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Re: Urgent help required [#permalink]
19 May 2011, 12:49
Useful formula "in any right angle triangle, the square of the altitude to the hypotenuse is equal to the product of two sectors it creates on hypotenuse"
Some useful properties of right triangles [#permalink]
26 Jul 2012, 08:22
1
This post was BOOKMARKED
Re (2): there are some properties of right triangles which can be useful. Please, refer to the attached drawing.
All three right triangles (the large one ABC and the two small ones, ABD and ADC) are similar (it can be easily seen that their angles are congruent).
From the similarity of the triangles ABD and CBA we obtain: \(\frac{AB}{BC}=\frac{BD}{AB}\), from which it follows that \(AB^2=BD*BC\). Similarly, using the similar triangles ACD and BCA, we get that \(AC^2=DC*BC\). We can call it the "leg property", as it expresses the leg of the right triangle in terms of its projection to the hypotenuse and the hypotenuse itself.
In our case, using the above property, we can find the two legs of the right triangle being \(\sqrt{5} = \sqrt{1*5}\) and \(2\sqrt{5}=\sqrt{4*5}\). Then the height of the right triangle is \(\frac{\sqrt{5}*2\sqrt{5}}{5}=2\) <--- leg * leg / hypotenuse.
But there is another useful property of the right triangle, which helps us find directly the height if we know the two projections of the legs to the hypotenuse. This property was already mentioned by "bellcurve". Using the similarity of the triangles ABD and CAD, we can write \(\frac{AD}{DC}=\frac{BD}{AD}\), from which \(AD^2=BD*DC\). So, in our case, the height of the right triangle is \(\sqrt{1*4}=2\). We can call this the "height property".
Well, we used to call the above properties "The leg Theorem" and "The height Theorem" back then in Romania. Although I have a PhD in Applied Math, I learned the above stuff in junior high. What I have learned during my graduate studies, cannot really help on the GMAT :O)
Attachments
RightTri-Prop.jpg [ 8.21 KiB | Viewed 3736 times ]
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: If vertices of a triangle have coordinates (-1,0), (4,0), [#permalink]
08 Oct 2012, 02:31
Hi Bunuel / Karishma -
Can you please confirm whether the below solution is OK?
For any triangle the max area is can be got by for a equilateral triangle. For any right angled triangle the max area can be got by forming an isosceles triangle.
Since statement B says the triangle is 90* degree triangle and we have the longest length (5) we can check what the max are will be?
isosceles triangle is in the ratio x : x : xroot2 so sides are in the ratio 5/root2 5/root2 and 5 Max area can be 1/2 (5 / root2) * (5 / root 2) = 6.25 (but this answer is not matching with Bunuel's). This will be the max area for any right angled triangle with the hypotenuse of 5.....
Similarly to get the max area of the rectangle we need the smallest perimeter and vice-versa!
Re: If vertices of a triangle have coordinates (-1,0), (4,0), [#permalink]
08 Oct 2012, 03:08
1
This post received KUDOS
Jp27 wrote:
Hi Bunuel / Karishma -
Can you please confirm whether the below solution is OK?
For any triangle the max area is can be got by for a equilateral triangle. For any right angled triangle the max area can be got by forming an isosceles triangle.
Since statement B says the triangle is 90* degree triangle and we have the longest length (5) we can check what the max are will be?
isosceles triangle is in the ratio x : x : xroot2 so sides are in the ratio 5/root2 5/root2 and 5 Max area can be 1/2 (5 / root2) * (5 / root 2) = 6.25 (but this answer is not matching with Bunuel's). This will be the max area for any right angled triangle with the hypotenuse of 5.....
Similarly to get the max area of the rectangle we need the smallest perimeter and vice-versa!
For any triangle the max area is can be got by for a equilateral triangle. - This is not a correct statement. You can have a "tiny" equilateral triangle or a "huge" right or any other triangle. The correct statement is "For a given perimeter, the maximum area is that of an equilateral triangle."
The second statement isn't correct either: For any right angled triangle the max area can be got by forming an isosceles triangle. I think what you meant is "For a right triangle inscribed in a given circle (which means given constant diameter), the maximum area is that of an isosceles right triangle."
