If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A)

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
(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.

Bunuel · Accepted Answer

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:
Approach 1:
 
 If the vertices of a triangle have coordinates  (-1, 0) ,  (4, 0) , and  (0, A) , is the area of the triangle greater than 15? 
 
(1)  A \lt 3 
 
Statement (1) alone is insufficient. If  A  is close to 0, the area of the triangle is small, but if  A  is a large negative number (for example,  A = -1000 ), the area of the triangle is large.
 
(2) The triangle is a right triangle
 
Statement (2) alone is sufficient. Only the angle at vertex  (0, A)  can be a right angle. As  |A|  increases, the value of the angle decreases. At some point, the angle becomes equal to 90 degrees, and we can snapshot the triangle at this moment. After that, we can find the area of the triangle and answer the question. Please see the image below.
 
 https://gmatclub.com/gmat-practice-tests/resources/import/img/M16-15.png 
 
Approach 2:
 
 If the vertices of a triangle have coordinates  (-1, 0) ,  (4, 0) , and  (0, A) , is the area of the triangle greater than 15? 
 
(1)  A \lt 3 
 
Two vertices lie 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), resulting in an area less than 15, OR the third vertex could be at (0, -100), resulting in an area greater than 15. Therefore, this statement alone is not sufficient.
 
(2) The triangle is a right triangle
 
It is evident that the third vertex, which is located on the y-axis, forms a right angle. Consequently, the hypotenuse, which is on the x-axis, measures 5 units. We can consider this hypotenuse as the diameter of a circle as shown below:
 
 https://gmatclub.com/gmat-practice-tests/resources/import/img/M16-15-2.png 
 
Any point on the circumference of that circle will create a right angle with the diameter/hypotenuse. This is because a right triangle inscribed in a circle must have its hypotenuse as the diameter of the circle. Conversely, if the diameter of the circle is also the triangle's hypotenuse, then that triangle is a right triangle.
 
Since we know that the third vertex is on the y-axis, it must be one of the two intersection points of the circle with the y-axis. Thus, we obtain two symmetrical triangles with equal areas. Both of these triangles are fixed and defined, allowing us to determine whether their area is greater than 15 without calculating the exact value. Therefore, this statement alone is sufficient.
 
If we wish to understand how the area can be calculated with the help of statement 2, here is an explanation:
 
One approach is to use the equation of a circle:  (x - a)^2 + (y - b)^2 = r^2 , where  (a, b)  represents the center and  r  represents the radius.
 
In this case, we know that  r = 2.5  because the hypotenuse measures 5 units. The center of the circle, denoted by  (a, b) , lies on the x-axis at the point  (1.5, 0) , which is halfway between (-1, 0) and (4, 0).
 
To find the intersection points of the circle with the y-axis, we need to determine the y-coordinate  y  when  x = 0 . Substituting these values into the equation of the circle, we have:
 
 (0 - 1.5)^2 + (y - 0)^2 = 2.5^2 .
 
Simplifying, we get  y^2 = 4 , which yields two possible values for  y :  y = 2  and  y = -2 . Consequently, the third vertex can be located at either (0, 2) or (0, -2). In either case, the area can be calculated using the formula  Area = \frac{1}{2}*	ext{base }*	ext{height} . Plugging in the values, we have  Area = \frac{1}{2}*5*2 = 5 .
 
Therefore, the area of the triangle is 5, which is less than 15.

Answer: B

pratikdas007 · Answer

I appreciate the below solution but to solve in this lengthy method will need GMAC to provide 150 mins for Q instead of 75 mins

The concept to note here is product of slopes of 2 perpendicular lines = -1

Since the triangle can only be right angle at 0,A , then - A/4*A/1 = -1

Solve for A and then find out the max possible area of the triangle
(B) is the answer

If you like my method , kindly provide comment

gmat620 · Answer

Great explanation by Bunuel !! I wish Gmat wouldn't throw such questions on me. :wink:

ramgmat · Answer

Hi Bunuel,

Please tell me where I am going wrong.
Calculated using the matrix formula to solve the area of the triangle.
\(1/2 [-1 (1-A) + 4 (A-1) +0(1-1)] >15\)
\(A>7\)

Option 1 says A< 3
Hence Statement A is sufficient.

