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The perimeter of a triangular field is 240dm. If two of its si

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The perimeter of a triangular field is 240dm. If two of its sides are 78dm and 50dm, find the length of the perpendicular on the side of length 50dm from the opposite vertex.

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The perimeter of a triangular field is 240dm. If two of its sides are 78dm and 50dm, find the length of the perpendicular on the side of length 50dm from the opposite vertex.

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Text solutionVerified

Solution: AB+BC+AC=240
78+50+AC=240
AC=240-128
AC=112
AD=9
area

(∠ A B C) = \frac{1}{2} \cdot A D \cdot 50 = 25 A D

s = \frac{a + b + c}{2} = \frac{240}{2} = 120

area

= s (s - a) (s - b) (s - c)

= 120 (120 - 50) (120 - 112) (120 - 75)

= 120 \cdot 70 \cdot 8 \cdot 42

= 1680

25 A D = 1680

A D = 67.2 d m

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Question Text	The perimeter of a triangular field is 240dm. If two of its sides are 78dm and 50dm, find the length of the perpendicular on the side of length 50dm from the opposite vertex.
Answer Type	Text solution:1
Upvotes	150

The perimeter of a triangular field is 240dm. If two of its sides are 78dm and 50dm, find the length of the perpendicular on the side of length 50dm from the opposite vertex.

Text solutionVerified

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