



 
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 You can also see the right triangles and notice that Pythagoras' Theorem must work here, so 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 Substitute this into the Pythagorean equation we wrote earlier: 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 x^{2}  100x + 2304 = 0 where x = b^{2}. Which we can solve using "completing the square" method: 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} + h^{2} = 3^{2} + 4^{2} = 9 + 16 = 25. And that's the same as 5^{2}, so it works. Now check b = 8 cm: h = 24 / b = 24 / 8 = 3 cm. (b/2)^{2} + h^{2} = 4^{2} + 3^{2} = 16 + 9 = 25. That's right. Cheers,  


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