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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