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In a Triangle Abc Right Angled at C, P and Q Are Point of Sides Ca and Cb Respectively, Which Divide These Sides the Ratio 2 : 1. Prove That: 9aq2 = 9ac2 + 4bc2 - Mathematics

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Question

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: 9AQ2 = 9AC2 + 4BC2 

Sum
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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)

In  ΔACQ
Using Pythagoras Theorem we have,
AQ2 + AC2 + CQ2

⇒ AQ2 = `"AC"^2 + (4)/(9)"BC"^2`    ...(using (ii))

⇒ 9AQ2 = 9AC2 + 4BC2.    ......(iii)

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Chapter 17: Pythagoras Theorem - Exercise 17.1

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Frank Mathematics [English] Class 9 ICSE
Chapter 17 Pythagoras Theorem
Exercise 17.1 | Q 20.1

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