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655-705 Level|   Geometry|                                 
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Sarabjeets746
I think there is a faster way to evaluate statement 1.
Area=25 => b*h=50 => b and h are factors of 50. Possible pairs: (1,50); (2,25); (5,10).
Remember: this triangle is a right triangle, which means the hypotenuse is supposed to be the largest side of this triangle.
So out of these three pairs only one pair (5,10). Sufficient. Bunuel VeritasKarishma what do you think?

It is not necessary that the legs should be integer values. Also, of the three pairs you suggested, none will work because (5, 10) gives a leg length of 10 which cannot be the same as hypotenuse.
The legs will be 5*sqrt(2) each.
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The hypotenuse of a right triangle is 10 cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimeters.
(2) The 2 legs of the triangle are of equal length.

We are given that the hypotenuse of a right triangle is 10 cm and we need to determine the perimeter of the triangle. We can let the other two sides (i.e., the two legs) of the right triangle be a and b. Since it’s a right triangle, by the Pythagorean theorem, we have a^2 + b^2 = 10^2, or a^2 + b^2 = 100. If we can find the values of a and b, then we can determine the perimeter of the triangle, since it will be a + b + 10.

Statement One Alone:

The area of the triangle is 25 square centimeters.

We are given a right triangle and we’ve let the two non-hypotenuse sides be a and b. Recall that the two non-hypotenuse sides of a right triangle are actually the base and height of the triangle, so the area of this triangle is A = ab/2. From the aforementioned equation a^2 + b^2 = 100, we can solve b as b = √(100 - a^2). Since we are given that A = 25, and substituting b = √(100 - a^2) in A = ab/2, we can say:

25 = [a√(100 - a^2)]/2

We see that we can solve for a (though we don’t have to actually solve for it). And once we’ve solved for a, we can determine the value of b, since b = √(100 - a^2). Therefore, we can determine the perimeter of the triangle, since we can determine the values of both a and b. Statement one alone is sufficient to answer the question.

Statement Two Alone:

The 2 legs of the triangle are of equal length.

Since a^2 + b^2 = 100 and we are given that a = b, we can say:

a^2 + a^2 = 100

We see that we can solve for a (though we don’t have to actually solve for it). And once we’ve solved for a, we can determine the value of b, since b = a. Therefore, we can determine the perimeter of the triangle, since we can determine the values of both a and b. Statement two alone is sufficient to answer the question.

Answer: D


Something I always struggle with when given an equation with exponents such as the one Scott solved for in statement (1) (25 = [a√(100 - a^2)]/2) is how do we know that there aren't multiple solutions to the equation? (i.e. like when you're factoring the form of aX^2+bx+c). Is there any indication that there will NOT be multiple possible solutions from the equation without fully factoring out/solving?

Response:

The equation b^2 = 100 - a^2 actually has two solutions, but we can disregard the negative solution since a and b denote lengths (and therefore, they must be positive). That is the reason b = √(100 - a^2) is the only possible value of b.

For the equation 25 = [a√(100 - a^2)]/2, we pretty much have to solve it before we know there aren’t multiple solutions. For instance, if the equation were 20 = [a√(100 - a^2)]/2, then the equation would have two solutions and if it were 30 = [a√(100 - a^2)]/2, then it would have no solutions.
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The hypotenuse of a right triangle is 10 cm. What is the perimeter, in centimetres, of the triangle?

Lets assume that the height and base of the right angled triangle be a and b respectively and the hypotenuse is given as 10 cm.
We are asked to find the perimeter of the triangle i.e. a + b + 10
Value of a+ b is required to find the perimeter. Individual values of a and b are not necessary here.
Since hypotenuse is 10 cm, \(a^2 + b^2 = 10^2\)

(1) The area of the triangle is 25 square centimetres.

Area of a triangle = (1/2 )* base * height = (1/2) * a* b = 25
ab = 50
\((a+b)^2 = a^2 + b^2 + 2ab\)
= \(10^2 +2*50 = 200\)
\(a+b =\sqrt{200}\)
The value of a+ b is enough to find the perimeter.
Hence ,statement 1 is sufficient

(2) The 2 legs of the triangle are of equal length.
From this statement we can conclude that its a right angled isosceles triangle. Since the hypotenuse is given, we will be able to find the length of the legs and also the perimeter.
Statement 2 is sufficient.

Option D is the answer.

Thanks,
Clifin J Francis,
GMAT SME
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Few things to remember when approaching DS problems:

1. In GMAT Data Sufficiency questions, avoid calculations as much as possible
2. In GMAT Data Sufficiency questions, When asked to find a value, just knowing that the value can be found with the information is enough, you don’t have to calculate all the way through
3. In GMAT Data sufficiency questions, the two answer choices never contradict each other. If they are both sufficient, they must give the same value

Statement 1: Since the base and perpendicular are equal and hypotenuse = base^2 + perpendicular^2 we can calculate perimeter of the right angled triangle
Statement 2: This is the tricky one - and the solution comes from understanding that the sum of squares (a^2 + b^2) are related to product in two ways: (a+b)^2 or (a-b)^2

Once you realize this, since area is given you can find ab. Hypotenuse is given, you have (a^2+b^2) and that should yield perimeter. We do not really need to calculate the perimeter because just knowing it can be found it enough [perimeter cannot be negative - in case you take the square root]

Here is a video solution for those who feel better visualizing the solution:

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Stmt 1:
Lets say the other 2 sides are 'x' and 'y'
\(\frac{1}{2}\)xy=25
xy=50
\(x^{2}+ y^{2}=10^{2}\) (Pythagoras theorem)
\((x+y)^{2}=x^{2}+ y^{2}+2xy\)
\((x+y)^{2}=200\)
\(x+y=\sqrt{200}\)
perimeter= x+y+10
sufficient

Stmt 2:
Lets say the other 2 sides are 'x' and 'x'
\(2x^{2}=100\)
\(x=\sqrt{50}\)
perimeter= \(\sqrt{50}+ \sqrt{50}+ 10\)
sufficient

Ans D
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Video solution from Quant Reasoning:
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1
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marcusaurelius
The hypotenuse of a right triangle is 10 cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimeters.
(2) The 2 legs of the triangle are of equal length.

This didn't strike me quick enough, but the solution is not very tedious.
by Pythagoras,
a^2+b^2=c^2
here c=hypotenuse=10cm
Therefore,
B^2+H^2=100...(i)

Statement 1:
A=25
(1/2)BH=25
BH=50...(ii)
from eqn i and ii, we can get individual sides. so sufficient.

Statement 2:
from info given in statement 2 and eqn i
B=H
therefore,
B^2+B^2=100
2B^2=100
B^2=50

Moreover, imagine a triangle where one side is fix and the opposite angle is also fixed as 90.
If we constrain the two sides to be equal, then there can only be one such triangle. Hence, sufficient.

Ans-D
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