Question

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

Answer 1

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Answer 2

$\Large{\textbf{\underline{\underline{According\;to\;the\;Question}}}}$

Perimeter = 240 dm

Sides = 78 dm , 50 dm

Assume,

a = 78 dm

Also,

b = 50 dm

Perimeter of Triangle = a + b + c

240 = 78 + 50 + c

240 = 128 + c

c = 240 - 128

c = 112 dm

Now,

Semi Perimeter of the Triangle,

s = (a + b + c)/2

s = (78 + 50 + 112)/2

s = 240/2

s = 120 dm

Now,

Using Heron formula :-

A = √s(s - a)(s - b)(s - c)

A = √120(120 - 78)(120 - 50)(120 - 112)

A = √120 × (42) × (70) × (8)

A = √(10 × 12) (6 × 7) × (7 × 10) × (2 × 2 × 2)

A = √(10 × 10) × (6 × 2) × (6) × (7 × 7) × (2 × 2 × 2)

A = √(10 × 10) × (6 × 6) × (7 × 7) × (2 × 2 × 2 × 2 )

A = 10 × 6 × 7 × 2 × 2

A = 10 × 168

A = 1680 dm²

So,

Now,

Area of triangle,

A = ½ x Base x altitude

1680 = ½ × 50 × altitude

altitude = (1680 × 2)/50

altitude = 1680/25

altitude = 67.2 dm