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In a ΔABC, AB = 15 cm, BC = 13 cm and AC = 14 cm. Find the area of ΔABC and hence its altitude on AC ?

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#### Solution

Now ,

`2s+a+b+c`

`⇒S=1/2(a+b+c)`

`⇒s=((15+13+14)/2)cm`

`⇒=21cm`

∴area of a triangle= `sqrt(s(s-a)(s-b)(s-c))`

`=sqrt(21(21-15)(21-14)(21-13)) cm^2`

`=sqrt(21xx6xx8xx7)cm^2`

`=84cm^2`

Let BE be perpendicular (⊥er) to AC

Now, area of triangle = `84 cm^2`

`⇒1/2xxBExxACxx84`

`⇒BE=(84xx2)/(AC)`

`BE=168/14=12cm`

∴Length of altitude on AC is 12 cm

Concept: Area of a Triangle

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