Question

In a triangle if the square of one side is equal to the sum of the squares of other two sides. Prove that the angle opposite to first side is right angle. Use the above theorem to find the measure of angle PKR.

Answer 1

HEY GUYS .............

Given : AC2 = AB2 + BC2

To prove : angle B = 90°

Proof : In triangle PQR

PR2 = PQ2 + QR2 ( Pythagoras theorem ) (1)

AC2 = AB2 + BC2 (2)

From (1) { Cons. AB = PQ, BC = QR}

PR2 = AB2 + BC2 (3)

From (2) and (3)

AC2 = PR2

AC = PR (4)

In triangle ABC And triangle PQR

AC = PR

AB = PQ

BC = RQ

Triangle ABC ~= PQR [ SSS]

CPCT Angle B = Q

Angle Q = 90°

Angle B = 90°

HENCE PROVED

THANKS. .........

I HOPE U UNDERSTAND THIS ........

Answer 2

Answer:

Given:- ABC is a triangle

AC²=AB²+BC²

To prove:- ∠B=90°

Construction:- Construct a triangle PQR right angled at Q such that, PQ=AB and QR=BC

Proof:-

In △PQR

PR² =PQ²+QR² (By pythagoras theorem

⇒PR²=AB²+BC² .....(1)(∵AB=PQ and QR=BC)

AC²=AB²+BC² .....(2) (Given)

From equation (1)&(2), we have

AC²=PR²

⇒AC=PR.....(3)

Now, in △ABC and △PQR

AB=PQ

BC=QR

AC=PR(From (3))

∴△ABC≅△PQR(By SSS congruency)

Therefore, by C.P.C.T.,

∠B=∠Q

∵∠Q=90°

∴∠B=90°

Hence proved.