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anonymous
 one year ago
What is the area of the rectangle? Round to the nearest square inch.
A.
27 in2
B.
31 in2
C.
45 in2
D.
47 in2
http://static.k12.com/calms_media/media/1579000_1579500/1579373/2/4ac8cba629d8de89098f07b616577fd1aa6b2747/MS_IMC150126130912.jpg
anonymous
 one year ago
What is the area of the rectangle? Round to the nearest square inch. A. 27 in2 B. 31 in2 C. 45 in2 D. 47 in2 http://static.k12.com/calms_media/media/1579000_1579500/1579373/2/4ac8cba629d8de89098f07b616577fd1aa6b2747/MS_IMC150126130912.jpg

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iGreen
 one year ago
Best ResponseYou've already chosen the best response.2Use the Pythagorean Theorem to find the Length of the rectangle.

iGreen
 one year ago
Best ResponseYou've already chosen the best response.2\(\sf a^2 + b^2 = c^2\) 'a' and 'b' are the two legs, in this case the length and the width, and 'c' is the hypotenuse, in this case the diagonal of the rectangle.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so its \[5^2+b^2=8^2\] ??

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0how do i solve for b again?

iGreen
 one year ago
Best ResponseYou've already chosen the best response.2Then after you find 'b', multiply it to 5 to get the Area.

iGreen
 one year ago
Best ResponseYou've already chosen the best response.2\(\sf 5^2 + b^2 = 8^2\) Subtract 5^2 to both sides: \(\sf b^2 = 8^25^2\) Find the square root of both sides: \(\sf b = \sqrt{8^25^2}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i got \[b=\sqrt{39}\]

iGreen
 one year ago
Best ResponseYou've already chosen the best response.2Yep, now find the square root of 39

iGreen
 one year ago
Best ResponseYou've already chosen the best response.2Yep, now multiply it to 5.

iGreen
 one year ago
Best ResponseYou've already chosen the best response.2Yep, that's our answer.
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