer 1

Answer:

Here, only one angle is 90 degrees and the sum of other triangles is equal to 90 degrees, which are acute angles. ... 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.

Answer 2

Answer:

this theorem is called Pythagoras 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 AC2 = AB2 + BC2

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

PR2 = PQ2+ QR2 (According to Pythagoras theorem,as ∠Q = 90°)

or, PR2 = AB2 + BC2 (By construction) …… (1)

We know that;

AC2 = AB2+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.

Step-by-step explanation:

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