For the given question, in fact we don't have to go the either of the above properties. Lets' denote by B the point with the coordinates (-1,0) and by C the point with the coordinates (4,0). Since BC lies on the X-axis and A is on the Y-axis, the area of the triangle ABC is given by Base*Height/2, which in this case is 5*Lenght of OA/2, regardles whether A is above or below the X-axis (\(A\) or \(A_1\) in the attached drawing). So, the area of triangle ABC is 5*A/2, where A is the length of OA (consider A>0). Therefore, the question in fact is "Is 5A/2>15?" or"Is A>6?"
(1) Not sufficient, as \(A_1\) can be below -6, in which case A>6, but if A is above 0 and below 3, obviously the area is less than 15. (2) If the triangle ABC is a right triangle, then necessarily \(\angle{BAC}\) is a right angle and BC is the diameter of the circle in which this triangle can be inscribed. Even if the triangle is isosceles, the height AO cannot be greater than the radius of the circle \(AO_1,\) which is 5/2. So, A is definitely smaller than 6. Sufficient.
Answer B.
Attachments
CoordTriangleArea.jpg [ 18.11 KiB | Viewed 3702 times ]
_________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: If vertices of a triangle have coordinates (-1,0), (4,0), [#permalink]
08 Oct 2012, 03:50
EvaJager wrote:
Jp27 wrote:
Hi Bunuel / Karishma -
Can you please confirm whether the below solution is OK?
For any triangle the max area is can be got by for a equilateral triangle. For any right angled triangle the max area can be got by forming an isosceles triangle.
Since statement B says the triangle is 90* degree triangle and we have the longest length (5) we can check what the max are will be?
isosceles triangle is in the ratio x : x : xroot2 so sides are in the ratio 5/root2 5/root2 and 5 Max area can be 1/2 (5 / root2) * (5 / root 2) = 6.25 (but this answer is not matching with Bunuel's). This will be the max area for any right angled triangle with the hypotenuse of 5.....
Similarly to get the max area of the rectangle we need the smallest perimeter and vice-versa!
For the given question, in fact we don't have to go the either of the above properties. Lets' denote by B the point with the coordinates (-1,0) and by C the point with the coordinates (4,0). Since BC lies on the X-axis and A is on the Y-axis, the area of the triangle ABC is given by Base*Height/2, which in this case is 5*Lenght of OA/2, regardles whether A is above or below the X-axis (\(A\) or \(A_1\) in the attached drawing). So, the area of triangle ABC is 5*A/2, where A is the length of OA (consider A>0). Therefore, the question in fact is "Is 5A/2>15?" or"Is A>6?"
(1) Not sufficient, as \(A_1\) can be below -6, in which case A>6, but if A is above 0 and below 3, obviously the area is less than 15. (2) If the triangle ABC is a right triangle, then necessarily \(\angle{BAC}\) is a right angle and BC is the diameter of the circle in which this triangle can be inscribed. Even if the triangle is isosceles, the height AO cannot be greater than the radius of the circle \(AO_1,\) which is 5/2. So, A is definitely smaller than 6. Sufficient.
Answer B.
Great Explanation EvaJager. I completely get it now.
EvaJager wrote:
For any triangle the max area is can be got by for a equilateral triangle. - This is not a correct statement. You can have a "tiny" equilateral triangle or a "huge" right or any other triangle. The correct statement is "For a given perimeter, the maximum area is that of an equilateral triangle."
The second statement isn't correct either: For any right angled triangle the max area can be got by forming an isosceles triangle. I think what you meant is "For a right triangle inscribed in a given circle (which means given constant diameter), the maximum area is that of an isosceles right triangle."
