In a Triangle Abc Right Angled at C, P and Q Are Points of Sides Ca and Cb Respectively, Which Divide These Sides the Ratio 2 : 1. Prove that : 9(Aq2 + Bp2) = 13ab2 - Mathematics

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Sum

In a triangle ABC right angled at C, P and Q are points of sides CA and CB respectively, which divide these sides the ratio 2 : 1.
Prove that : 9(AQ2 + BP2) = 13AB2 

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Solution


P divides AC in the ratio 2 : 1

So C.P. = `(2)/(3) "AC"` .......(i)

Q divides BC in the ratio 2 : 1

QC = `(2)/(3)"BC"` ......(ii)

Adding (iii) and (iv), we get
9(AQ2 + BP2) = 13(BC2 + AC2)
⇒ 9(AQ2 + BP2) = 13AB2.

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Chapter 17: Pythagoras Theorem - Exercise 17.1

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Frank Class 9 Maths ICSE
Chapter 17 Pythagoras Theorem
Exercise 17.1 | Q 20.3

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