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 Question from tahrima, a student: find the base of an isosceles triangle whose area is 12sq.cm and the length of 1 of the equal sides is 5 cm.

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.

Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.