Question

The areas of: FGHM is 12, MHIJ is 36, EFNL is 24, LNJK is 48. EGIK is a rectangle cut into the rectangles with given areas. What is the area of the shaded quadrilateral EMIN?

Answer 1

5

Overall area of quadrilateral EGIK is 12+36+24+48 = 120. I'll call the length of EK to be w, and EG to be h, representing width and height.

The line FJ splits EGIK into EFJK and FGIJ which have areas of 72 and 48 respectively. Since EFJK and FGIJ have the width, their heights have to be in the ratio of 72/48 = 3/2. So GF = 2h/5 and EF = 3h/5

Now FGHM and MHIJ share the same height, so their widths have to be in the ratio 12/36 = 1/3, so GH = w/4 and HI = 3w/4

And EFNL and EMIN also share the same height, so their widths have to be in the ratio 24/48 = 1/2, so EL = w/3 and  LK = 2w/3.

So the length of MN will be FN - FM. And FN is the same as EL which is w/3, and FM is the same as GH which is w/4. So
w/3 - w/4 = 4w/12 - 3w/12 = w/12

If you look at quadrilateral EMIN, you'll see that it's area is the sum of triangles MNI and MNE. And the area of a triangle is 1/2 bh. And conveniently, we have MN as the base with a length of w/12 for both triangles.

area MNI = 1/2 MN * GF = 1/2 * w/12 * 2h/5 = 2hw/120
area MNE = 1/2 MN * EF = 1/2 * w/12 * 3h/5 = 3hw/120

And their sums are 2hw/120 + 3hw/120 = 5hw/120.
And hw is the total area of EGIK which is 120. So we get:
5 * 120/120 = 5 * 1 = 5

So the area of quadrilateral EMIN is 5.

Answer 2

Area of EMIN = 5 The shaded quadrilateral EMIN is composed of two triangles MNI and MNE. Since the area of a triangle is 1/2bh where b is the base and h the height, the area of the quadrilateral is A = 0.5 * MN * MH + 0.5 MN * EF A = 0.5 MN (MH + EF) Now the large rectangle GIKE has an area of 12+36+24+48 = 120. We'll call the overall width "w" and the overall height "h", so wh = 120. (1) Consider that rectangle GIKE has been divided into 2 rectangles by line FJ. The area above the line is 12+36 = 48, and the area below the line is 24+48 = 72. So the ratio of their heights is 2/3. Therefore we'll call the height of the top rectangle 2h/5 and the bottom rectangle 3h/5. (2) Considering rectangles FGHM and MHIJ, their heights are the same and their areas are in the ratio 12/36=1/3. So we'll consider the width of FGHM to be w/4 and the width of MHIJ to be 3w/4. (3) Now consider rectangles EFNL and LNJK. They also share the same height, so their ratios of width is 24/48 = 1/2. So their widths will be w/3 and 2w/3 respectively. (4) This means that the line segment MN will be w/3 - w/4 = 4w/12 - 3w/12 = w/12 (See (2) and (3) above) Now taking the formula at the very beginning for the area of the quadrilateral EMIN, we have A = 0.5 MN (MH + EF) MN = w/12 ; See (4) above MH = 2h/5 ; See (1) above EF = 3h/5 ; See (1) above So A = 0.5 MN (MH + EF) A = 0.5 w/12 (2h/5 + 3h/5) A = 0.5 w/12 (5h/5) A = 0.5 w/12 h A = w/24 h A = wh/24 And since the overall area of the entire rectangle is wh = 120, we get A = wh/24 A = 120/24 A = 5 As for the actual width and height of the overall rectangle? Don't know and don't care. Any value of w and h will work as long as their product is 120. But regardless of the actual dimensions, the area of quadrilateral EMIN will be 5.