Question 1071495: I have a tough geometry question. Here it is: L, M, and N are the midpoints of the sides of triangle ABC. Find the ratio of the perimeters and areas of triangles LMN and ABC.
Found 2 solutions by Boreal, rothauserc:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The line joining the midpoints of two sides can be shown to be parallel to and half the length of the third side. This being the case, all the sides of triangle LMN are half that of triangle ABC, so the perimeter is half as much.
Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! We know the following
:
AM = AB/2, NC = BC/2, BN = BC/2
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We use proportionality of similar triangles
:
AB/BC = AM/ML
ML = (BC * AM) / AB
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BC/AB = NC/NL
NL = (AB * NC) / AB
:
BC/AC = BN/MN
MN = (AC * BN) / BC
:
ML+NL+MN = [(AB * BC) / (2 * AB)] + [(AB * BC) / (2 * BC)] + [(AC * BC) / (2 * BC)] = (BC / 2) + (AB / 2) + (AC / 2), therefore
:
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ratio of perimeters of LMN to ABC is (1/2)
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:
We can use Heron's formula to determine the ratio of the areas
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The area(A) of ABC is
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A = square root[P*(P-AB)*(P-BC)*(P-AC)] where P = (AB + BC + AC) / 2
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We know that 2 = AB/ML = BC/NL = AC/MN, therefore
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P(ABC) = [(AB+BC+AC)/2] = [(2*ML)+(2*NL)+(2*MN)/2] = 2*P(LMN)
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Substitute P(ABC) by 2*P(LMN) and AB = 2ML, BC = 2NL, AC = 2MN
:
P(ABC)*(P-AB)*(P-BC)*(P-AC) = 2^4*P(LMN)*(P(LMN)-ML)*(P(MNL)-NL)*(P(MNL)-MN)
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take square roots and we get
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ratio of areas of LMN to ABC is 1/4
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