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Question

# In ΔABC, AD is the median, find the length of AD if AC =7 cm, 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 AE⊥DC Given: AB = 9 cm BD =DC =4 cm [AD is median] AC =7 cm Area of Δ=√s(s−a)(s−b)(s−c) where s=a+b+c2 Here, a=7 b=8 c=9, s=7+8+92=12 ∴ Area (ΔABC)=√12(12−7)(12−8)(12−9) =√12×5×4×3 =√12×12×5 =12√5 cm2 So, area (ΔADB) =area (ΔADC)=6√5cm2 [Since median of a triangle divides it in two equal parts]. Area (ΔADC)=6√5cm2 12×DC×AE=6√5 12×4×AE=6√5 AE=3√5cm In ΔAEC, using Pythagoras theorem, AE2+EC2=AC2 ∴EC2=AC2−AE2 ⇒EC=√72−(3√5)2 =√49−45 =√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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