A field is shaped like an equilateral triangl | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For an equilateral triangle with side length $a = 120 \, m$,the semi-perimeter $s$ is calculated as:$s = \frac{a + a + a}{2} = \frac{3a}{2} = \fra

Vedclass · Accepted Answer

$3600 \sqrt{3} \, m^2$

Vedclass · Answer

(A) For an equilateral triangle with side length $a = 120 \, m$,the semi-perimeter $s$ is calculated as:$s = \frac{a + a + a}{2} = \frac{3a}{2} = \frac{3 	imes 120}{2} = 180 \, m$.Using Heron's formula,Area $= \sqrt{s(s-a)(s-b)(s-c)}$.Since $a = b = c = 120 \, m$,we have:Area $= \sqrt{180(180-120)(180-120)(180-120)}$Area $= \sqrt{180 	imes 60 	imes 60 	imes 60}$Area $= \sqrt{180 	imes 60^3} = \sqrt{3 	imes 60 	imes 60^3} = \sqrt{3 	imes 60^4}$Area $= 60^2 \sqrt{3} = 3600 \sqrt{3} \, m^2$.

$A$ field is shaped like an equilateral triangle. Its length is $120 \, m$ on each side. Calculate its area using Heron's formula.

Similar Questions

Area of a square $= (........... )^{2}.$

In $\Delta ABC$,$AB = 20 \, cm$,$BC = 48 \, cm$,and $AC = 52 \, cm$,find the area of $\Delta ABC$. (in $, cm^2$)

An isosceles triangle has a perimeter of $50\,cm$ and each of the equal sides is $17\,cm$. Find the area of the triangle. (in $,cm^2$)

In a quadrilateral $ABCD$,$AB = 9 \, cm$,$BC = 40 \, cm$,$CD = 28 \, cm$,$DA = 15 \, cm$,and $\angle B = 90^{\circ}$. Find its area. (in $, cm^2$)

For an equilateral triangle,each of whose side is $a$,the length of any altitude of that triangle is ...........

Vedclass Test Series

Exam Paper Generator

Online Exam Module