if the area of an isosceles triangle is 60cm m2 and length of each of its equal sides is 13cm find base
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10 cm will be the answer
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length of the equal sides equal to 13 cm, the width of the base and the altitude equal to 2.x and h respectively, then we may write for the area A of the triangle
A = h.x = 60 cm²
so that h = 60/x
and from Pythagoras applied to either of the right-angled triangles
x² + h² = 13²
Substituting for h from above gives
x² + (60/x)² - 13² = 0
Multiplying through by x² and rearranging
x^4 - 169.x² + 3600 = 0
which is a quadratic in z = x² with solutions
z = [169 ±√(169² - 4*3600)]/2
giving z = [169 ± 119]/2 = 25 and 144
so that the solutions for x are 5 and 12, both of which are valid,
The base of the triangle may therefore be either 2.x = 10 or 24 cm in length.
A = h.x = 60 cm²
so that h = 60/x
and from Pythagoras applied to either of the right-angled triangles
x² + h² = 13²
Substituting for h from above gives
x² + (60/x)² - 13² = 0
Multiplying through by x² and rearranging
x^4 - 169.x² + 3600 = 0
which is a quadratic in z = x² with solutions
z = [169 ±√(169² - 4*3600)]/2
giving z = [169 ± 119]/2 = 25 and 144
so that the solutions for x are 5 and 12, both of which are valid,
The base of the triangle may therefore be either 2.x = 10 or 24 cm in length.
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