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Congruency of Triangles

In the adjoin...

Question

In the adjoining figure, ABCD and PQRC are rectangles, where Q is the midpoint of AC. Then DP is equal to

DP < PC

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DP > PC

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DP = PC

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DP = ¹/₃ DC

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Solution

The correct option is C
DP = PC

CRQ = ∠CBA = 90^o => QR || AB.

Therefore R is the midpoint of BC.

Similarly P is the midpoint of DC.

Therefore DP = PC.

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Similar questions

P R = \frac{1}{2} A C

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In the figure, ABCD and PQRC are rectangles and Q is the mid-point of AC.

Prove that (i) DP = PC (ii) $P R = \frac{1}{2} A C$ .

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