ABCD is rectangle.E is the midpoint of 𝐵𝐶,F is the midpoint of 𝐴𝐵 and G is the point of intersection of 𝐴𝐸 and 𝐶𝐹.If the area of ABCD is 1260 sq units,find the area of the quadrilateral BFGE.

Answer 1

Consider ΔABC (draw diagonal AC). It's area is 1/2 the area of the rectangle. Segment CF is a median of that triangle. A median of a triangle divides the "big" triangle into two triangles of equal area. So ΔFBC has an area that is 1/2 (1/2) (1260).

Segment AE is a 2nd median for ΔABC. There is another theorem (property) for medians that the 2nd median divides the areas of the two triangles created by one median in the ratio 2:1. This means that quadrilateral FBEG = 2(area ΔGEC); if x = areaΔGEC, then 2x + x = 315 (area ΔFBC).. Solve for x and then 2x.

All of the above assumes you have theorems available on the properties of the medians of a triangle. If not, then finding the coordinates of point G (coordinate geometry), then dividing quadrilateral FBEG into 2 triangles and a rectangle (or a triangle and a trapezoid) is another and perhaps most obvious) option.

Answer 2

Hi Lasya V.,

This question appears very difficult -- how can you possibly know where point G is, or determine the area of the quadrilateral in question?

But, patience gives you everything! Draw your rectangle, and section it off into halves each direction. Let AB be the top horizontal side, and C and D successive corners going clockwise around. You can start to see some quantitative relationships: the segment AE cuts the vertical bisector (of rectangle ABCD) at a point (3/4) of the way up, and the segment FC also cuts the horizontal bisector of rectangle ABCD (3/4) of the way towards the right.

You are now in a position to do a little algebra! Assign the distance DA to the quantity y , and the distance DC to the quantity x . Note that I say "quantity", since they are fixed with respect to how you have drawn your rectangle, although you could have made them other sizes by drawing your rectangle differently! So, in one sense, you could regard them as a variable quantity, but think of them as a fixed quantity here.

OK, now you can assign point C to be the Cartesian-graph point (0,0) (the origin). Then write equations for the lines including segments AE and FC respectively. Do this yourself, then check vs. the following result:

y = (-2y/x) x + 2 y

y = (-y/2x) x + y

Now you can solve these simultaneous equations for the point of intersection G.

Pause and note what you have done here so far: you set up a figure, you assigned quantitites to important distances, and you used algebra to figure additional information. This is a good general approach to such problems!!!

Do this then check vs. the following point:

G = ((2/3)x,(2/3)y)

From there the path to the solution should be evident: divide the target quadrilateral into pieces and conquer.

If you need to check, you should have obtained an inset rectangle of (1/9)xy area, and two triangles each of (1/36)xy area, for a total of (1/6)xy area. Since you were given the area of ABCD, you should be able to calculate that?

--- Cheers, -- Mr. d.