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Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]
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18 Jul 2015, 16:53
luisnavarro wrote:
Engr2012 wrote:
luisnavarro wrote:
Hi Bunuel, I also have the same doubt, Knowing 2 of the 3 (Diagonal, perimeter, or area), we can find the third one por Triangles, Squares and rectangles? Is that correct. Could you help us?
Thanks a lot.
Regards.
Luis Navarro Looking for 700
Let me try to answer this question.
The answer to your question (in general) is as follows:
1. For ANY triangle, perimeter = P = a+b+c ---> so given 2 of the 3 sides and P, you can definitely calculate the 3rd side. For triangles that are not right angled or equilateral or isosceles, you would 1 angle or some sort of a relation between the sides and the area or a combination of these to sufficiently answer DS questions for these triangles.
i) For right triangles you also have the hypotenuse ^2 = base^2 + perpendicular ^2 as another equation. Thus for a right angled triangle, you have 2 equations.
ii) Similarly for isosceles and equilateral triangles (by drawing an a right angle, as shown in picture attached), you divide these 2 triangles into 2 same or congruent right angled triangles. So, for these 2 triangles as well you have 2 equations as mentioned above.
FYI, also remember that for right angled triangles and isosceles/equilateral triangles, you can calculate the 3rd side or the perimeter if the area is given as shown in the question discussed in this thread (by applying the relation, \((a+b)^2 = a^2+b^2+2ab\) and noticing that 0.5*ab represents the area of the triangle).
2. For rectangles and squares, the diagonals of these 2 shapes, divide these rectangles and squares into 2 congruent or same right angled triangles. The points mentioned above in 1) thus also apply these shapes as well.
Hope this answers your question. Let me know if there are any other questions.
That mean that this kind of question (Given area and diagonal of the examples above mentioned) in DS are "automatic", for instance; No matter system equation is quadratic always is going to be one answer and would be sufficient (I do not need to check the specific roots of the quadratic). Is it correct?
Regards.
Luis Navarro Looking for 700
I would be very skeptical of making sweeping statements like the one you mentioned above and remember to follow a series of steps to arrive at an answer. The way you are looking at it will lead you to memorize the questions and not understand the correct logic behind it. But for the sake of this conversation and this particular question, yes you will 1 value of the side if the area and the hypotenuse (both need to be there!) of the right angled triangle are given.
Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]
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27 Aug 2016, 08:33
Yes, I know that. My confusion was because, I thought 50 could be either 1/2 (Leg 1 x leg 2) Or 1/2 (Hypotenuse x height from hypotenuse). Realized the latter isn't possible because the height comes out to 10.
In case the height was not given to be 50, but was given as a figure less than 50, say 40, will my assumption be wrong in this scenario? Because, either 1/2 (Leg 1 x leg 2) Or 1/2 (Hypotenuse x height from hypotenuse) can give this area.
abhimahna wrote:
ameyaprabhu wrote:
Statement 1 says area is 50.
But how can we assume that the product of the 2 legs is 50, and not of the hypotenuse and the height?
Triangle given is a right angled triangle. So, area of the Right Triangle is given as 1/2 * base * height.
Please go through the basics of geometry once before attempting the questions.
Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]
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13 Jan 2017, 02:18
ameyaprabhu wrote:
Yes, I know that. My confusion was because, I thought 50 could be either 1/2 (Leg 1 x leg 2) Or 1/2 (Hypotenuse x height from hypotenuse). Realized the latter isn't possible because the height comes out to 10.
In case the height was not given to be 50, but was given as a figure less than 50, say 40, will my assumption be wrong in this scenario? Because, either 1/2 (Leg 1 x leg 2) Or 1/2 (Hypotenuse x height from hypotenuse) can give this area.
abhimahna wrote:
ameyaprabhu wrote:
Statement 1 says area is 50.
But how can we assume that the product of the 2 legs is 50, and not of the hypotenuse and the height?
Triangle given is a right angled triangle. So, area of the Right Triangle is given as 1/2 * base * height.
Please go through the basics of geometry once before attempting the questions.
Hi Ameyprabhu,
Both of the formulas that you have mentioned for the area of right angle triangle are correct. I don't see any problem. Given hypotenuse = 10, 1/2 * hypotenuse * height =25 ==> height = 5 cm.
But the question is asking for perimeter, so height(corresponding to hypotenuse) information is not going to help. You have to look for the required information and then apply the relevant formula.
Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]
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16 Apr 2017, 04:26
Bunuel wrote:
marcusaurelius wrote:
The hypotenuse of a right triangle is 10cm. 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.
I know that (2) is sufficient but I am having difficulty with (1).
For (1): Let the legs be \(x\) and \(y\).
Given: \(h=10\) and \(\frac{xy}{2}=25\) (area of the right triangle \(area=\frac{leg_1*leg_2}{2}\)) --> \(h^2=100=x^2+y^2\) (Pythagoras) and \(xy=50\).
Q: \(P=x+y+10=?\) So we have to calculate the value \(x+y\).
Is it not enough to realise that we have 2 equations (50=bh; h^2+b^2=100) with two unknowns (namely b and h) and therefore we could solve the problem ?
_________________
The hypotenuse of a right triangle is 10cm. 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.
I know that (2) is sufficient but I am having difficulty with (1).
For (1): Let the legs be \(x\) and \(y\).
Given: \(h=10\) and \(\frac{xy}{2}=25\) (area of the right triangle \(area=\frac{leg_1*leg_2}{2}\)) --> \(h^2=100=x^2+y^2\) (Pythagoras) and \(xy=50\).
Q: \(P=x+y+10=?\) So we have to calculate the value \(x+y\).
Is it not enough to realise that we have 2 equations (50=bh; h^2+b^2=100) with two unknowns (namely b and h) and therefore we could solve the problem ?
Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]
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06 Jul 2017, 11:05
Bunuel wrote:
marcusaurelius wrote:
The hypotenuse of a right triangle is 10cm. 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.
I know that (2) is sufficient but I am having difficulty with (1).
For (1): Let the legs be \(x\) and \(y\).
Given: \(h=10\) and \(\frac{xy}{2}=25\) (area of the right triangle \(area=\frac{leg_1*leg_2}{2}\)) --> \(h^2=100=x^2+y^2\) (Pythagoras) and \(xy=50\).
Q: \(P=x+y+10=?\) So we have to calculate the value \(x+y\).
But I solved this question using a rather long approach: How to notice such small elements like you noticed?
In triangle ABC, AB is the altitude, AC is the Hypotenuse, and BC is the base.
Given: AC= 10 Angle ABC is 90 degrees.
To find: AB+BC+AC
1) AB*BC= 50 Let AB= a and BC= b
So we can say: ab=50
a^2 + b^2= 10^2 a^2 +b^2 =100
a= 50/b
(50/b)^2 +b^2= 100
The equation now is in one variable and I am not really sure how to solve it further. But I keep A because the equation can be solved using this one variable b.
2) It is a 45-45-90 triangle.
So sides are in the ratio 1:1: sq. root 2 sq.root 2*x= 10 x= 10/sq.root 2
The hypotenuse of a right triangle is 10cm. 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.
I know that (2) is sufficient but I am having difficulty with (1).
For (1): Let the legs be \(x\) and \(y\).
Given: \(h=10\) and \(\frac{xy}{2}=25\) (area of the right triangle \(area=\frac{leg_1*leg_2}{2}\)) --> \(h^2=100=x^2+y^2\) (Pythagoras) and \(xy=50\).
Q: \(P=x+y+10=?\) So we have to calculate the value \(x+y\).
But I solved this question using a rather long approach: How to notice such small elements like you noticed?
In triangle ABC, AB is the altitude, AC is the Hypotenuse, and BC is the base.
Given: AC= 10 Angle ABC is 90 degrees.
To find: AB+BC+AC
1) AB*BC= 50 Let AB= a and BC= b
So we can say: ab=50
a^2 + b^2= 10^2 a^2 +b^2 =100
a= 50/b
(50/b)^2 +b^2= 100
The equation now is in one variable and I am not really sure how to solve it further. But I keep A because the equation can be solved using this one variable b.
2) It is a 45-45-90 triangle.
So sides are in the ratio 1:1: sq. root 2 sq.root 2*x= 10 x= 10/sq.root 2
Can be solved.
D
Is this approach fine?
Thanks
Yes, it's fine. For (1) you can denote b^2 as say m and solve the quadratics.
_________________
I know that (2) is sufficient but I am having difficulty with (1).
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.
Let a and b be two legs of the right triangle and c be its hypotenuse. Then we have \(a^2 + b^2 = c^2\) and \(c = 10\). Thus there are 3 variables and 1 equation. By using VA method, D is the answer most likely.
