As shown in the figure, angle a plus angle B plus angle C plus angle D plus angle e equals

As shown in the figure, angle a plus angle B plus angle C plus angle D plus angle e equals

Let AE intersect CD in G and CD intersect AB in H, then
DGE=A+AHG=A+DHG=A+B+C
And DGE + e + D = 180 degrees, so a + B + C + D + e = 180 degrees

As shown in the figure, the areas of the two squares are 16 and 9 respectively, and the areas of the two shadow parts are a and B (a > b), then A-B is equal to
That is, two squares have a rectangular part overlapped at a corner, and the overlapped part is not a shadow. The shadow part of a large square is a, and the shadow part of a small square is B

Where is the picture?
Let the area of coincidence part be C, then a + C = 16, B + C = 9, so A-B = 16-9 = 7

As shown in the figure, ∠ a + B + C + D + e equals ()
A. 180°B. 360°C. 540°D. 720°

∫∠ 1 is the outer angle of △ CEF, ∫∠ 1 = ∠ C + ∠ E; ∫∠ 2 is the outer angle of △ BDG, ∫ 2 = ∠ B + ∠ D, ∫ a + ∠ 1 + ∠ 2 = 180 ° and ∫ a + ∠ B + ∠ C + ∠ D + ∠ e = 180 °. Therefore, a

As shown in the figure, angle a plus angle B plus angle C plus angle D plus angle e equals 180 degrees