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Question

P and Q are points on sides CA and CB respectively of Δ ABC right angled at C. Prove that AQ2 + BP2 = AB2 + PQ2 [4 MARKS]

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Solution

Concept : 1 Mark
Application : 1 Mark
Proof : 2 Marks




In right- angled triangle ACQ,

AQ2 = AC2 + CQ2....(1)

In right- angled triangle PCB,

PB2 = PC2 + CB2......(2)

In right- angled triangle ABC,

AB2=AC2+BC2.......(3)

In right- angled triangle PQC,

and PQ2=PC2+QC2......(4)

Adding (1) and (2)

AQ2 + BP2 = (AC2 + CQ2) + (PC2 + CB2)

AQ2 + BP2 = (AC2 + BC2) + (PC2 + QC2)

AQ2 + BP2 = AB2 + PQ2

[By (3) and (4)]


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