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Question

In ΔABC, AD is the median, find the length of AD if AC =7cm, BC =8 cm and AB =9 cm.


A

75cm

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B

7 cm

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C

85cm

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D

8 cm

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Solution

The correct option is B

7 cm


Construction: Draw AEDC

.... Given: AB = 9 cm

BD =DC =4 cm [AD is median] AC =7 cm Area of Δ=s(sa)(sb)(sc)

whre s=a+b+c2

Here, a=7 b=8 c=9, s=7+8+92=12

Area (ΔABC)=12(127)(128)(129)

=12×5×4×3

=125cm2

So, area (ΔADB) =area (ΔADC)=65cm2

[since, median of a triangle divides it in two equal parks]. Area (ΔADC)=65cm2

12×DC×AE=65

12×4×AE=65

AE=35cm

In ΔAEC, using Pythagoras theorem,

AE2+EC2=AC2

EC2=AC2AE2

EC=72(35)2

=4945

=4

=2

or EC =2 cm.

DC =4 cm, EC =2 cm DE=2cm.

This means, ΔADC must be an isosceles triangle. Since, perpendicular AE bisects the base CD.

Hence AC =AD =7 cm.


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