Q29 of 95 Page 5

In the given figure, 0 is a point inside a ∠PQR such that ∠POR = 90°, OP = 6 cm and OR= 8 cm. If PQ = 24 cm and QR = 26 cm, prove that ΔPQR is right angled. P

Given: ∠POR = 90°, OP = 6 cm and OR= 8 cm

and PQ = 24 cm and QR = 26 cm


To Prove: PQR is right angled at P


In ∆POR, using Pythagoras theorem, we get


(Perpendicular)2 + (Base)2 = (Hypotenuse)2


⇒ (PO)2 + (OR)2 = (PR)2


⇒ (6)2 + (8)2 = (PR)2


⇒ 36 +64 = (PR)2


⇒ (PR)2= 100


⇒ PR =√100


⇒ PR = 10 [taking positive square root]


In ∆PQR


Using Pythagoras theorem, i.e. if the square of the hypotenuse is equal to the sum of the other two sides. Then, the given triangle is a right angled triangle, otherwise not.


Here, (PR)2 + (PQ)2


⇒ (10)2 + (24)2


= 100 + 576


= 676


= (26)2 = (QR)2


∴ given sides 10cm, 24cm and 26cm make a right triangle right angled at P.


Hence Proved


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