Explanation:
15.
the area of the whole region is the sum of the area of the rectangle BCDE and the triangle ABE.
the area of a rectangle is
length × width
the area of a triangle is
baseline × height / 2
as a nice coincidence (that reduced the amount of work we have to do) the points of the rectangle are on the same x and y lines parallel to the x-axis and y-axis.
so, we can directly see, the distance BC = 5 (only the x- difference matters, as both have the same y- coordinate).
the distance CD = 4 (only the y- difference matters, as they have the same x-coordinate).
DE = 5, EB = 4. so, a true rectangle.
for the triangle we need the length of the baseline (EB = 4) and the length of the height to that baseline (FA = 2 from -2 to 0 of the x-values, and again they have the same y- coordinate).
so, the area of the rectangle is
5×4 = 20 units²
the area of the triangle is
4×2/2 = 4 units²
therefore, together, the shaded area is 20+4 = 24 units².
24.
there is no shaded region (all is white), but I assume we are taking about the inner regions with all sides being continuous lines.
that area is the area of the total rectangle AGDH minus the 2 right-angled triangles BGC and EFH.
the same formulas for the area of rectangle and a triangle apply as before.
right-angled triangle have a big advantage, as their legs can be used as baseline and corresponding height.
again, the points are sharing x- and y- coordinates, so that we only need to calculate the difference of either the x- or the y coordinates.
AG = 5
BG = 2
GC = 5
GD = 14 (6+8)
EH = 2
FH = 11 (6+5)
so, the area of the large rectangle is
14×5 = 70 units²
the area of BGC is
5×2/2 = 5 units²
the area of EFH is
11×2/2 = 11 units²
the total inner area is then
70 - 5 - 11 = 54 units²