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?

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.

0=6x+2y+10 is equivalent to the simplified form: y= -3x - 5• parallel lines always have the same slope, different y intercept knowing this, we can say the slope, or m of the line parallel to y= -3x - 5 is -3.m= -3 hope this could !