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