The perimeter of a triangular field is 240 dm. If two of its sides are 78 dm and 50 dm,
Question:

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.

Solution:

Given,

In a triangle ABC, a = 78 dm = AB, b = 50 dm = BC

Now, Perimeter = 240 dm

Then, AB + BC + AC = 240 dm

78 + 50 + AC = 240

AC = 240 – (78 + 50)

AC = 112 dm = c

Now, 2s = a + b + c

2s = 78 + 50 + 112

s = 120 dm

Area of a triangle $\mathrm{ABC}=\sqrt{s \times(s-a) \times(s-b) \times(s-c)}$

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

$=1680 \mathrm{dm}^{2}$

Let AD be a perpendicular on BC

Area of the triangle ABC = 1/2 × AD × BC

$1 / 2 \times \mathrm{AD} \times \mathrm{BC}=1680 \mathrm{dm}^{2}$

AD = 67.2 dm