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

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