Question

find the area of triangular park whose sides are of length 120m 80m 50m ?

Answer 1

semi-perimeter=120+80+50÷2=125m
area=√s(s-a)(s-b)(s-c)
answer=150√675m

Answer 2

Answer:

• Sides - 120m, 80m, 50m

To Find:

• Area of the Triangle ?

We have Heron's Formula to find the area of the Triangle;

$\bullet{\boxed{\sf{ \sqrt{s(s-a)(s-b)(s-c) } }}}$

Where,

$\\ {\boxed{\sf{ s = \dfrac{a+b+c}{2} }}}$

${\sf{ s = \dfrac{120+80+50}{2} }}$

${\sf{ s = \dfrac{ \cancel{250}^{ \: \: 125}}{ \cancel{2}} }}$

${\sf{ s = 125 }}$

Solution:

Substitute the values, we get:

$\\ \colon\implies\sf{ \sqrt{ 125(125-120)(125-80)(125-50)} }$

$\\ \colon\implies\sf{ \sqrt{ 125(5)(45)(75)} }$

$\\ \colon\implies\sf{ \sqrt{ 2109375}}$

$\\ \colon\implies\sf{ 375 \sqrt{15} m^2}$

Hence,

Area of the Triangular Park is 375√15 m².