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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 A

3.75 & 15 cm respectively


In ΔAME & ΔANC

A=A (common)

AEM=ACM (corresponding angle)

So ΔAMEANC (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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