The area of the shaded sector is: 245.7 square units.
That depends on what area you choose to shade.
In math, when a fraction is shaded, it typically refers to the portion of a shape or region that has been colored in or highlighted. This visual representation helps to understand the concept of fractions as parts of a whole. The shaded area represents the numerator of the fraction, while the total area of the shape represents the denominator. By visually seeing the shaded portion in relation to the whole, students can grasp the concept of fractions more concretely.
The answer depends on whether it is the smaller square that is shaded or the bits that are left. The area of the smaller square is 56% of the larger square.
What do you call shaded part?
The area is 0 square units since no shaded area is visible.
The area of the square is 98 square cm. Assuming the shaded area is the remainder of the circle, its area is 55.9 square cm (approx).
The area of the shaded sector is: 245.7 square units.
Area = pi*122 = 144pi square units Shaded area = (260/360)*144pi = 104pi square units
The approximate area of the shaded region of 10 cm is 100 square centimeters.
Typically, when a mathematical problem wants you to find the value of a shaded area, it wants you to calculate the area. If the shaded area is a circle, the area can be found by multiplying pi by the square of the radius. If the shape is a triangle, the area is base times height, divided by 2. If the shape is a square or rectangle, the area is length times width.
A circle with a radius of 135 units has an area of 57,255.53 square units.
Since there is no shaded part, the answer must be that its area is 0 square units.
You find the area of the whole square first. Then you find the area of the circle inside of it And then subtract the area of the circle from the area of the square and then you get the shaded area of the square
There is no shaded space (indeed, no triangle) so the answer must be 0.
(Length of side of square)^2 - Pi * radius^2
I wanted to know how to solve the problem.