Question

find the area of isoceles triangle whoes equal sides is 5 cm each and the third side is 8cm​

Answer 1

Answer:

The answer will be 12cm^2.

Step-by-step explanation:

We have given;

Measure of two sides of triangle = 5cm

measure of third side = 8cm

We know that a perpendicular bisector in an isosceles triangle bisects the third side in two equal part.

So, here the measure of bisected third side will be;

= 8/2;

= 4cm.

Now, measure of the perpendicular bisector will be;

(5)^2 = (4)^2 + (PB)^2; ( using pythgoreaus theorem and PB is the perpendicular bisector)

25 = 16 + (PB)^2;

(PB)^2 = 9;

PB = 3cm

Thus, the area of triangle equals;

= 1/2 × (base) × (height)

= 1/2 × 8 × 3

= 4×3

= 12cm^2.

Here, we can even use Heron's formula which states;

area of triangle = √s(s-a)(s-b)(s-c);

where, s is the semi-perimeter

a = side 1

b= side 2

c = side3.

That's all.

Answer 2

Answer:

$\red\bigstar$ Area of the triangle whose equal sides is 5 cm each and the third side is 8cm​ = 12 cm².

SOLUTION

Equal Sides of the isosceles triangle = 5cm.

The Third side of the isosceles triangle = 8cm.

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(1, 0)(1,0)(3,3)\qbezier(5,0)(5,0)(3,3)\qbezier(5,0)(1,0)(1,0)\put(2.85,3.2){ \bf A }\put(0.5,-0.3){ \bf C }\put(5.2,-0.3){ \bf B }\end{picture}$

Let us Assume that AC and AB are the equal sides , then

AC = AB = 5cm.

Also,

BC = 8cm.

$\sf Area \; of \: a \: triangle = \sqrt{s(s-a)(s-b)(s-c)}$

Here,

s is the semi - perimeter of the triangle.

Let us Assume that ,

a = AC = 5cm.

b = AB = 5cm.

c = BC = 8cm.

$\sf s = \dfrac{a+b+c}{2}$

$\implies$

$\sf s = \dfrac{5+5+8}{2}$

$\implies$

$\sf s = \dfrac{18}{2}$

$\implies$

Semi - Perimeter (s) = 9cm.

Then,

(s-a) = 9 - 5 = 4cm.

(s-b) = 9 - 5 = 4cm.

(s-c) = 9 - 8 = 1cm.

$\implies$

$\sf Area \; of \: the \: triangle = \sqrt{9*4*4*1}$

$\implies$

$\sf Area \; of \: the \: triangle = \sqrt{144}$

$\implies$

Area of the triangle = 12cm².