And yes I didn't phrase it correctly. What i meant was exactly this... "For given fixed perimeter the maximum area is that of an equilateral triangle"
ANd for the 2nd one -> The right triangle with the largest area will be an isosceles right triangle (where both the base and height are of equal length). Therefore, given the length of longest leg = 5, we can determine the largest possible area of triangle ABC by making it an isosceles right triangle
Re: If vertices of a triangle have coordinates (-1,0), (4,0), [#permalink]
08 Oct 2012, 04:32
Jp27 wrote:
EvaJager wrote:
Jp27 wrote:
Hi Bunuel / Karishma -
Can you please confirm whether the below solution is OK?
For any triangle the max area is can be got by for a equilateral triangle. For any right angled triangle the max area can be got by forming an isosceles triangle.
Since statement B says the triangle is 90* degree triangle and we have the longest length (5) we can check what the max are will be?
isosceles triangle is in the ratio x : x : xroot2 so sides are in the ratio 5/root2 5/root2 and 5 Max area can be 1/2 (5 / root2) * (5 / root 2) = 6.25 (but this answer is not matching with Bunuel's). This will be the max area for any right angled triangle with the hypotenuse of 5.....
Similarly to get the max area of the rectangle we need the smallest perimeter and vice-versa!
For the given question, in fact we don't have to go the either of the above properties. Lets' denote by B the point with the coordinates (-1,0) and by C the point with the coordinates (4,0). Since BC lies on the X-axis and A is on the Y-axis, the area of the triangle ABC is given by Base*Height/2, which in this case is 5*Lenght of OA/2, regardles whether A is above or below the X-axis (\(A\) or \(A_1\) in the attached drawing). So, the area of triangle ABC is 5*A/2, where A is the length of OA (consider A>0). Therefore, the question in fact is "Is 5A/2>15?" or"Is A>6?"
(1) Not sufficient, as \(A_1\) can be below -6, in which case A>6, but if A is above 0 and below 3, obviously the area is less than 15. (2) If the triangle ABC is a right triangle, then necessarily \(\angle{BAC}\) is a right angle and BC is the diameter of the circle in which this triangle can be inscribed. Even if the triangle is isosceles, the height AO cannot be greater than the radius of the circle \(AO_1,\) which is 5/2. So, A is definitely smaller than 6. Sufficient.
Answer B.
Great Explanation EvaJager. I completely get it now.
EvaJager wrote:
For any triangle the max area is can be got by for a equilateral triangle. - This is not a correct statement. You can have a "tiny" equilateral triangle or a "huge" right or any other triangle. The correct statement is "For a given perimeter, the maximum area is that of an equilateral triangle."
The second statement isn't correct either: For any right angled triangle the max area can be got by forming an isosceles triangle. I think what you meant is "For a right triangle inscribed in a given circle (which means given constant diameter), the maximum area is that of an isosceles right triangle."
And yes I didn't phrase it correctly. What i meant was exactly this... "For given fixed perimeter the maximum area is that of an equilateral triangle"
ANd for the 2nd one -> The right triangle with the largest area will be an isosceles right triangle (where both the base and height are of equal length). Therefore, given the length of longest leg = 5, we can determine the largest possible area of triangle ABC by making it an isosceles right triangle
I hope the above is correct.
ANd for the 2nd one -> The right triangle with the largest area will be an isosceles right triangle (where both the base and height are of equal length). In our case, not correct. Or rather, not clearly formulated. BC can be only the hypotenuse of the right triangle, as A must be on the Y-axis. So, we cannot have an isosceles right triangle with both legs 5. If A is not on the Y-axis, with no other restrictions, the area can be as large as we wish. In this case, the right triangle is uniquely determined (except its mirror image), because the diameter is fixed. A must be at the intersection point of the circle with the Y-axis. _________________
PhD in Applied Mathematics Love GMAT Quant questions and running.
Re: If vertices of a triangle have coordinates (-1,0), (4,0), [#permalink]
06 Nov 2012, 12:16
If an altitude (from the vertex opposite the hypotenuse) divides the hypotenuse in the ratio p:q then the altitude= (p*q)^1/2 i.e. the square root of p*q. Applying to the question above A= (1*4)^1/2 = 2 Hence, area will be smaller than 15. Hence, B is sufficient.
