take the square root of that which is 14.0028. the radius is approximately 14
Area of rectangle = 2 × 1 = 2 ft²
Area of rectangle = area of circle = 2 ft²
πr² = 2
22/7 × r² = 2
r² = 7/11
r = 0.7977 ft
We that the area of a circle is Pi times r^2 where r is the radius
Let A be the area of the first circle and S the area of the second one and T the total one
T = Pi( 4^2 +6^2)
= Pi ( 16+36) = Pi× 52
52 is the radius square
So r = root square 52 = 2 root square 13
The radius of the circle is, ft
Area of the circle(A) and Area of rectangle(A') is given by
where, r is the radius of the circle
A' = lw ....
where, l is the length and w is the width of the rectangle.
As per the statement:
Area of circle equals the area of a rectangle.
⇒A = A'
It is also given that: a measure of rectangle is 2 ft. by 11 ft.
⇒l = 2ft and w = 11 ft
Substitute in  we have;
Since, A = A'
Divide both sides by 22 we have;'
Multiply both sides by 7 we have;
Therefore, the radius of the circle is, ft
616 = 3.1416 * r^2, r^2 = 616/3.1416 = 196.07, r = sqrt(196.07) = 14.002 meters.
Looks like it was written to be almost 14. The answer is 14.002 m, maybe rounded to just 14 m
616 = pi r^2
r^2 = 616/ pi = 196.08 meters
r = sqrt 196.08 = 14 meters to nearest meter
A(r) = A(c) = W*L = 2*11 = 22 ft²
A(c) = πr²
π = 22/7
22/7*r² = 22
Multiply both sides for 7/22 to get rid of the fraction.
7/22 * 22/7 * r² = 22 * 7/22
r² = 7
r = √7
rectangle area=length. width
r=square root of 7
an advantage of the standard deviation is that it increases as the dispersion of the data increases. e. the interquartile range is preferred when the data are not skewed or no have outliers. an advantage of the standard deviation is that it uses all the observations in its computation.
answer: i guess 5 units?