Question

the sides of a triangle are 35, 54,and 61. find it area without height​

Answer 1

Answer:

$\sf The\; Area \; of \;the \;Triangle = 420 \sqrt{5} \ cm\²$

Step-by-step explanation:

Let the sides  of  triangle  be a , b and  c

Here,

a = 35 cm

b= 54 cm

c = 61 cm

Let the semiperimeter of the triangle be S

∴ S = a+b+c​  

            2

S = 35+ 54+ 61  

                2

S = 75 cm

By Heron's Formula,

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

A = $\sqrt{75(75-35)(75-54)(75-61)}$

A = $\sqrt{75(40)(21)(14)}$

A =$\sqrt{882000$

A = 420√5 cm²