Question

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

Brainliest if you solve in a book

Answer 1

check the attachment

Answer 2

Answer:

Step-by-step explanation:

Perimeter = 240 m

semi-perimeter (s) =$\frak{240}{2}$

=120 m

120 m = $\frac{a+b+c}{2}$

240 = a+b+c

240=78+50+c

240=128 + c

240-128 = c

112 = c

So, other side = 112 m.

Now, by Heron's formula

area = $\sqrt{s(s-a)(s-b)(s-c)}$

=$\sqrt{120(120-128)(120-78)(120-50)}$

=$\sqrt{120×-8×42×70}$

=$\sqrt{-2822400}$

=1680 (here - sign Will be removed as we take it out from the root)

NOW,

$\textbf{Area of square }$=1680 ${m}^{2}$

We need to find the corresponding altitude on 50 m side.

Area of triangle= $\frac{1}{2} ×\:base\:×\:height}$

1680=$\frac{1}{2}$ × 50 × h

$\frac{1680×2}{50}$=h

67.2 m = h

$\textbf{so, height of altitude on 50 m side is 67.2 m }$

$\huge{\textbf{Hope it helps :)}}$