Question

find the area of an isosceles triangle whose perimeter is 32cm and non-equal side is 12 cm respectively.​

Answer 1

Given:-

In an isosceles triangle,

• Perimeter = 32 cm
• Non-equal side = 12 cm

The sum of the remaining sides is $32-12=20$ cm.

Each equal side is $\dfrac{20}{2}=10$ cm.

The isosceles triangle has three sides 12 cm, 10 cm, 10 cm.

If we draw a leg from a vertex angle, the triangle is divided into two congruent right triangles.

By Pythagorean theorem,

$\rightarrow h^2=10^2-6^2$

$\rightarrow h^2=16\times 4$

$\rightarrow h^2=8^2$

$\rightarrow h=8(\because h>0)$

Hence, base and height are 12 cm and 8 cm.

The area of the triangle is,

½ × base × height

= ½ × 12 × 8

= 48 cm².

Answer 2

$\bf\fbox\red{Answer}$

The area of the isosceles triangle is 48cm^2

$\bf\fbox\green{explanation: }$

It is given that,

• The perimeter of the isosceles triangle= 32 cm
• The base of the triangle = 12 CM

Therefore,

The sum of the equal sides of the triangle

=(32-12)cm

= 20 Cm

Therefore,

$\sf{The \:length\: of\: each \: equal \: side \:( \frac{20}{2}) cm} \\ = 20cm$

Now, imagine a perpendicular through the mid point of the triangle such that it bisects the base.

Therefore,

we get 2 right angled triangles.

For each equal triangle:

$\sf{Hypotenuse \:=10cm}$

$\sf{Base\:= \frac{12}{2} cm} \\ = 6 \: cm$

therefore,

$\sf\pink{Height\: = \sqrt{hypotenuse {}}^{2} - base {}}^{2}$

[Pythagoras theorem]

$\sf{ = \sqrt{10} {}^{2} - 6 {}^{2} }$

$\sf{ = \sqrt{100} - 36}$

$\sf{ = \sqrt{64} } \\ \sf{ = 8 \: cm}$

Therefore,

For the isosceles triangle we get,

$\sf{height = 8cm}$

Therefore,

$\sf{ \: area \: = \frac{(base \times height)}{2} }$

$\sf{ = \frac{(12 \times 8) {} }{2} cm {}}^{2}$

$\sf{ = \frac{96}{2} \: cm {}^{2} }$

$\sf{48 \: cm {}^{2} }$

Therefore,

$\small\bf\purple{The\:area\:of\:the\:triangle\:is \: 48 {}^{2} }$