Question

In A ABC, AD and BE are altitudes. If AD = 6 cm, BE = 9 cm and BC = 15 cm, then find AC.​

Answer 1

Step-by-step explanation:

In triangle ADC

by pythagoras theorem

H²= P²+B²

AC² = AD² +DC²

AC² = 6² + 7.5²

= 36.+ 15

= 51

AC. = √51

Answer 2

Answer:

In ∆ABC, AB = 15 cm , BC = 13 cm and AC = 14 cm.

By Heron's formula ,

S = ( A + B + C ) / 2

S = ( 15 + 13 + 14 ) / 2

S = 42 / 2

S = 21

Now, area of triangle

= √S (S–A) (S–B) (S–C)

= √21 (21–15) (21–13) (21–14)

= √21 (6) (8) (7)

= √7056

= 84 cm2

Again,area of triangle = 1/2×base×altitude=84 cm2

1/2×14×altitude = 84

7 × altitude=84

altitude = 84/7

Altitude = 12cm

Therefore, the area of ∆ABC is 84cm² and its altitude on AC is 12 cm.