Question

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

Answer 1

hope it's helpful to you. . . . . .

Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Here we have,

AB = 15 cm

BC = 13 cm

AC = 14 cm

Now,

Assume,

Sides :-

p = 15 cm

q = 13 cm

r = 14 cm

Now,

Semi Perimeter,

s = (p + q + r)/2

s = (15 + 13 + 14)/2

s = 42/2

s = 21 cm

Now

Using Heron formula :-

A = √s(s - a)(s - b)(s - c)

A = √21(21 - 15)(21 - 13)(21 - 14)

A = √21 × 6 × 8 × 7

A = √(3 × 7) × (2 × 3) × (2 × 2 × 2) × (7)

A = √(3 × 3) × (7 × 7) × (2 × 2 × 2 × 2)

A = 3 × 7 × 2 × 2

A = 84 cm²

So,

Area of triangle = ½ x base × altitude

Putting the values we get,

84 = ½ × 14 × altitude

altitude = (84 × 2)/14

altitude = 6 × 2

altitude = 12 cm