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Question

# In the given figure, DE || BC, AE = 15cm, EC = 9cm, NC = 6 cm & BN = 24 cm. The value of ME & DM are

A

3.75 & 15.75 cm respectively

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B

3 cm & 15 cm respectively

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C

3.75 & 15 cm respectively

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D

3 cm & 15.75 cm respectively

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Solution

## The correct option is C 3.75 & 15 cm respectively In ΔAME & ΔANC ∠A=∠A (common) ∠AEM=∠ACM (corresponding angle) So ΔAME∼∠ANC (AA similarity) AMAN=AEAC=MENC (corresponding sides) ------- (1) AEAE+EC=MENC1524=ME6 ME = 3.75 cm Now in ΔADM & ΔABN ∠A=∠A (common) ∠ADM=∠ABN (corresponding angle) DMBN=AMAN but from (1) AMAN=AEAC So DMBN=AEAC DM24=1524 ⇒ DM=15cm

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