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.

Given:

The perimeter of a triangular field is $240\ dm$. Two of its sides are $78\ dm$ and $50\ dm$.

To do:

We have to find the length of the perpendicular on the side of length $50\ dm$ from the opposite vertex.

Solution:

Perimeter of the triangular field $= 240\ dm$
Two sides of the triangle are $a=78\ dm$ and $b=50\ dm$

Third side of the triangle $c= 240 - (78 + 50)$

$= 240 - 128$

$= 112\ dm$

$s=\frac{\text { Perimeter }}{2}$

$=\frac{240}{2}$

$=120$

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

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

$=\sqrt{120 \times 42 \times 70 \times 8}$

$=\sqrt{2 \times 2 \times 2 \times 3 \times 5 \times 7 \times 2 \times 3 \times 7 \times 2}{\times 5 \times 2 \times 2 \times 2}$

$=2 \times 2 \times 2 \times 2 \times 3 \times 5 \times 7 \mathrm{dm}^{2}$

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

Length of the perpendicular on $50 \mathrm{~m} =\frac{\text { Area } \times 2}{\text { Base }}$

$=\frac{1680 \times 2}{50}$

$=\frac{336}{5}$

$=65.2 \mathrm{dm}$

The length of the perpendicular on the side of length $50\ dm$ from the opposite vertex is $65.2\ dm$.

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Updated on: 10-Oct-2022

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