Question #61142

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Question #61142

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Answer 1 · 2017-11-05T20:55:58

$8.58 s q . u n i t s$ $8.58 sq. units$

Explanation:

The area of the square with the length $2 \sqrt{10}$ $2sqrt10$ is:
${(2 \sqrt{10})}^{2} = 40$ $(2sqrt10)^2 = 40$

The diameter of the inscribed circle is $2 \sqrt{10}$ $2sqrt10$ .
The area of the circle can be computed as:
$A = \frac{π d^{2}}{4}$ $A=(pid^2)/4$
$A = \frac{π {(2 \sqrt{10})}^{2}}{4}$ $A=(pi(2sqrt10)^2)/4$
$A = 10 π = 31.42$ $A=10pi=31.42$

Subtracting the area of the circle from the area of the square.
$40 - 31.42$ $40 -31.42$
$8.58 s q . u n i t s$ $8.58 sq. units$

Answer 2 · 2017-11-05T21:01:58

Given, that the side of the square is $s = 2 \sqrt{10}$ $s = 2sqrt10$ ft, then the radius of the circle is $r = \sqrt{10} ft$ $r = sqrt10" ft"$ .

The area of the square is:

$A_{square} = s^{2}$ $A_"square" = s^2$

$A_{square} = {(2 \sqrt{10} ft)}^{2}$ $A_"square" = (2sqrt10" ft")^2$

$A_{square} = 40 {ft}^{2}$ $A_"square" = 40" ft"^2$

The area of the circle is:

$A_{circle} = π r^{2}$ $A_"circle" = pir^2$

$A_{circle} = π {(\sqrt{10} ft)}^{2}$ $A_"circle" = pi(sqrt10" ft")^2$

$A_{circle} = 10 π {ft}^{2}$ $A_"circle" = 10pi" ft"^2$

The area not covered is:

$A = A_{square} - A_{circle}$ $A = A_"square" - A_"circle"$

$A = 40 {ft}^{2} - 10 π {ft}^{2}$ $A = 40" ft"^2 - 10pi" ft"^2$

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Question #61142

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