Question

the base of an isosceles triangle is 12 centimetre and its perimeter is 32 cm find its area

Answer 1

let the other 2 side be x
x + x + 12 = 32
2x = 20
x=10

let AD perpendicular to BC is drawn in triangle ABC..
AD^2= AC^2-CD^2
=10^2-6^2
= 100-36
=64
AD=ROOT UNDER 64
AD=8CM
therefore
area of the triangle=1/2 * BC * AD
=1/2 * 12 * 8
=48CM^2

Answer 2

Hey friend,
Here is the answer you were looking for:
Given,
base of isosceles triangle is 12cm
Now we know that perimeter of isosceles triangle= a+b+c
c= 12 cm
a = b ( as two opposite sides of isosceles triangle are equal)
P = a + b + c
32= 2a + 10. ( as a = b )
32-10 = 2a
22 = 2a
22/2 = a
11 = a
a= 11 cm
Now,
a=11 cm
b=11 cm
So by heron's formula,
A = √ s(s - a) (s - b) (s - c)
2s = P
2s = 32
s = 16
A = √ 16 (16 - 11) (16 - 11) (16 - 12)
A = √ 16 × 5 × 5 × 4
A = √ 1600
A = 40 cm^2
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