Official Solution:

What is the area of a region enclosed by \(|\frac{x}{3}|+|\frac{y}{9}|=10\)?

A. \(675\)
B. \(1350\)
C. \(2700\)
D. \(5400\)
E. \(10800\)


Find the \(x\) and \(y\) intercepts.

When \(y=0\), then \(x=30\) or \(x=-30\).

When \(x=0\), then \(y=90\) or \(x=-90\).

So, we have 4 points: (30, 0), (-30, 0) (0, 90), (0, -90). When joining these points we get the rhombus:



The area of a rhombus is \(\frac{d_1*d_2}{2}\) (where \(d_1\) and \(d_2\) are the lengths of the diagonals), thus the area of the enclosed figure is \(\frac{60*180}{2}=5,400\).


Answer: D
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If you refer to the image you'd see that the figure is NOT a square (the diagonals are not equal).
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Hi Karan,

More reliable way to find out whether this is square or not is to know the fact that, when the Rise/Run is equal i.e 10/10 or 25/25 in otherwords where the Slope is 1, we end up with 45-45-90 triangle. we add up all the four 45-45-90 triangle we end up with square, in this example clearly the Rise/Run is not 1. Hence it is a Rhombus and not a square.