Given that the area of the triangle ABC is 12 square centimeters, calculate the area of the shadow part?

Let the right sides of the isosceles right triangle ABC be a, 12a2 = 12, A2 = 24; the area of sector abd: 18 π A2 = 3 π = 9.42 (square centimeter), the area of blank part BCD: 12-9.42 = 2.58 (square centimeter), the area of semicircle: 12 π (a △ 2) 2 = 12 × 3.14 × 14a2 = 18 × 3.14 × 24 = 9.42 (square)

In the right figure, the area of the right triangle [shadow part] is 12 square centimeters, and the area of the circle is () square centimeters

Dark style
12*2=24
3.14 * 24 = 75.36 cubic centimeter, so
The area of the right triangle [shadow part] is 12 square centimeters, and the area of the circle is (75.36) square centimeters

Shadow part is square, area is 12 square centimeter, seek circle area

3.14 * 12 = 37.68 square centimeter (because the side length of the square is exactly the radius of the circle)

As shown in the figure, two identical right triangles are overlapped, and the area of the shadow part is calculated according to the known conditions in the figure. (unit: cm)

According to the stem analysis, we can get the following conclusions
Area of shadow part = area of red trapezoid = (120-40 + 120) × 20 △ 2
=200×20÷2
=2000 (square centimeter),
A: the shadow area is 2000 square centimeters

As shown in the figure below, two identical right triangles overlap together. What is the area of the shadow part? There should be a problem-solving process, such as this: ∵S=XXXXX The inverted triangle is because and the positive triangle is so

Let AC and EF intersect with n
∵ the two triangles are the same
The two triangles are similar
∴NF/AB=FC/BC
5 / 8 = FC / (FC + 6)
∵FN=5
∴FC=10
∴BC=BF+CF=16
∵AB=8 EN=3 FN=5
ν s = area of triangle EFG - area of NFC = 64-25 = 39

As shown in the figure, two right triangles overlap together to find the area of shadow part

Because the areas of the two triangles are equal, the shadow area is equal to the trapezoid area on the left
(8+8-3)/2*5=32.5

As shown in the figure on the right, two identical right triangles overlap. The shadow area in the figure is () square centimeter

Thirty

Two identical right triangles, partially overlapped, as shown in the figure. What is the area of the shadow? (unit: cm)

（9-3+9）×4÷2，
=15×2，
=30 (square centimeter),
A: the shadow area is 30 square centimeters

As shown in the figure, if two identical right triangles are overlapped together, the shadow area in the figure is () square centimeter A. .24 B. .30 C. 60

（12-4+12）×3÷2，
=20×3÷2，
=30 (square centimeter);
Answer: the area of shadow in the picture is 30 square centimeter
Therefore, B

Δ ABC and △ def are two isosceles right triangles stacked together (as shown in the figure). Given BC = 10, CF = 1, de = 7, what is the area of the shadow?

According to the analysis, because △ ABC and △ def are two isosceles right angle triangles stacked together, so ∠ FMB = 90 °, FCG = 90 °, beh = 90 °, the height of △ FBM is equal to half of FB; because Fe = de = 7, CF = 1, CE = 7-1 = 6; because BC = 10, be = 10-6 = 4; FB = FC + BC = 1 + 1

Given that the area of the triangle ABC is 12 square centimeters, calculate the area of the shadow part?