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How much long are the sides of an isosceles triangle given that it's perimeter is 190 mm and it's altitude is 70 mm?
If the perimeter of an isosceles triangle is 190 mm, its sides has these measures:
a + 2b = 190 mm, where a represents the base and b represents the congruent sides.
b = (190 - a)/2 = 95 - a/2
b^2 = (95 - a/2)^2
h = 70
h^2 = 4,900
The altitude, h, is equal:
h^2 = b^2 - (a/2)^2 substitute what you know:
4,900 = (95- a/2)^2 - (a/2)^2 Solve for a:
4,900 = 9,025 - 95a + (a^2)/4 - (a^2)/4
4,900 = 9,025 - 95a add 95a and subtract 4,900 to both sides;
95a = 9,025 - 4,900
95a = 4,125 divide 95 to both sides;
a ≈ 43.42 mm
b = 95 - a/2 = 95 - 21.71
b = 73.29
Answer: The sides of the triangle are 73.29 mm, 73.29 mm, and 43.42 mm.
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