Am I missing something here?

Bunuel · Answer

This is a valid approach if you are familiar with the formula which gives the area based on the coordinates of the three vertices of a triangle.

If the vetices of a triangle are:  A(a_x, a_y) ,  B(b_x, b_y)  and  C(c_x,c_y)  then the area of ABC is:

area=|\frac{a_x(b_y-c_y)+b_x(c_y-a_y)+c_x(a_y-b_y)}{2}| .

So if we consider:  A(-1,0) ,  B(4,0) , and  C(0,A)  then the area would be:  area=|\frac{-1(0-A)+4(A-0)+0(0-0)}{2}|  -->  area=|\frac{5A}{2}| .

Question: is  area=|\frac{5A}{2}|>15  --> is  |A|>6 .

Statement (1) says A>3, which is not sufficient to say whether  |A|>6 .

P.S. You made some errors in calculation and also didn't put the area formula in ||.

Hope it helps.

ramgmat · Answer

Thanks a lot Bunuel! I really appreciate this.

I need to be careful of both my calculation as well as my silly mistakes

pratikdas007 · Answer

The answer is B As in (1) , A<3 , which means A = -3 , -100 , -200 anything --> Not suffcient From (2) We conclude that the triangle is rt. angled at (0,A) We can set up the equation as (A-0)/(0-(-1)) * (A-0)/(0-4) = -1 as product of slopes of two perpendicular lines of a rt angled triangle is -1 A= +/- 2 which gives area of the rt angled traingle as 5 sq units < 15 sq units ---> sufficient Hence Answer is B

Raths · Answer

Great explanation Bunuel...

muralimba · Answer

from the given info. the base of the triangle is 5 i.e the disantce between
-1 and 4.

for the area to be greater than 15 , 0.5*5*hiegt > 15 ==> hiegt > 6

note that the third vertex(0,A) decides the hieght of the triangle.

Stmnt1. A < 3 ==> A could be 2 (hieght = 2 and answer to the question is NO)
or -10 (hieght = 10 and the anser to the qtn is YES)...hence NOT suff.

stmnt2: the triangle is right

VERY IMPORTANT, USEFUL AND GMAT'S FAVORITE property: the angle on the semi circle is 90, means, if two vertices of
the triangle are on the extreme sides of the diameter and the third is on
the semi circle, then at the third vertex the angle is 90.

Using the above property and the stmnt 2 , we can say that the third verthex
(0,A) is on a semi circle having the diameter connected between (-1,0) and
(4,0) ==> radius of the circle is 5 (distance b/w -1 and 4) / 2 = 2.5

hence the trianlge (right angle) in this semicircle would have the maximum
area = 0.5 * 5 * 2.5 (height=radius) = 6.25 that is < 15 ==> the max area is
< 15 ==> any other possible right angle triangle areas would be < 15 ==>
answer to the question is "NO".

Answer "B".

KarishmaB · Answer

First, your question - No, you cannot assume that the sides are 3 and 4 if you have 5 as hypotenuse. lagomez is right. What if it is 45-45-90 triangle? The sides will not be 3 and 4. But in this question, you have something more. You know the line on which your third vertex of the triangle will fall. Ques2.jpg There will be only 1 such right triangle so you will be able to say whether the area is greater than 15 or not. You don't need to find the triangle but you know this statement alone is sufficient.

Bunuel · Answer

Little correction here: actually there will be 2 such right triangles, the second one will be the mirror image of the first (urgent-help-required-87344.html#p656628).

KarishmaB · Answer

Yes, that's true. But since we are only concerned with area, we can pretty much ignore the mirror images for this statement.

Bunuel · Answer

Yes, that's also true.

EvaJager · Answer

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)

Jp27 · Answer

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!

EvaJager · Answer

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.

Jp27 · Answer

Great Explanation EvaJager. I completely get it now.

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.

EvaJager · Answer

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.

wiseman55555 · Answer

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?

If vertices of a triangle have coordinates (-1,0), (4,0), and (0,A)

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