My ans will be A - 19. But the approach to the solution is different.
With the the vertices of the triangle, I noticed that one point is on the X axis (6,0) and one on y axis (0,2) and the other point in 1 Quandrant.
I tried to construct a Trapeziod with the points A(0,0), B(6,0), C(4,7)(these are two of the vertices of the trianle in question)and D(0,7). Here CD is parallel to AB, AB and CD are the bases and AD is the height of the trapeziod. Now area of the trapoziod is 1/2(sum of the bases) * height = 1/2(6+4) * 7 = 35.
Let the E(0,2) be the third vertex which lies on AD. I tried finding the area of traingle AEB and area of triangle ECD which are the right triangles.
Area of triangle AEB = 1/2 * 6 * 2 = 6
Area of Triangle ECD = 1/2 * 4 * 5 = 10.
So Area of traingle BCE(which is our question) is Area of trapeziod ABCD - area of traingle AEB and area of triangle ECD.
35-10-6 = 19.
So my Ans is 19.
My apologies for not showing in a diagram. I don't know how can I put that in the diagram format. But if you try to draw the figure, you should be able to understand the points.
Brent Hanneson,
Thank you for posting some good questions. When you said that you are not aware of the formula that other posters have made, I realised there should be a way. And here is my way. Taking the inspiration of Thinking Out of the Box. [/img]
You will be surprised!
