Question

7. P and Q are the mid points of the sides CA and CB respectively of a AABC right
angled at C. Prove that
(i) 4AQ2 = 4AC2 + BC2​

Answer 1

Answer:

Step-by-step explanation:In ACB, C = 90o, using Pythagoras theorem,

AB2 = AC2 + BC2 ………………… (i)

In ACQ, C = 90o, using Pythagoras theorem, AQ2 = AC2 + CQ2  

4AQ2 = 4AC2 + BC2 ………………..(ii)

In PCB, C = = 90o, using Pythagoras theorem,

PB2 = PC2 + BC2

PB2 =  

4PB2 = AC2 + 4BC2 ………………(iii)

From (ii) and (iii), 4(AQ2 + PB2) = 5AC2 + 5BC2 = 5(AC2 + BC2) = 5AB2

[Using (i)]

Answer 2

Question

P and Q are the mid points of the sides CA and CB respectively of a AABC right

angled at C. Prove that

4AQ² = 4AC² + BC²

Theorem to be used ( Pythagorean triplet)

•In a right angled triangle

Hypotenus² = Base² + perpendicular ²

Construction

AQ and PQ are joined

Proof

Given data

• P and Q are the mid points of the sides CA and CB

• CQ = BQ = (1/2)BC

•AP = CP = (1/2 )AC

• angle ACB = 90°

From triangle ACQ we have

${AQ}^{2} = {AC}^{2} + {CQ}^{2} \\ \implies AQ {}^{2} = {AC}^{2} + (\frac{1}{2} {BC)}^{2} \\ \implies AQ {}^{2} = {AC}^{2} + \frac{1}{4} {BC}^{2} \\ \implies {AQ}^{2} = \frac{4 {AC}^{2} + {BC}^{2} }{4} \\ \implies 4{AQ}^{2} = 4 {AC}^{2} + {BC}^{2}$

$Hence \: \: Proved$