As shown in the figure, two identical right triangles are stacked together to find the area of the shadow part. (unit: cm)

As shown in the figure, two identical right triangles are stacked together to find the area of the shadow part. (unit: cm)

Because the area of Figure 1 + the area of Figure 2 - the area of Figure 2 + the area of Figure 3, so: the area of Figure 3 = the area of Figure 1, figure 1 is a trapezoid, the upper bottom is 12 cm, the lower bottom is 12-3 = 9 (CM), the height of the trapezoid is 6 cm, so the shadow area of Figure 1 is: (12 + 9) × 6 △ 2 = 21 × 6 △ 2 = 126 △ 2 = 63 (square cm). A: the area of the shadow part is 63 square cm

It is known that the height of trapezoid ABCD is h, and the two diagonals form 30 degrees and 45 degrees with the bottom respectively, so as to calculate the trapezoid area
Such as the title

Let the trapezoid ABCD, AB > CD, ab ∥ CD be de ⊥ AB, CF ⊥ AB, the perpendicular foot e, F, AC and BD are diagonals, < cab = 30 °, DBA = 45 °, the quadrilateral DCFE is rectangle, CD = EF, de = h, CF / AF = tan30 °, AF = CF / tan30 ° = H / (√ 3 / 3) = √ 3h, de / be = tan45 °, be = de / tan45 ° = h, AF + be = AE + E

It is known that the right angled trapezoid is 30 cm high,

It can be seen from the figure that AE = ad, ∠ BEC = 45 ° and then ∠ BCE = 45 ° so EB = BC
Trapezoid height AB = 30cm, that is AE + EB = 30cm, so AD + BC = 30cm
Trapezoid area s = (AD + BC) * AB / 2
=30*30/2
=450 (square centimeter)

It is known that in trapezoidal ABCD, angle a is 90 degrees, ad is 3, DC is 6, and angle c is 60 degrees

Make de ⊥ BC, perpendicular e, abed as rectangle, be = ad = 3,
Angle c is 60 degrees, angle CDE is 30 degrees, EC = DC / 2 = 6 / 2 = 3; [right side of right triangle opposite 30 degrees = hypotenuse / 2]
DE²=DC²-EC²=6²-3²=27,
DE=3√3,
BC=BE+EC=3+3=6,
Trapezoidal ABCD area
=(AD+BC)*DE/2
=(3+6)*3(√3)/2
=27(√3)/2