The perimeter of a triangular field …
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Rashmi Bajpayee 8 years, 3 months ago
Third side of given triangle = 240 - (78 + 50) = 240 - 128 = 112 dm
Semiperimeter of triangle (s) = {tex}{{240} \over 2}{/tex} = 120 dm
Now, Area of triangle = {tex}\sqrt {s\left( {s - a} \right)\left( {s - b} \right)\left( {s - c} \right)} {/tex}
= {tex}\sqrt {120\left( {120 - 78} \right)\left( {120 - 50} \right)\left( {120 - 112} \right)} {/tex}
= {tex}\sqrt {120 \times 42 \times 70 \times 8} {/tex}
= {tex}\sqrt {10 \times 10 \times 7 \times 7 \times 6 \times 6 \times 4 \times 4} {/tex}
= 10 x 7 x 6 x 4
= 1680 sq. dm
Now, again Area of triangle = {tex}{1 \over 2}{/tex}x Base x Height (Perpendicular)
=> 1680 = {tex}{1 \over 2}{/tex}x 50 x Perpendicular
=> Perpendicular = {tex}{{1680 \times 2} \over {50}}{/tex}
=> Perpendicular = 67.2 dm
0Thank You