Can any expert comment n this problem please ?

vineeta wrote: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.

Sure!

If two triangles have the same angles, their sides follow the same ratio, whatever that ratio is. (You need trigonometry to actually find it.)

For example, if I have two 30-60-90 triangles, their sides follow the ratio x :: x √3 :: 2x, irrespective of what x actually is. (One of the triangles might be 4 :: 4√3 :: 8 and the other might be 10 :: 10√3 :: 20.)

For this test, you need to know that these ratios exist, even if you don't know what they actually are. (You're expected to know the 30-60-90, but few others.)

S1::

We have two similar triangles. Let's say KLM has sides of a :: b :: 10, and ABC has sides of c :: d :: e, where 10 and e are the hypotenuses of the two triangles.

Since they're similar triangles, a/c = b/d.

Since they're right triangles, their areas are (ab)/2 and (cd)/2, respectively.

If (cd)/2 = 4 * (ab)/2, or cd = 4ab, then c = 2a and d = 2b. This gives us the ratio BETWEEN the triangles: ABC has sides that are each DOUBLE the corresponding sides of KLM. So the hypotenuse of ABC is double 10, or 20.

S2::

LM is 6, so KLM is a 6-8-10 right triangle. We don't know that the triangles are similar, however, other that they're both right triangles, so ABC could have any sides such that (cd)/2 = 96, or four times the area of KLM. Since the hypotenuse depends on the lengths of the legs, this is insufficient.

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NOTE: you won't see this on the real exam, but the two statements contradict each other. S1 says the triangles have 55 degree angles, S2 says they're 3-4-5's (which have angles of ~53 degrees, not 55). This is a bug in the question, though, not a feature meant to influence your answer.

Hi Matt,

Thanks for the explanation. I was just wondering why, once we know an area ratio between two similar triangles, do we always assume each side is proportionally the same...

i.e. Why is it that we know when ab=4xy, that a =2x and b=2y? Why can't we have a=(4/3)x and b=3y?

Thanks in advance!

Nvm, apologies for post above, I see that using ratio a/c = b/d, and the equation cd=4ab, we can rearrange to get d^2=4b^2, which means d=2b since they're both positive

Similar triangle's property.
So, Answer is A?

goyalsau wrote:

bharathh wrote:IMO The answer is A

I Since both triangles have the same angles, they are similar.

If ABC has 4 times the area of KLM, each side of ABC is twice the size of KLM.

So the hyp of ABC = 2*10 = 20

So I is sufficient

II Knowing one side and the hypt I can get the area of KLM sqrt(10^2-6^2) = 8. Thus, I can get the area of KLM = 24

Area of triangle ABC = 24*4 = 96

But I can't do much else with this.. So II is insufficient

So answer is A

Great explanation, Simple and Easy....

knowing ABC = 4 times KLM may not help

it can be possible that 1 side AC is 4 times KM or 2 times.

Got it,

Answer is A

First, recall that in a right triangle, the two shorter sides intersect at the right angle. Therefore, one of these sides can be viewed as the base, and the other as the height. Consequently, the area of a right triangle can be expressed as one half of the product of the two shorter sides (i.e., the same as one half of the product of the height times the base). Also, since AB is the hypotenuse of triangle ABC, we know that the two shorter sides are BC and AC and the area of triangle ABC = (BC × AC)/2. Following the same logic, the area of triangle KLM = (LM × KM)/2.

Also, the area of ABC is 4 times greater than the area of KLM:
(BC × AC)/2 = 4(LM × KM)/2
BC × AC = 4(LM × KM)

(1) SUFFICIENT: Since angle ABC is equal to angle KLM, and since both triangles have a right angle, we can conclude that the angles of triangle ABC are equal to the angles of triangle KLM, respectively (note that the third angle in each triangle will be equal to 35 degrees, i.e., 180 - 90 - 55 = 35). Therefore, we can conclude that triangles ABC and KLM are similar. Consequently, the respective sides of these triangles will be proportional, i.e. AB/KL = BC/LM = AC/KM = x, where x is the coefficient of proportionality (e.g., if AB is twice as long as KL, then AB/KL = 2 and for every side in triangle KLM, you could multiply that side by 2 to get the corresponding side in triangle ABC).

We also know from the problem stem that the area of ABC is 4 times greater than the area of KLM, yielding BC × AC = 4(LM × KM), as discussed above.

Knowing that BC/LM = AC/KM = x, we can solve the above expression for the coefficient of proportionality, x, by plugging in BC= x(LM) and AC = x(KM):
BC × AC = 4(LM × KM)
x(LM) × x(KM) = 4(LM × KM)
x2 = 4
x = 2 (since the coefficient of proportionality cannot be negative)

Thus, we know that AB/KL = BC/LM = AC/KM = 2. Therefore, AB = 2KL = 2(10) = 20

(2) INSUFFICIENT: This statement tells us the length of one of the shorter sides of the triangle KLM. We can compute all the sides of this triangle (note that this is a 6-8-10 triangle) and find its area (i.e., (0.5)(6)(8) = 24); finally, we can also calculate that the area of the triangle ABC is equal to 96 (four times the area of KLM). We determined in the first paragraph of the explanation, above, that the area of ABC = (BC × AC)/2. Therefore: 96 = (BC × AC)/2 and 192 = BC × AC. We also know the Pythagorean theorem: (BC)2 + (AC)2= (AB)2. But there is no way to convert BC × AC into (BC)2 + (AC)2 so we cannot determine the hypotenuse of triangle ABC.

The correct answer is A.