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 In triangle ABC, MN  || BC and MN and MN bisects the area of triangle ABC. If AD =...

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electrika | Student, College Freshman | eNoter

Posted June 14, 2011 at 6:19 PM via web

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 In triangle ABC, MN  || BC and MN and MN bisects the area of triangle ABC. If AD = 10, find ED, if AE is the height of triangle AMN.

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giorgiana1976 | College Teacher | Valedictorian

Posted June 14, 2011 at 7:20 PM (Answer #1)

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Since triangles ABC and AMN are similar and area of AMN is half of the area of ABC, we'll have;
Area of AMN/Area of ABC = 1/2 (1)
Let ED = x => AE = AD - ED = 10-x
Area of AMN/Area of ABC = (10-x)^2/10^2 (2)
We'll equate (1) and (2) and we'll get:
(10-x)^2/10^2 = 1/2
We'll cross multiply:
2(10-x)^2 = 100
We'll expand the binomial and we'll multiply by 2:
200 - 40x + 2x^2 - 100 = 0
x^2 - 20x + 50 = 0
We'll apply quadratic formula:
x1 = [20+sqrt(400 - 200)]/2
x1 = 10+5sqrt2
x2 = 10-5sqrt2
Since x has to be smaller than 10, we'll keep only the second value, namely x = ED = 10-5sqrt2.

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mustafanw | Student, Undergraduate | Honors

Posted June 14, 2011 at 7:21 PM (Answer #2)

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Since triangles ABC and AMN are similar and area of AMN is half of the area of ABC, we'll have;
Area of AMN/Area of ABC = 1/2 (1)
Let ED = x => AE = AD - ED = 10-x
Area of AMN/Area of ABC = (10-x)^2/10^2 (2)
We'll equate (1) and (2) and we'll get:
(10-x)^2/10^2 = 1/2
We'll cross multiply:
2(10-x)^2 = 100
We'll expand the binomial and we'll multiply by 2:
200 - 40x + 2x^2 - 100 = 0
x^2 - 20x + 50 = 0
We'll apply quadratic formula:
x1 = [20+sqrt(400 - 200)]/2
x1 = 10+5sqrt2
x2 = 10-5sqrt2
Since x has to be smaller than 10, we'll keep only the second value, namely x = ED = 10-5sqrt2.

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