The area of rectangle ABCD is closest to which of the following?

Triangle ABC or ACD are 30-60-90 triangle, thus, AD= \(a\sqrt{3} = 10\sqrt{3} = 17 => Area = 10*17 = 170\)

Hence Answer is C

MAGOOSH OFFICIAL SOLUTION:

Suppose we don't know the math to answer this question. We are told it's a rectangle, so we know the angles must be right angles, and we know the area must be length (AD) times height (AB). We know the height is 10. We know AD is drawn to scale. It definitely is longer than AB, so the area is definitely larger than 10 x 10 (answer (A) is out). AD doesn't look as long as twice AB, so the area is definitely less than 10 x 20 (answers (D) & (E) are out). Notice, with pure spatial estimation, we eliminated three of the five answer choices, so it will be to our advantage to guess randomly from the remaining two if we can't decide between them. Estimating from size can be a huge help if you don't remember the way to solve the problem.

BTW, the real math solution to that question: from the properties of the 30-60-90 triangle (ACD), we know that \(AD = 10*\sqrt{3}\), and since \(\sqrt{3}\) is approximately 1.7, AD is approximately 17, and the area is approximately 170.

Answer = C.

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Let the length of the diagonal be h.
h*cos 60 = 10 (=b)
h*(1/2)=10
h=20

l*l + b*b =20*20
since b =10
=> l = 10 * sqrt(3)
Area =l*b = 10 * sqrt(3) * 10
=100 * sqrt(3)
=100 * 1.732
=173.2 closest to 170

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