Condition 1) : \(\frac{1}{2}ab = 25\) or \(ab = 50\) \((a+b)^2 = a^2 + b^2 + 2ab = c^2 + 2*50 = 100 + 100 = 200\). Thus \(a + b = \sqrt{200} = 10\sqrt{2}\). Hence \(a + b + c = 10 +10\sqrt{2}\). It is sufficient.
Condition 2) : \(a = b\) We have \(a^2 + b^2 = 2a^2 = c^2 = 100\). Then \(a^2 = 50\) or \(a = 5\sqrt{2}\). Thus \(a + b + c = 2 \cdot 5\sqrt{2} + 10 = 10 + 10\sqrt{2}\). It is also sufficient.
D is the answer as expected.
-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
_________________
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.
Aside: Plugging in numbers might be challenging, because we'd need to find values that satisfy BOTH the given information (hypotenuse = 10) AND the information in the statements.
IMPORTANT: For geometry DS questions, we are typically checking to see whether the statements "lock" a particular angle or length into having just one value. This concept is discussed in much greater detail the video below.
So, for this question, if a statement FORCES our right triangle into having ONE AND ONLY ONE shape and size, then that statement is sufficient. Moreover, we NEED NOT find the actual perimeter of the triangle. We need only recognize that we could find its perimeter (finding the perimeter will just waste time).
Okay, onto the question....
Target question:What is the perimeter of the right triangle?
Given: The hypotenuse of the triangle has length 10 cm.
Statement 1: The area of the triangle is 25 square centimeters. Let's let x = length of one leg Also, let y = length of other leg So, if the area is 25, we can write (1/2)xy = 25 [since area = (1/2)(base)(height)] Multiply both sides by 2 to get xy = 50 Multiply both sides by 2 again to get 2xy = 100[you'll soon see why I performed this step]
Now let's deal with the given information (hypotenuse has length 10) The Pythagorean Theorem tells us that x² + y² = 10² In other words, x² + y² = 100
We now have two equations: 2xy = 100 x² + y² = 100
Since both equations are set equal to 100, we can write: 2xy = x² + y²
Rearrange this to get x² - 2xy + y² = 0 Factor to get (x - y)(x - y) = 0 This means that x =y, which means that the two legs of our right triangle HAVE EQUAL LENGTH.
So, the two legs of our right triangle have equal length AND the hypotenuse has length 10. There is only one such right triangle in the universe, so statement 1 FORCES our right triangle into having ONE AND ONLY ONE shape and size.
This means that statement 1 is SUFFICIENT
Statement 2: The 2 legs of the triangle are of equal length We already covered this scenario in statement 1. So, statement 2 is also SUFFICIENT
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
_________________
Scott Woodbury-Stewart Founder and CEO
GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions
Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]
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21 Jan 2018, 04:23
2
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For 1 there is a much easier approach:
Think the hypotenuse as the diameter of a circle. In this case the area will be hy*height/2=25 --> height = 5. height = 5 --> half of hy --> half of diameter = radius --> therefore the height fall in the middle point, and so leg1 = leg2. 1 is SUFF.
One of the interesting things about the GMAT is that most questions can be approached in a few different ways, so it is beneficial to be a "strong thinker" on Test Day. Knowing how to do math is sometimes what's needed (although sometimes the math can be done in several different ways); in other situations, the ability to figure out patterns is what's needed.
The prompt tells us that the hypotenuse of a RIGHT triangle is 10. The question asks us for the perimeter, so we're going to need the lengths of the other two sides of the triangle.
Since we have a right triangle, we can use the Pythagorean Theorem:
X^2 + Y^2 = 10^2
**Note: triangles can't have "negative" sides, so both X and Y MUST be positive.**
We have 2 variables, but only 1 equation, so we can't figure out the exact values of X and Y
Fact 1: Triangle area = 25
Since Area = (1/2)(B)(H) we know that….
(1/2)(X)(Y) = 25 (X)(Y) = 50
Now we have ANOTHER equation. Combined with what we were given in the beginning, we have…
Two variables AND two UNIQUE equations, which means we have a "system" of equations and we CAN solve for X and Y. No more math is needed. Fact 1 is SUFFICIENT.
Fact 2: The two legs of the triangle are equal.
Now we know that X = Y.
Fact 2 also gives us a SECOND equation to work with. Just as in Fact 1, we have a "system" and we can solve for X and Y. Fact 2 is SUFFICIENT
Be on the lookout for "system" questions. You'll see a couple on Test Day and they're built around some useful math "patterns" that you can use to avoid doing certain 'math work.'
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