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