Find the area of the triangle in which $a = 91 m, b = 98 m, c = 105 m$ , where $a, b, c$ are sides of triangle.

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Solution

Given:-
$a = 91 c m$
$b = 98 c m$
$c = 105 c m$
Semi-perimeter of triangle $(s) = \frac{a + b + c}{2}$
$\Rightarrow s = \frac{91 + 98 + 105}{2} = 147 c m$

Now, by using Heron's formula-

Area of triangle
$= \sqrt{s (s - a) (s - b) (s - c)} = \sqrt{147 (147 - 91) (147 - 98) (147 - 105)} = \sqrt{147 \times 56 \times 49 \times 42} = 4116 {c m}^{2}$

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

Assertion: Area of a triangle
$= \sqrt{s (s - a) (s - b) (s - c)}$ where a, b and c are the sides of the triangle and $s = \frac{a + b + c}{2}$

Reason: Area of a triangle $= \frac{1}{2} \times b a s e \times h e i g h t$
Which of the following is correct?

\sqrt{(b^{2} + c^{2})}, \sqrt{(c^{2} + a^{2})}, \sqrt{(a^{2} + b^{2})}

\sqrt{b^{2} + c^{2}}, \sqrt{c^{2} + a^{2}}, \sqrt{a^{2} + b^{2}}

a, b, c > 0

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