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555-605 (Medium)|   Geometry|               
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Bunuel
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Answer should be D. Both together are sufficient.

The pictures shows its a right angled triangle

1. Z= 13. Since its right angled, \sqrt{\(x^2+y^2\)} =13 => \(x^2+y^2\) = 169

Alone not Enough

2. A= 5y/2

Area of triangles = 1/2bh => SInce its right angled, in this case A= x*y/2

xy/2 = 5y/2 => x=5. We still need value of X. Hence this option is alone not enough

Together used, we can determine value of "Y" by substituting "X" value in Eq1 => \(y^2\) = 169-25 = 144
=> y= 12. Hence A= 30
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A is sufficient alone. first of all 144 + 25 = 169. secondly, we have to find A it does not matter whether we are able to find the X and Y.

12*5*(1/2)=30

regardless of the height and length.
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The Logical approach to this question would start off with the pythagorean theorem: a²+b²=c². Since this is one equation with three variables, statement (1) on its own is not enough to solve it. Answer choices (A) and (D) are eliminated.
Statement (2) relates to the formula of the area of a right triangle, which is the product of its legs divided by 2. Since we know that one leg is y and the area is 5y/2, we have enough information to find x, but not for any of the other sides. Answer choice (B) is also eliminated.
Using both statements, if we have both x and z, we can find y, and thus have enough information to calculate the area. The correct answer is (C).

Posted from my mobile device

Hey David... According to each statements, is it wrong to conclude that it's a 5-12-13 right angle traingle... Is it a trap??
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Hi! I am confused. Why can't the answer be A? If the hypotenuse is 13 then the right angle triangle has to be a 5-12-13. We have the Base and Height to figure the area.
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Hi! I am confused. Why can't the answer be A? If the hypotenuse is 13 then the right angle triangle has to be a 5-12-13. We have the Base and Height to figure the area.

Hi Spatel1992 / All ,

Finally I got the answer , here it goes

Initially , I also thought that the only answer that goes with hypotenuse 13 would be 5,12,13 but later on , I found something interesting.

There is another pythagoras triplet with 13 and that’s 13,84,85 ......

Hope this helps :)

Posted from my mobile device
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Bunuel

If A is the area of a triangle with sides of lengths x, y, and z as shown above, what is the value of A ?

(1) z = 13
(2) A = 5y/2


Attachment:
2019-04-26_1842.png


Hey guys!

Hope this explanation proves to be valuable for all!

1. Statement 1 says that z=13,
Yes, we do know that there is a Pythagorean triplet for this question- '12,5,13'.
But we cannot conclude on this as there can be alternate measures as well.
So, Drop Statement 1 for now

2. Statement 2 says that \(Area(A)= \frac{ 5y}{2 }\)
So, we can conclude saying that the value of x=5,
Because, \(Area of Triangle= \frac{1}{2} * x * y \)
So, A= 2.5y

We cannot go further on this, as the information is insufficient.

Combining statements 1 and 2,
We get to know the 2 values, i.e., x=5 and z=13
We finally accomplish the values for this triplet!
z^2= x^2 + y^2

And, we shall find the Area of triangle using the standard formula.

Official Answer:- Option C

Hope this helps you friends! saarthakkhanna04 SPatel1992 manu11 thyagi anilesh10 LeenaSai UNSTOPPABLE12 digvijayk1


Thank you!

Regards,
Raunak Damle :cool:
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Bunuel

If A is the area of a triangle with sides of lengths x, y, and z as shown above, what is the value of A ?

(1) z = 13
(2) A = 5y/2

DS20602.01
Quantitative Review 2020 NEW QUESTION


Attachment:
2019-04-26_1842.png


Area = \(\frac{1}{2}*x*y\)

1) z=13
So x^2+y^2=13^2
As we are not told x and y are integers, we will have numerous (x,y) fitting in.
For every real value of x<13, there will be a value for y. Although integer value will be just (5,12) or (12,5)
x=1.....\(y^2=169-1=168\)....\(y=\sqrt{168}\)
Insuff

2) A=5y/2
A=xy/2=5y/2......x=5
But nothing about y

Combined
x=5.....\(5^2+y^2=13^2......y=12\)
Area = 5*12/2=30
Suff

C
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Bunuel GMATBusters please could you confirm the following notes that I have made so far on right-angled triangles:

"1. If we have a triangle with sides in the ratio 3: 4: 5 (of any triplet) THEN by default the triangle is 90°

2. If we have a triangle with two sides in the ratio 3: 4 and even if we know it is a 90° triangle WE CANNOT assume the third side to be a multiple of 5 (or any such triplet sequence)

3. If we are told that the triangle is 90° AND one of its legs is 1/2 the hypotenuse, THEN the triangle is 30°- 60° - 90° "
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"1. If we have a triangle with sides in the ratio 3: 4: 5 (of any triplet) THEN by default the triangle is 90°.

