Find the area of the figure. Round to the nearest tenth if necessary.
D(0,2), E(4, -2), F(8,2)
instead of d and f connecting with a straight line, they are connected with half a circle/curve
The highest point this semicircle part reaches is (4,6)...if im describing it in a confusing way imagine the shape of a sno-cone...its like that
If I understand you correctly, DF is the diameter of a half-circle
radius is clearly 4 units, so that part = π(4^2) = 16π units^2
slope EF = +1, slope DE = -1, so obviously you have a right-angled triangle
Find the length o EF and DE and use the area of a triangle formula to find the area.
Add up the two parts
To find the area of the figure formed by the points D(0,2), E(4,-2), F(8,2) connecting with a semicircle, you can break it down into two parts: the triangle formed by D, E, and F, and the semicircle.
First, calculate the area of the triangle formed by D, E, and F using the formula for the area of a triangle:
Area of triangle = (1/2) * base * height
The base of the triangle is the distance between points D and F, which can be calculated using the distance formula:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Distance between D and F = sqrt((8 - 0)^2 + (2 - 2)^2) = sqrt(64) = 8
The height of the triangle is the vertical distance between point E and the highest point of the curve, which is (4,6):
Height = 6 - (-2) = 8
Now, you can calculate the area of the triangle:
Area of triangle = (1/2) * 8 * 8 = 32 square units
Next, calculate the area of the semicircle:
The radius of the semicircle is the distance between the highest point of the curve (4,6) and either point D or F. Let's use point D:
Radius = sqrt((4 - 0)^2 + (6 - 2)^2) = sqrt(20) ≈ 4.47
The area of a semicircle can be calculated using the formula:
Area of semicircle = (1/2) * pi * radius^2
Area of semicircle = (1/2) * 3.14159 * 4.47^2 ≈ 31.4 square units
Finally, add the area of the triangle and the area of the semicircle to find the total area of the figure:
Total area = Area of triangle + Area of semicircle = 32 + 31.4 ≈ 63.4 square units
Therefore, the area of the figure is approximately 63.4 square units.