Question

The radius of circle A is three feet less than twice the diameter of circle B. If the sum of the diameters of both circles is 49 feet, find the area and circumference of circle A?

Answer 1

For this case we have the following system of equations according to the statement:

$r_ {a} = 2d_ {b} -3\\d_ {a} + d_ {b} = 49$

Taking into account that $d = 2r$, we can rewrite the second equation as:

$2r_ {a} + d_ {b} = 49$

We have the following system:

$r_ {a} = 2d_ {b} -3\\2r_ {a} + d_ {b} = 49$

We substitute the first equation in the second:

$2 (2d_ {3}) + d_ {b} = 49\\4d_ {b} -6 + d_ {b} = 49\\4d_ {b} + d_ {b} = 49 + 6\\5d_ {b} = 55\\d_ {b} = \frac {55} {5}\\d_ {b} = 11$

Thus, we have:

$r_ {a} = 2d_ {b} -3\\r_a = 2 (11) -3\\r_a = 22-3\\r_ {a} = 19$

Then the area of circle A is:

$A = \pi * (r_ {a}) ^ 2\\A = \pi * (19) ^ 2\\A = 1133.54 \ ft ^ 2$

The circumference is:

$C = 2 \pi * r\\C = 2 * 3.14 * 19\\C = 119.32 \ ft$

Answer:

$A = 1133.54 \ ft ^ 2\\C = 119.32 \ ft$