Response : YES, it will be a right angled triangle.

2. If we have a triangle with two sides in the ratio 3: 4 and even if we know it is a 90° triangle WE CANNOT assume the third side to be a multiple of 5 (or any such triplet sequence)

Response: yes coz the sides can be (3, 4, 5) when angle between 3 and 4 is 90 deg or (3, root 7, 4) when angle between 3 and root 7 is 90 deg.


3. If we are told that the triangle is 90° AND one of its legs is 1/2 the hypotenuse, THEN the triangle is 30°- 60° - 90° "[/quote]

Response: Yes, your understanding is correct.

Hoozan
Bunuel GMATBusters please could you confirm the following notes that I have made so far on right-angled triangles:

"1. If we have a triangle with sides in the ratio 3: 4: 5 (of any triplet) THEN by default the triangle is 90°.

Response : YES, it will be a right angled triangle.

2. If we have a triangle with two sides in the ratio 3: 4 and even if we know it is a 90° triangle WE CANNOT assume the third side to be a multiple of 5 (or any such triplet sequence)

Response: yes coz the sides can be (3, 4, 5) when angle between 3 and 4 is 90 deg or (3, root 7, 4) when angle between 3 and root 7 is 90 deg.


3. If we are told that the triangle is 90° AND one of its legs is 1/2 the hypotenuse, THEN the triangle is 30°- 60° - 90° "

Response: Yes, your understanding is correct.

Posted from my mobile device
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"1. If we have a triangle with sides in the ratio 3: 4: 5 (of any triplet) THEN by default the triangle is 90°.

Response : YES, it will be a right angled triangle.

2. If we have a triangle with two sides in the ratio 3: 4 and even if we know it is a 90° triangle WE CANNOT assume the third side to be a multiple of 5 (or any such triplet sequence)

Response: yes coz the sides can be (3, 4, 5) when angle between 3 and 4 is 90 deg or (3, root 7, 4) when angle between 3 and root 7 is 90 deg.


3. If we are told that the triangle is 90° AND one of its legs is 1/2 the hypotenuse, THEN the triangle is 30°- 60° - 90° "

Response: Yes, your understanding is correct.

Hoozan
Bunuel GMATBusters please could you confirm the following notes that I have made so far on right-angled triangles:

"1. If we have a triangle with sides in the ratio 3: 4: 5 (of any triplet) THEN by default the triangle is 90°.

Response : YES, it will be a right angled triangle.

2. If we have a triangle with two sides in the ratio 3: 4 and even if we know it is a 90° triangle WE CANNOT assume the third side to be a multiple of 5 (or any such triplet sequence)

Response: yes coz the sides can be (3, 4, 5) when angle between 3 and 4 is 90 deg or (3, root 7, 4) when angle between 3 and root 7 is 90 deg.


3. If we are told that the triangle is 90° AND one of its legs is 1/2 the hypotenuse, THEN the triangle is 30°- 60° - 90° "

Response: Yes, your understanding is correct.

Posted from my mobile device[/quote]


So following this thought process, if we were given that the triangle is 90° such that the hypotenuse is 5 and one of the sides is 4, now in this case unline in point 2, we can use the tripllet knowledge right? Since now we know that the 90° angle is between 3 and 4
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chetan2u
Bunuel

If A is the area of a triangle with sides of lengths x, y, and z as shown above, what is the value of A ?

(1) z = 13
(2) A = 5y/2

DS20602.01
Quantitative Review 2020 NEW QUESTION


Attachment:
2019-04-26_1842.png


Area = \(\frac{1}{2}*x*y\)

1) z=13
So x^2+y^2=13^2
As we are not told x and y are integers, we will have numerous (x,y) fitting in.
For every real value of x<13, there will be a value for y. Although integer value will be just (5,12) or (12,5)
x=1.....\(y^2=169-1=168\)....\(y=\sqrt{168}\)
Insuff

2) A=5y/2
A=xy/2=5y/2......x=5
But nothing about y

Combined
x=5.....\(5^2+y^2=13^2......y=12\)
Area = 5*12/2=30
Suff

C

chetan2u
I am so sorry to bother you with this very basic algebra question but to confirm when solving for .5bh=5y/2 --> I did xy=5y --> so then x=5 because the ys cancel out, right?

Also, the Official Answer says that at least two sides of a right triangle must be known to find the area. Doesn't this apply to ALL triangle types (in that you need at least two sides to find the area of any triangle)?

Thank you and Happy Holidays!
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assumes x & y, can't do that. it must be explicitly found
saarthakkhanna04
A is sufficient alone. first of all 144 + 25 = 169. secondly, we have to find A it does not matter whether we are able to find the X and Y.

12*5*(1/2)=30

regardless of the height and length.
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