Area of the quarter sector
Area of the entire circle = pi * r^2
Area of the quarter circle = pi r^2 / 4
r = 9
Area of the quarter circle = pi 9^2 / 4
Area of the quarter circle = pi 81/4 = 20.25 pi
Area of the triangle
Area pf the triangle = 1/2 b*h
b = h = r
r = 9
Area of the triangle = 1/2 9 * 9
Area of teh triangle = 1/2 81
Area of the triangle = 40.5
Area of the shaded area
The shaded area = Area of the 1/4 circle - area of the triangle
The shaded area = (20.24 pi - 40.5)ft^2
A <<<<<< Answer.
Answer:
7/50 (also 18% if u need it)
Step-by-step explanation:
18/100 ----- 7/50
Hope this helps!
First, we are given that the inscribed angle of arc CB which is angle D is equal to 65°. This is half of the measure of the arc which is equal to the measure of the central angle, ∠O.
m∠O = 2 (65°) = 130°
Also, the measure of the angles where the tangent lines and the radii meet are equal to 90°. The sum of the measures of the angle of a quadrilateral ACOB is equal to 360°.
m∠O + m∠C + m∠B + m∠A = 360°
Substituting the known values,
130° + 90° + 90° + m∠A = 360°
The value of m∠A is equal to 50°.
<em>Answer: 50°</em>
Answer: 
+ 
= 
Step-by-step explanation:
The standard form of equation of a circle is written in the form:
+ 
= 
where a and b are the coordinates of the center and r is the radius.
All we need to do is to covert the given equation into this form, how do we do that?
We will achieve this by using the completing method of solving quadratic equation.
Firstly re - write the equation , that is
- 6x + 
-4y = 12
The next thing is to Complete the square for each of x and y
How do we do this ?
i. multiply the coefficient of x and y by 1/2
ii. square the results
iii. Add the result to both sides of the equation , that is
- 6x + 
- 4y + 
= 12 ++
The next thing is to write the Left hand side of the equation as a perfect square binomial and simplify the Right hand side. That is
+ = 25 , that is
+ =
Which is now in standard form of the equation of a circle.
