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

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 \(Area=2*\frac{5}{2}=5\).

Bunuel,

Lets say base is - (-1,0), (4,0)
and
(0,A): Third vertices

Base =5, Height =A

1/2 X Base X height = 1/2 X 5 X A

IF A <3 Then Area, A < 7.5 This answers the questions. Hence A is the solution.

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.


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.

Dear Karishma, from the fogure above the triangle ABC is right triangle lets say that A=(-1,0) B=(4,0) C= (0,A) my

question is why we say that AB is the basic and CD is the hight (middle point of AB) while it shuld be in right triangle

that CA is the basic and CB is the hight I really confused here why the basic and the hight are different. For example if

the triangle ABC is 90-60-30 we will say that the basic is AC and the hight is CB.I mean why we try to find the CD

instead of CB Also if I have right triangle and I have both CD and CB what will be the hight in this case.

Dear Karishma, I apologize for the large number of questions but I really like your explanation.

Sorry but then should not y coordinate be 2.5+1.5 i.e. +4 or -4

Have you tried to draw it? Hope that the below image will help:

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.


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.

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.


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.

VeritasKarishma, why is the hypotenuse 5? I can't get this part