Question

Here is an attached question.

Answer 1

(19/2 - 19/6)*19/4*1/2*4?

Answer 2

60.167?

Answer 3

19/4 is the radius for Circle M and P.
Let r be the radius for circle and N and Q.
In the triangle OMN, (19/4)^2 + (19/2-r)^2 = (19/4 + r)^2 => r = 19/6
4*Area of the Triangle OMB = Area of Quad MNPQ
4*1/2*(19/2-19/6)*19/4 is the Area required.

Answer 4

Let x be the radius of one of the smallest circle and N be its ceneter. Let M be the center of the circle of diameter AO. Notice the:

ON = 19-x, OM=19/2 and MN = x + R/2 . In the right triangle MON , one can write:
\[
(19/2 + x)^2 = 19^2/4 + (19 - x)^2 .
\]
Solving for x gives

x = 19/3 .

Therefore

ON = 19 - 19/3 = (2/3) * 19 .

Now the area of the quadrilateral MNPQ is equal to four times the area of the triangle OMN . Hence this area is equal:

4 ( 1/2) OM*ON= 2 * (19/2)*(2/3)* 19 = 241 to the nearest integer.

Answer 5

Oh yeah 19/2 not 19/4 :/