Q2 of 65 Page 150

PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM×MR

Let, ∠MPR = x



In triangle MPR,


∠MRP = 180o – 90o – x


∠MRP = 90o – x


Similarly,


In triangle MPQ,


∠MPQ = 90o - ∠MPR


∠MPQ = 90o – x


In triangle QMP and PMR,


∠MPQ = ∠MRP


∠PMQ = ∠RMP


∠MQP = ∠MPR


Therefore,


ΔQMP ~ΔPMR (By Angle-Angle-Angle similarity)



PM2 = QM×MR

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Sides of two similar triangles are in the ratio 4: 9. Areas of these triangles are in the ratio

A. 2 : 3 B. 4 : 9


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Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.

(i) 7 cm, 24 cm, 25 cm


(ii) 3 cm, 8 cm, 6 cm


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(iv) 13 cm, 12 cm, 5 cm

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In Fig. 6.53, ABD is a triangle right angled at Aand AC ⊥ BD. Show that:

(i) AB2 = BC . BD


(ii) AC2 = BC . DC


(iii) AD2 = BD . CD


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ABC is an isosceles triangle right angled at C. Prove that