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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?
Wiki User
∙ 15y ago
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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∙ 15y ago
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