The area of the right triangle ABC is 4 times greater than the area of

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.

Hi Bunuel
i have a doubt in this question as it says ...The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM. so it implies that area ABC = 5* area of KLM

or the sentence should be 4 times of area of KLM

I want to clear this "greater than " dilemma

Nidhi

VeritasKarishma Bunuel , i have a similar doubt as Nidhi. Could you please help?

Yes, the answer to this isn't as straight forward as you would expect!

A is 10% greater than B obviously means A = B + 10% of B.
A is 10 times B again obviously means A = 10*B
Mathematically, A is 10 times greater than B "should" mean A = 10B + B BUT we use it to mean 10B so frequently in colloquial language that it has come to mean 10B.
Hence, if faced with this question in the actual exam, I will say that "A is 10 times greater than B" means A = 10B.
Though hopefully, official questions will use clearer language - A is 10 times B.

In a DS question on Geometry, do not make any assumptions and try to see if you can obtain a unique diagram. If you overlook any of these two rules, you may end up falling for the trap answer.

For example, in this question, although the triangles look like Right triangles, I’d only decide after reading the first sentence. Had the first sentence not mentioned that the triangle is a right triangle, we cannot assume that they are right angled triangles, just because they look like.
‘Looks-like’ has been the undoing on Geometry questions for a lot of students.

Also, I know that angles C and M are right angles in the respective triangles, only after I read the second statement in the question. Till this stage, I will not assume that C and M are right angles.

From the question statements, we know that,

Area of triangle ABC = 4 * Area of triangle KLM. Now, does this mean that triangles ABC and KLM are similar?? If you said yes, you have fallen for the trap. The answer is “Not necessarily”.

Hypotenuses of the respective right triangles are AB and KL, hence the right angles are at C and M respectively.

When you have 2 triangles with one set of angles equal, you should think of proving the triangles similar by finding another set of equal angles. Once this is done, you will be able to use the properties of similar triangles to solve the question.

Statement I alone says that angles ABC and KLM are equal. We now have the second set of angles that we were looking for. We can now say that the triangles ABC and KLM are similar. Therefore,

\(\frac{Area of triangle ABC }{ Area of triangle KLM}\) = \(\frac{{AB^2}}{{KL}}\). Since we know the ratio of the areas of the triangles and also KL, we can find out the length of AB from this data.
Statement I alone is sufficient. Possible answer options are A or D. Answer options B, C and E can be eliminated.

Statement II alone says that LM is 6 inches. This can only help us determine the other side of the triangle i.e. KM = 8 inches. We may also be able to establish the area of triangle KLM and hence the area of triangle ABC.
However, this is all we can do. We do not have any other link between the two triangles in terms of angles so that we could prove them similar. Hence, we will not be able to find out AB.
Statement II alone is insufficient. Answer option D can be eliminated.

The correct answer option is A.

Note that the question statement mentions '4 times greater than'; technically, this should mean ' Area of triangle ABC = 5* Area of triangle KLM'.

However, when we tie in this information with the other data like the lengths of the hypotenuse and the areas, we see that 'Area of triangle ABC = 4* Area of triangle KLM' makes more sense since we get Area of triangle KLM = 24 sq.units and Area of triangle ABC = 96 sq.units when we solve using statement I alone.


Hope that helps!

Bunuel

Regarding st1. We are only given 1 angle though? How do we know if the triangles are similar if we're only given 55? Or is it implicit given that the area of the larger triangle is 4x larger?

The way I thought about it is:

ABC / ABC/4 = 4 = AB^2 /100 --> AB = 20 <-- But this only works if they the triangles are similar.

St1 - ABC = KLM = 55
Confirms similarity I suppose?
Sufficient

St2 - LM = 6
Does nothing to confirm similarity
Insufficient

Notice that the stem says that both triangles are right triangles: "The area of the right triangle ABC is 4 times greater than the area of the right triangle KLM..." (1) says that angles ABC and KLM are each equal to 55 degrees. So, two angles are equal in ABC and KLM. Since the sum of the angles in a triangle is 180 degrees, then all three angles must be equal in ABC and KLM, which means that they are similar triangles.

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