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whats the answer of this one and why

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whats the answer of this one and why

by vineeta » Thu Aug 27, 2009 12:11 pm
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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by levocap » Thu Aug 27, 2009 1:11 pm
C is sufficient. Since this is a right triangle and you know the length of hypotenuse and a side, you can find out the third side. X^2 + 6^2 = 10^2
The third side works out to be 8. Using this, you know that area of KLM is 24. With this info, you know Area of ABC (4 time larger). Now, because the two triangles have the same angles, use ratios to figure out the length of all three sides
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by bharathh » Thu Aug 27, 2009 1:23 pm
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
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by levocap » Thu Aug 27, 2009 1:47 pm
i concurr with bharat
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by mmslf75 » Sun Jan 17, 2010 9:55 am
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.

Area of ABC = 5 Area of KLM

right ??

if x is 4 times of y

x = 4y

if x is 5 times greater than y

x = Y + 4y= 5y


Any1 pls confirm!!
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by gmatmachoman » Mon Jan 18, 2010 12:17 am
mmslf75 wrote:
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.

Area of ABC = 5 Area of KLM

right ??

if x is 4 times of y

x = 4y

if x is 5 times greater than y

x = Y + 4y= 5y


Any1 pls confirm!!
Dude...

what happend to u?? U need a break now...
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by Haaress » Thu May 27, 2010 2:18 pm
I know Trignometry is not covered in the GMAT curriculum, however, basic knowledge of relationships between sides and angles( Sine, Cosine, Tangent relationships - SOHCAHTOA) can help reduce the time taken on such a question.

From the info provided, the trangle is a right angled triangle and the length of KL = 10.

Stmt 1 provides that one of the other angles in KLM is 55 degree. That makes it sufficient to calculate the hypotenuse of KLM.

Stmt 2. One side of a right angle triangle is not sufficient to do the trick.

So the answer is A.
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by goyalsau » Fri Dec 24, 2010 3:03 am
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....
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by thebigkats » Wed Dec 29, 2010 11:59 am
I probably took a little long path, but in any case, her it is:

say smaller triangle is points KLM wherein KL=10 and KM=x, ML = y
say larger triangle is points ABC wherein AB = TBD and AC=p, CB = q
STEM - doesn't tell much about other angles in ABC or KLM, so we don't know the relationship of triangles to deduce anything except for that p.q / 2 = 4 . (x.y / 2) ==> area of KLM = x.y/2 and area of ABC = p.q/2 ==> this means that pq = 4*xy

STMT #1 -
<ABC = <KLM = 55 degrees
==> this means that AB and KL are parallel and all angles are same in triangles
==> x / y = p / q (trigonometry - tan theta)
==> (x*q) / y = (p * q) / q (multiply both sides with q)
==> xq / y = 4xy/q
==> cancel out x on both sides ==> q / y = 4.y / q
==> q^2 = 4 y^2
==> q^2 = (2.y) ^2
==> q = 2y

We can use the same method to come to p = 2x
Now AB^2 = p^2 + q^2
= (2x)^2 + (2y)^2
= 4.x^2 + 4.y^2
= 4 (x^2 + y^2)
==> 4 * 10^2 (remember from the stem - x^2+y^2 = 10^2) = 400
Sp AB = sqrt (400) = 20

STAT #2:
y = 6 ==> x^2 = 10^2 - 6^2 = 100-36 = 64 ==> x = 8
So area of triangle ABC = 4 . (8.6 / 2) = 24*4 = 96 = p.q / 2
This means that pq = 192, but it doesn;t tell us anything about p or q - becasue we don't know the relationship

So STAT#1 is sufficient but #2 is NOT

BTW - it looks like a long solution but actually it took me <1min to get the answer
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by Zerks87 » Mon Jan 10, 2011 2:18 pm
So we know that both of the these are right triangles and the side relationships for a right triangle are 1, root 3, and 2 where 1 is opposite the smallest angle, root 3 is opposite the middle angle and 2 opposite the right angle.

(1) angle KLM=ABC= 55 degrees. Since we know that one angle is 90, one is 55, the final one must be 35.

Since we know the hypotaneuse of KLM is 10 and ABC is four times bigger and they are identical triangles in terms of their angles, then the hypotaneuse of ABC must be 4 times bigger. SUFF

(2) LM is 6 inches. Using the same properties. We know that the hypotaneuse is 10 for KLM. Since the hypotaneuse is opposite the right angle it is the 2 in my relationship given above. Since 6 is not half of 10 it must be the root 3 portion of the triangle making the final side 5 or half of the hypotaneuse according to the properties of right triangles. This makes the area of KLM = 6(5)/2 or 15

Since we know that ABC has an area 4x greater than KLM its area must be 60.

The formula for the area of a right angle triangle is s^2(root 3)/4. If we take the area of ABC (60) and set it equal to this equantion than we can find the area of one of the sides. Then we can assume that ABC is a 45-45-90 triangle which would mean the side ratio is 1-1-root2. And at least conceptually we know that we can find the hypotaneuse without doing any calculations. SUFF

Therefore IMO the answer is D
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by gana » Thu Jan 27, 2011 5:51 pm
Expert is needed here. I am confused whose answer is right. is it answer A or C?
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by rohu27 » Thu Jan 27, 2011 6:23 pm
When two triangles are similar, the followng holds good:

Square of the ratio of corresponding sides=ratio of areas of the triangles

simple put (hypotenuse of abc/hypotenuse klm)^2=(area of abc/area of klm)

from the above we can easily get the hyp of abc.

so 1 is sufficient.

from 2 we can get the sides of KLM. in turn we get to know the bh (base*height) for ABC but we cant get the individual values. so insuficient.

A is the answer.
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by cyrwr1 » Fri Jan 28, 2011 10:49 am
(1) The triangles are similar with this. Area ratios 4to1, so sides : 2to1. Hypotenuse is 2(10)=20
(2) You know the smaller triangle is a 6-8-10 Triangle with area 24 and the larger is 96 but nothing else.

Hence, A
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by shresthaal » Wed Oct 19, 2011 11:20 am
https://www.cliffsnotes.com/study_guide/ ... 18815.html

Answer is A
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by prashant misra » Fri Oct 21, 2011 2:29 am
my official answer is A since the areas are in the ratio of 4 to 1 their corresponding sides will be in the ratio of 2 to 1 so the hypotenuse should be 2*10=20 please anyone just correct me if i am wrong
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