Question

The base of an isosceles triangle is 12cm and perimeter is 32cm.find its area

Answer 1

ANSWER:
Let x be the equal side.
==>Perimeter:2x+12=32.
x=10cm.
Height of the base=8 (by solving using Pythagorean theorem).
Area of triangle=1/2*b*h.
Area of triangle =1/2*12*8.
Area of triangle=48cm^2.

Answer 2

The area of the isosceles triangle is $\bold{48\ \mathrm{cm}^{2}}$.

Given:

Isosceles triangle’s base is, b=12 cm. (c side)

The Perimeter of the given isosceles triangle is, =32 cm

To find:

By using the given dimensions, to find the isosceles triangles area.

Solution:

For finding the length of the two equal sides, $x=\frac{\text { perimeter base}}{2}$

$\begin{aligned} x=& \frac{32-12}{2} \\ &=\frac{20}{2} \end{aligned}$

x=10cm

x is the length of the two equal sides a and b

By using Heron’s formula for finding area of triangle,

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

To find S,

$\begin{array}{l}{s=\frac{(a+b+c)}{2}} \\ {=\frac{10+10+12}{2}}\end{array}$

s=16 cm.

Then,  

$\begin{array}{c}{A=\sqrt{s(s-a)(s-b)(s-c)}} \\ {=\sqrt{16(16-10)(16-10)(16-12)}}\end{array}$

$\begin{array}{c}{=\sqrt{16 \times 6 \times 6 \times 4}} \\ {=\sqrt{2304}} \\ {=48 \mathrm{cm}^{2}}\end{array}$

The given isosceles triangle’s area is $\bold{48\ \mathrm{cm}^{2}}$.