In ΔABC, AD is the median, find the length of AD if AC =7cm, BC =8 cm and AB =9 cm.
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)
whre 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√5cm2
So, area (ΔADB) =area (ΔADC)=6√5cm2
[since, median of a triangle divides it in two equal parks]. 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.