It is known that the figure is an isosceles right triangle with a right angle side of 8 cm. Calculate the area of the shadow part in the figure. (unit: cm)

It is known that the figure is an isosceles right triangle with a right angle side of 8 cm. Calculate the area of the shadow part in the figure. (unit: cm)

According to the analysis of the question stem, we can get: 8 △ 2 = 4 (CM),
The area of shadow is (3.14 × 42 × 90)
360-4×4÷2）×2，
=（12.56-8）×2，
=4.56×2，
=9.12 (square centimeter),
A: the shadow area is 9.12 square centimeters

The hypotenuse length of an isosceles right triangle is 8 cm It's better to have a primary school equation

Because the hypotenuse AB = 8, AC = 8 uses the trigonometric function sine 30 degrees to be equal to the opposite side / hypotenuse = BD / AB = 1 / 2. And because AB = (8). The cross multiplication BD = 4

The hypotenuse length of an isosceles right triangle is 8 cm

Let the area of an isosceles right triangle be x and the length of its right angle side be y, then x = 1 / 2 * y * y. according to Pythagorean theorem, y * y + y * y = 8 * 8, y * y = 32. X = 32 / 2 = 16

We know that the side length of the big square is 12 cm and the side length of the small square is 8 cm

8×8÷2+12×12-（12+8）×12÷2-（12-8）×12÷2
=32+144-120-24
=32 (square centimeter)
A: the shadow area is 32 square centimeters

We know that the side length of the big square is 12 cm and the side length of the small square is 8 cm

8×8÷2+12×12-（12+8）×12÷2-（12-8）×12÷2
=32+144-120-24
=32 (square centimeter)
A: the shadow area is 32 square centimeters

As shown in the figure, if the side length of the large square is 8cm and the side length of the small square is 6cm, then the area of the shadow part is______ .

S shadow = s trapezoidal ebcd + s △ aed-s △ ABC,
=（6+8）×6÷2+8×8÷2-（6+8）×6÷2，
=8×8÷2，
=32 (square centimeter);
A: the shadow area is 32 square centimeters
So the answer is: 32 square centimeter

In the picture, the side length of the big square is 12 cm, and the side length of the small square is 8 cm. What is the area of the painted part

No picture, no way

As shown in the figure, the three sides of RT △ ABC are known to be 6, 8 and 10 respectively. Take its three sides as the diameter, make three semicircles upward, and calculate the area of the shadow part in the graph

The area of the semicircle with the diameter of AC is π× (6 ￣ 2) 2 × 12 = 92 π = 4.5 π, the area of semicircle with BC diameter is π× (8 ￣ 2) 2 × 12 = 8 π, the area of semicircle with ab diameter is π× (10 ￣ 2) 2 × 12 = 252, π = 12.5 π, the area of triangle ABC is 6 × 8 × 12 = 24, the surface of shadow part is

As shown in the figure, we know that the two right angles of the right angle △ ABC are 6 and 8 respectively. Take the three sides as the diameter to make a semicircle, and calculate the area of the shadow part in the graph

∵ the two right angle sides of ᙽ right angle ᙽ ABC are 6,8, ᙽ AB = 62 + 82 = 10, ∵ the area of semicircle with BC as diameter is ᙽ 12 π (82) 2 = 8 π, the area of semicircle with AC as diameter is ᙽ 12 π (62) 2 = 9 π 2, and the area with ab as diameter is

The length of the hypotenuse of a right triangle is 10 cm, and the difference between the two right sides is 2 cm. To find the length of the longer right angle side x, the equation listed is to convert it into a general form

The side length of the shorter right angle side is X-2. The square of Pythagorean theorem x plus the square of (X-2) is equal to the square of 10
The square of x minus 2x minus 48 equals 0