Re: If vertices of a triangle have coordinates (-1,0), (4,0), [#permalink]
02 Jan 2013, 04:56
Expert's post
gmat620 wrote:
If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) , is the area of the triangle greater than 15 ?
1. A < 3
2. The triangle is right
Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not. Please help me i have my test 4 days away. Thanks.
Another approach to solve this question: Since the area of the triangle is \(1/2( Base * Perpendicular height)\), therefore the question can be interpreted as IS A>6? #Base=5, # questionable area=15 Hence IS \(\frac{(5*A)}{2}> 15\)? or IS A>6?
Statement 1 is clearly insufficient as A can be 1 or -100.
Staement 2 is sufficient because its given that vertex at y-axis forms a right angle. Hence the base is the hypotenuese. Now if we draw a circle in such a way that (0,A) is at the circle, then the radius will be 2.5. Anything beyond 2.5 units from the centre of the circle will not form the right angle. Hence A is always less than 2.5 units from the centre of the circle. Sufficient. _________________
Re: If vertices of a triangle have coordinates (-1,0), (4,0), [#permalink]
09 Mar 2013, 16:03
3
This post received KUDOS
I thought of the problem in a more simple way, but I'm not sure if its the right thought process...
Basically when I assessed statement 2 I took given that the hypotenuse of the triangle was 5. This means that regardless of what the other 2 sides are, both must be less than 5. So even if the sides were let's say 4.9 * 4.9 ...the multiplication of these two values divided by 2 is always going to give a value less than 15 (especially since 5*5= 25/2 is less than 15).
Re: If vertices of a triangle have coordinates (-1,0), (4,0), [#permalink]
11 Mar 2013, 00:35
Expert's post
wiseman55555 wrote:
I thought of the problem in a more simple way, but I'm not sure if its the right thought process...
Basically when I assessed statement 2 I took given that the hypotenuse of the triangle was 5. This means that regardless of what the other 2 sides are, both must be less than 5. So even if the sides were let's say 4.9 * 4.9 ...the multiplication of these two values divided by 2 is always going to give a value less than 15 (especially since 5*5= 25/2 is less than 15).
Bad thought process?
Certainly not! The thought process is perfectly fine. In fact, it's quite good. _________________
Re: If vertices of a triangle have coordinates (-1,0), (4,0), [#permalink]
11 Mar 2013, 23:19
Expert's post
gmat620 wrote:
If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) , is the area of the triangle greater than 15 ?
1. A < 3
2. The triangle is right
If you plot the given points, we have a a triangle with area as \(\frac{1}{2}*5*A.\)
From F.S 1, we have A<3. As A is any point on the y-axis, we can have A = 1, and then the area would be less than 15. However, we can also have A = -8 and that would make the area more than 15. Insufficient.
From F.S 2, we know that the triangle is a right angle triangle. Thus, it can only be at the point(0,A). The slope between the points (-1,0) and (0,A) = A = m1
Slope between the point (0,A) and (4,0) = A/-4 = m2.
m1*m2 = \(A^2/-4\)= -1. Thus, A = 2 or -2. In any case, we can answer the question. Sufficient. _________________
Re: Urgent help required [#permalink]
31 Jan 2014, 13:47
Bunuel wrote:
gmat620 wrote:
If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) , is the area of the triangle greater than 15 ?
1. A < 3
2. The triangle is right
Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not. Please help me i have my test 4 days away. Thanks.
First of all right triangle with hypotenuse 5, doesn't mean that we have (3, 4, 5) right triangle. If we are told that values of all sides are integers, then yes: the only integer solution for right triangle with hypotenuse 5 would be (3, 4, 5).
To check this: consider the right triangle with hypotenuse 5 inscribed in circle. We know that a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.
So ANY point on circumference of a circle with diameter \(5\) would make the right triangle with diameter. Not necessarily sides to be \(3\) and \(4\). For example we can have isosceles right triangle, which would be 45-45-90: and the sides would be \(\frac{5}{\sqrt{2}}\). OR if we have 30-60-90 triangle and hypotenuse is \(5\), sides would be \(2.5\) and \(2.5*\sqrt{3}\). Of course there could be many other combinations.
