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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. Thanks a lot Engr2012. 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.
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The hypotenuse of a right triangle is 10 cm. What is the [#permalink]
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20 Dec 2015, 16:16
With in statement: type of triangle: right angle Hypotenuse, c = 10 cm P = a+b+c c^2 = a^2 + b^2 Statement I: \(A = 1/2 ab = 25\) \(a=25x2/b = 50/b\) i.e. \(25 = a^2 + b^2\) \(25 = \frac{50^2}{b^2} + b^2\) Statement is sufficinet to find values of a & b, we already know value of c. All set statement I alone is sufficient. Statment II: \(a=b\) \(10^2 = a^2 + b^2\) a and b sides are equal, we can write above equation as \(10^2 = a^2 + a^2\) \(100 = 2a^2\) So value of a and b can be determined by statement II alone as well. Answer is D.
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Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]
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22 Jan 2016, 09:58

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Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]
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27 Aug 2016, 06:30
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?

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Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]
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27 Aug 2016, 07:23
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.
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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.

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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. Hope it helps.

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Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]
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13 Jan 2017, 02:33
My Solution: From question stem, we have following: Hypotenuse = 10, let the other two sides of right angle triangle is \(x\) and \(y\).
==> \(x^{2} + y^{2} = 100\)
From statement 1:
Area = 25 ==> \(\frac{1}{2} \times x \times y = 25 \Rightarrow xy = 50\)
\((x  y)^{2} = x^{2} + y^{2}  2xy = 100  2*50 = 0 \Rightarrow x = y\)
We have isosceles right angle triangle, so we can find the other two sides, and hence the perimeter. Sufficient.
From Statment 2: Triangle is isosceles right angle triangle, hence sufficient.
Answer (D).
Thanks.

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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 \(x+y\). Square \(x+y\) > \((x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200\) > \(x+y=\sqrt{200}\). Thus \(P=x+y+10=\sqrt{200}+10\). Sufficient. hey just a question: 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 ?
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Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]
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16 Apr 2017, 05:21
daviddaviddavid wrote: 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 \(x+y\). Square \(x+y\) > \((x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200\) > \(x+y=\sqrt{200}\). Thus \(P=x+y+10=\sqrt{200}+10\). Sufficient. hey just a question: 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 ? Check here: https://gmatclub.com/forum/thehypotenu ... l#p1051143
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The hypotenuse of a right triangle is 10 cm. What is the [#permalink]
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16 Apr 2017, 12:31
Check here: https://gmatclub.com/forum/thehypotenu ... l#p1051143[/quote] thx so i guess in this case it should be ok since were dealing with a triangle and therefore there will only be one solution
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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 \(x+y\). Square \(x+y\) > \((x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200\) > \(x+y=\sqrt{200}\). Thus \(P=x+y+10=\sqrt{200}+10\). Sufficient. Hi BunuelYour approach is great and to the point. 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 454590 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

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Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]
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06 Jul 2017, 11:21
Shiv2016 wrote: 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 \(x+y\). Square \(x+y\) > \((x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200\) > \(x+y=\sqrt{200}\). Thus \(P=x+y+10=\sqrt{200}+10\). Sufficient. Hi BunuelYour approach is great and to the point. 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 454590 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.
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Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]
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06 Jul 2017, 11:53
BunuelLet b^2= m m^2 100m+2500=0 m=50 b^2= 50 b= 5* sq. root 2
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The hypotenuse of a right triangle is 10 cm. What is the [#permalink]
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07 Jul 2017, 11:52
marcusaurelius wrote: 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. 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.
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