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The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. If the hypotenuse KL is 10 inches, what is the length of the hypotenuse AB?
(1) Angles ABC and KLM are each equal to 55 degrees.
(2) LM is 6 inches.
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The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. If the hypotenuse KL is 10 inches, what is the length of the hypotenuse AB?
(1) Angles ABC and KLM are each equal to 55 degrees.
(2) LM is 6 inches.
S1. Tells you that the two triangles ABC and KLM are similar traingles A1/A2 = AB^2/KL^2 4/1 = AB^2/10^2 AB = SQ.RT 400 = 20 SUFFICIENT
S2. Tells you the lenght of LM. This does not really help us since it does not prove that the triangles are similar
1) tells us the two triangles are similar hence hyp of abc can be found sufficient 2) we can only find the area of ABC doesn t tell anything else ---insufficient
It should be A. (1) Since all angles are same these are similar triangles. We know that ratio of area of similar triangles is square the ratio of corresponding length. Thus Ab can be found. Sufficient (2) we can knw the product of BC and AC but othing can be known abt AB. Insufficient
S1. Tells you that the two triangles ABC and KLM are similar traingles A1/A2 = AB^2/KL^2 4/1 = AB^2/10^2 AB = SQ.RT 400 = 20 SUFFICIENT
What is the concept behind this?
why do we need to use the squares of the hyp's.? Are we some how using the pythagorean theorem?
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Let there be two similar triangles ABC and PQR. Let us call the ratio of corresponding sides as scale factor (SF). PPTY(1) The perimeters of two similar triangles are in the ratio of their scale factor. PPTY (2) The areas of two similar triangles are in the same ratio as the square of their scale factors.
According to PPTY (2), in the given problem, the scale factor (SF) happens to be the ratio of the hypotenuse. It doen't have to be the hypotenuse always (could be any side). Hence, A1/A2 = AB^2/KL^2
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