Q. 29

# In a right triangle ABC right-angled at B, if P and Q are points on the sides AB and AC respectively, then

A. AQ^{2} + CP^{2} = 2 (AC^{2}+ PQ^{2})

B. 2 (AQ^{2} + CP^{2}) = AC^{2} + PQ^{2}

C. AQ^{2} + CP^{2} = AC^{2} + PQ^{2}

D. AQ + CP=(AC + PQ).

Answer :

Given in right triangle ABC right-angled at B, P and Q are points on the sides AB and BC respectively.

Applying Pythagoras Theorem,

In ΔAQB,

⇒ AQ^{2} = AB^{2} + BQ^{2} … (1)

In ΔPBC,

⇒ CP^{2} = PB^{2} + BC^{2} … (2)

Adding (1) and (2),

⇒ AQ^{2} + CP^{2} = AB^{2} + BQ^{2} + PB^{2} + BC^{2} … (3)

In ΔABC,

⇒ AC^{2} = AB^{2} + BC^{2} … (4)

In ΔPBQ,

⇒ QP^{2} = PB^{2} + BQ^{2} … (5)

From (3), (4) and (5),

∴ AQ^{2} + CP^{2} = AC^{2} + PQ^{2}

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