Hi Tahrima.

Draw the triangle with the shared vertex at the top and you can drop a bisector from that to the opposite side. This divides the triangle into two congruent right triangles, with that new line the height of the triangle.

The area of a triangle is (1/2) base times height. Let b = the base of the isosceles triangle and h be its height. Then
12 = (1/2)bh

You can also see the right triangles and notice that Pythagoras' Theorem must work here, so
5² = (b/2)² + h².

So you have two equations with two unknown values h and b (which you want to know). So get rid of the h by using the substitution method:

12 = (1/2)bh
24/b = h
576/b² = h²

Substitute this into the Pythagorean equation we wrote earlier:
5^2 = (b/2)² + h²
25 = b²/4 + 576/b²
(1/4)b⁴ - 25b² + 576 = 0
b⁴ - 100b² + 2304 = 0

Which is a quadratic in b^2 (this is just a formal way of saying a quadratic where the variable's value is a square of what we really want). It is the same as

x² - 100x + 2304 = 0 where x = b².

Which we can solve using "completing the square" method:
x² - 100x + 2500 = 2500 - 2304
(x - 50)² = 196
x - 50 = +/- 14
x = 50 +/- 14 = { 36, 64 }

But recall that we want b, not x. and b = sqrt(x). Both values of x are positive, therefore we have to carry two possibilities into this calculation:

b = sqrt ( { 36, 64 } ) = { 6, 8 }

Now I need to check these values. First I will check b = 6 cm:

h = 24 / b = 24 / 6 = 4 cm.

(b/2)² + h² = 3² + 4² = 9 + 16 = 25.

And that's the same as 5², so it works. Now check b = 8 cm:

h = 24 / b = 24 / 8 = 3 cm.

(b/2)² + h² = 4² + 3² = 16 + 9 = 25. That's right.

Cheers,
Stephen La Rocque.