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

1.	The perimeter of a triangular field is 240 dm. If two of its 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» Let the third side of the triangle is x Since the perimeter of a triangle is given by: a+b+c = perimeter 78+50+x = 240 x = 240 - 128 x = 112^dm Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by: A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(\frac{a+b+c}{2}\)[Heron’s Formula] a = 78, b = 50, c = 112 s = \(\frac{a+b+c}{2}\) = \(\frac{78+50+112}{2}\) = 120 A = \(\sqrt{120(120-78)(120-50)(120-112)}\) A = \(\sqrt{120\times42 \times70 \times8}\) = 1680 dm² Area of triangle = \(\frac{1}2\)(Base x Altitude) 1680 = \(\frac{1}2\)(50 x Altitude) Altitude = 67.2 dm

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

Let the third side of the triangle is x

Since the perimeter of a triangle is given by:

a+b+c = perimeter

78+50+x = 240

x = 240 - 128

x = 112^dm

Let a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

A = \(\sqrt{s(s-a)(s-b)(s-c)}\)where s = \(\frac{a+b+c}{2}\)[Heron’s Formula]

a = 78, b = 50, c = 112

s = \(\frac{a+b+c}{2}\) = \(\frac{78+50+112}{2}\) = 120

A = \(\sqrt{120(120-78)(120-50)(120-112)}\)

A = \(\sqrt{120\times42 \times70 \times8}\) = 1680 dm²

Area of triangle = \(\frac{1}2\)(Base x Altitude)

1680 = \(\frac{1}2\)(50 x Altitude)

Altitude = 67.2 dm