# Algebra 2

The base of an isosceles triangle is 1/4 as long as the two equal sides. Write the area of the triangle as a function of the length of the base.

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1. Let x be the length of the two equal sides.

x/4 is length of the base of triangle.

let height = h

x^2-(x^2/64)=h^2 (using pythagoras theorem, half of base = x/8)

(64x^2-x^2)/64 = h^2

h=(sqrt63*x)/8

Area = 1/2 * base* height

Area = 1/2 * (x/4) * (sqrt 63 * x)/8

Area=(x^2 / 32) * sqrt 63

Area = (x^2 / 32) * 3* sqrt 7

but x=4*b (where b is the base)

Area = ((4b)^2/32) * 3sqrt7

=16b^2/32 * 3sqrt7

= (b^2)/2 * 3sqrt7

=(3/2)sqrt7 * b^2

=3.96 b^2

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