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