Question

GUYS.... Plzz ... anyone help me doing this ques......
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Find the area of a triangle whose two sides are 18cm, 10cm, & the perimeter is 42 cm.

Answer 1

length of first side of a triangle (a) = 18cm
length of 2nd side of the triangle (b)= 10cm
let the length of 3rd side be c cm
perimeter = 42 cm
=> sum of all sides of triangle = 42
=> c+ 18 + 10 = 42
=> c = 42- 28
=> c = 14
so, length of 3rd side of triangle = 14 cm
now, s = 42/2
s = 21 cm
by using Heron's formula
area = √s(s-a)(s-b)(s-c)
= √21(21-18)(21-10)(21-14)
=√21(3)(11)(7)
= √21*231 = √4851 = 69.6 cm²

Answer 2

first side of triangle is = 18cm
second side of triangle is = 10cm
let the third side be = x
sum of three sides is equal to its perimeter
18 + 10 + x = 42
28 + x = 42
x = 42 - 28
x = 14
so the third side of triangle is equal to 14cm