Question

... In ABCD, I(AB) = 14
cm, I(BC) = 24 cm. I(CD)
= 15 cm, l(AD) = 13 cm
and I(DB) = 15 cm. Find
the area of ABCD


​

Answer 1

$\huge\star\mathfrak\blue{{Answer:-}}$

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 cm^2

↪ Again,area\:of\:triangle,

1/2×base×altitude=84 cm^2

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.

Answer 2

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 cm^2

↪ Again, area\:of\:triangle,

1/2×base×altitude=84 cm^2

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.