Back to the original question: If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) , is the area of the triangle greater than 15? (1) A < 3 --> two vertices are on the X-axis and the third vertex is on the Y-axis, below the point (0,3). The third vertex could be at (0,1) and the area would be less than 15 OR the third vertex could be at (0,-100) and the area would be more than 15. So not sufficient.
(2) The triangle is right. --> Obviously as the third vertex is on the Y-axis, the right angle must be at the third vertex. Which means the hypotenuse is on X-axis and equals to 5. Again if we consider the circle, the radius mus be 2.5 (half of the hypotenuse/diameter) and the third vertex must be one of two intersections of the circle with Y-axis. We'll get the two specific symmetric points for the third vertex, hence the area would be fixed and defined. Which means that it's possible to answer the question whether the area is more than 15, even not calculating actual value. Sufficient.
Answer: B.
If we want to know how the area could be calculated with the help of statement 2, here you go:
One of the approaches:
The equation of a circle is \((x - a)^2 + (y-b)^2 = r^2\), where \((a,b)\) is the center and \(r\) is the radius.
We know: \(r=2.5\), as the hypotenuse is 5. \(a=1.5\) and \(b=0\), as the center is on the X-axis, at the point \((1.5, 0)\), half the way between the (-1, 0) and (4, 0). We need to determine intersection of the circle with Y-axis, or the point \((0, y)\) for the circle.
So we'll have \((0-1.5)^2 + (y-0)^2 =2.5^2\)
\(y^2=4\) --> \(y=2\) and \(y=-2\). The third vertex is either at the point \((0, 2)\) OR \((0,-2)\). In any case \(Area=2*\frac{5}{2}=5\).
Great explanation as usual. Now one additional thing, if we know that the hypothenuse = diameter of circle = 5 than the largest area we will have for the right triangle that is inscribed will be when the two other legs are perpendicular to each other, is this correct?
So could calculate the 45-45-90 triangle and for both legs using the radius and then have the maximum possible area of the triangle?
Re: Urgent help required [#permalink]
01 Feb 2014, 05:00
Expert's post
jlgdr wrote:
Bunuel wrote:
gmat620 wrote:
If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) , is the area of the triangle greater than 15 ?
1. A < 3
2. The triangle is right
Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not. Please help me i have my test 4 days away. Thanks.
First of all right triangle with hypotenuse 5, doesn't mean that we have (3, 4, 5) right triangle. If we are told that values of all sides are integers, then yes: the only integer solution for right triangle with hypotenuse 5 would be (3, 4, 5).
To check this: consider the right triangle with hypotenuse 5 inscribed in circle. We know that a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. The reverse is also true: if the diameter of the circle is also the triangle’s hypotenuse, then that triangle is a right triangle.
So ANY point on circumference of a circle with diameter \(5\) would make the right triangle with diameter. Not necessarily sides to be \(3\) and \(4\). For example we can have isosceles right triangle, which would be 45-45-90: and the sides would be \(\frac{5}{\sqrt{2}}\). OR if we have 30-60-90 triangle and hypotenuse is \(5\), sides would be \(2.5\) and \(2.5*\sqrt{3}\). Of course there could be many other combinations.
Back to the original question: If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) , is the area of the triangle greater than 15? (1) A < 3 --> two vertices are on the X-axis and the third vertex is on the Y-axis, below the point (0,3). The third vertex could be at (0,1) and the area would be less than 15 OR the third vertex could be at (0,-100) and the area would be more than 15. So not sufficient.
(2) The triangle is right. --> Obviously as the third vertex is on the Y-axis, the right angle must be at the third vertex. Which means the hypotenuse is on X-axis and equals to 5. Again if we consider the circle, the radius mus be 2.5 (half of the hypotenuse/diameter) and the third vertex must be one of two intersections of the circle with Y-axis. We'll get the two specific symmetric points for the third vertex, hence the area would be fixed and defined. Which means that it's possible to answer the question whether the area is more than 15, even not calculating actual value. Sufficient.
Answer: B.
