Question

find thearea of a triangular field whose sides are 91m,98 and 105 in length .find the height corresponding to the longest side

Answer 1

Here, a = 91 m, b = 98 m and c =105 m

Therefore, S=91+98+1052=147

Area = S(S−a)(S−b)(S−c)−−−−−−−−−−−−−−−−−−√

= 147(147−91)(147−98)(147−105)−−−−−−−−−−−−−−−−−−−−−−−−−−−−√

= 147×56×49×42−−−−−−−−−−−−−−−√

= 49×3×7×2×2×2×49×7×3×2−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√

= 49×3×2×2×7 = 4116 m2

Longest side = 105 m and base = 105 m

Let, h be the height corresponding to the longest side.

Area of the triangle = (12×base×height) sq units

⇒12×base×height=4116m2 ⇒105×height=2×4116

Therefore, height (h)  = 78.4 m

Answer 2

let a equal to 91 and B equal to 98 and equal to 105 then as equal to 1 by 2 into a + b + c