Question

The perimeter of a traingular field 240dm if two of its side are 78dm and 50dm find the length of the perpendicular on the side of length 50dm from the opposite vertex?

Answer 1

$\large\underline{Given:-}$

• Perimeter of triangle ⇢ 240 dm
• Two sides ⇢ 78 dm and 50 dm

$\large\underline{To find:-}$

• length of perpendicular.

$\large\underline{Solutions:-}$

• 1st side ⇢ 78 dm
• 2nd side ⇢ 50 dm
• 3rd side ⇢ ?

⇢ Perimeter of triangle ⟹ 240 dm

$\: \: \: \: \: \therefore Area \: \: of \: \: perimeter \: \: \leadsto \: \: a \: + \: b \: + \: c$

$\: \: \: \: \: \leadsto \: \: a \: + \: b \: + \: c \: \: = \: \: {240}$

$\: \: \: \: \: \leadsto \: \: {78} \: + \: {50} \: + \: c \: \: = \: \: {240}$

$\: \: \: \: \: \leadsto \: \: {128} \: + \: c \: \: = \: \: {240}$

$\: \: \: \: \: \leadsto \: \: c \: \: = \: \: {240} \: - \: {128}$

$\: \: \: \: \: \leadsto \: \: c \: \: = \: \: {112} \: dm$

$\: \: \: \: \: \therefore Area \: \: of \: \: triangle \: \: \leadsto \: \: \sqrt{s \: {(s \: - \: a)} {(s \: - \: b)}{(s \: - \: c)}}$

$\: \: \: \: \: \leadsto \: \: \sqrt{{120} \: {({120} \: - \: {78})} {({120} \: - \: {50})}{({120} \: - \: {112})}}$

$\: \: \: \: \: \leadsto \: \: \sqrt{{120} \: \times \: {42} \: \times \: {70} \: \times \: {8}}$

$\: \: \: \: \: \leadsto \: \: \sqrt{{5040} \: \times \: {560}}$

$\: \: \: \: \: \leadsto \: \: \sqrt{{2822400}}$

$\: \: \: \: \: \leadsto \: \: {1680} \: dm$

$\: \: \: \: \: \therefore Area \: \: of \: \: triangle \: \: \leadsto \: \: \frac{1}{2} \: \times \: b \: \times \: h$

$\: \: \: \: \: \leadsto \: \: {1680} \: \: = \: \: \frac{1}{2} \: \times \: {50} \: \times \: h$

$\: \: \: \: \: \leadsto \: \: {1680} \: \: = \: \: {25} \: \times \: h$

$\: \: \: \: \: \leadsto \: \: {h} \: \: = \: \: {1680}{25}$

$\: \: \: \: \: \leadsto \: \: {h} \: \: = \: \: {67.2} \: dm$

$\: \: \: \: \: So, \: \: The \: \: length \: \: perpetual \: \: is \: \: {67.2} \: dm.$

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