If we want to know how the area could be calculated with the help of statement 2, here you go:
One of the approaches:
The equation of a circle is \((x - a)^2 + (y-b)^2 = r^2\), where \((a,b)\) is the center and \(r\) is the radius.
We know: \(r=2.5\), as the hypotenuse is 5. \(a=1.5\) and \(b=0\), as the center is on the X-axis, at the point \((1.5, 0)\), half the way between the (-1, 0) and (4, 0). We need to determine intersection of the circle with Y-axis, or the point \((0, y)\) for the circle.
So we'll have \((0-1.5)^2 + (y-0)^2 =2.5^2\)
\(y^2=4\) --> \(y=2\) and \(y=-2\). The third vertex is either at the point \((0, 2)\) OR \((0,-2)\). In any case \(Area=2*\frac{5}{2}=5\).
Great explanation as usual. Now one additional thing, if we know that the hypothenuse = diameter of circle = 5 than the largest area we will have for the right triangle that is inscribed will be when the two other legs are perpendicular to each other, is this correct?
So could calculate the 45-45-90 triangle and for both legs using the radius and then have the maximum possible area of the triangle?
Please advice Cheers J
I guess you mean "equal" instead of "perpendicular" above. If yes, then it's correct. _________________
Re: If vertices of a triangle have coordinates (-1,0), (4,0), [#permalink]
14 May 2014, 21:15
This seems to be a very old post...but I will try something different. Instead of drawing circles & thinking of properties of circles all that this question is asking is (based on statement 2 ruling out 1) are there 2 sides of a right angled triangle with lengths x, y such that x^2 + y^2 = 25 & 1/2*x*y > 15 or x*y > 30? No restrictions on values of x,y (integer or non-integer as long as they are real numbers). Remember, in a right angled triangle area can be found out by using the sides opposite to the hypotenuse (you just have to turn the triangle or your head to find that out)
To solve this, let us input some values of x to find y: if x = 1, y = 24^0.5 if x = 2, y = 21^0.5 if x = 3, y = 16^0.5 = 4 if x = 4, y = 9^0.5 = 3 if x = 5, y = 0 which is not possible.
So you have 4 different combinations that can satisfy the given condition (not sure if they can really form such a triangle with the given co-ordinates of the other two vertices, which isn't the issue here). None of them can form a right angled triangle whose area is < 15. Choice B can answer that & hence is the right answer.
Re: If vertices of a triangle have coordinates (-1,0), (4,0), [#permalink]
25 Aug 2014, 09:38
gmat620 wrote:
If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) , is the area of the triangle greater than 15 ?
(1) A < 3 (2) The triangle is right
Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not. Please help me i have my test 4 days away. Thanks.
Using the area formula derived from matrices,we find the area to be 5A/2. From (1)if A<3,then Area<15/2 which is <15.Sufficient From (2) using right angle in a circle we can get the answer.Sufficient. So,answer should be D. Guys,pls let me know if I'm going wrong somewhere.
If vertices of a triangle have coordinates (-1,0), (4,0), [#permalink]
25 Aug 2014, 20:20
2
This post received KUDOS
Expert's post
DebWenger wrote:
gmat620 wrote:
If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A) , is the area of the triangle greater than 15 ?
(1) A < 3 (2) The triangle is right
Hello friends, I solved this problem but I have a doubt. if we know that triangle is right triangle and hypotenus is 5, can we always safely assume other two sides as 3, 4 without knowing whether other sides are integer or not. Please help me i have my test 4 days away. Thanks.
Using the area formula derived from matrices,we find the area to be 5A/2. From (1)if A<3,then Area<15/2 which is <15.Sufficient From (2) using right angle in a circle we can get the answer.Sufficient. So,answer should be D. Guys,pls let me know if I'm going wrong somewhere.
Statement 1 tells you that A is less than 3. Since A is just the y coordinate of the 3rd vertex, it can be negative, right? Say, A = -100
Area = |5A/2| = 250 (Mind you, area is the absolute value of 5A/2) In this case, area is more than 15
If A = 1, Area = 2.5 In this case, area is less than 15
Statement 1 is not sufficient alone. _________________
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