I have an SAT II question that asks: What is the area of the polygon formed by the points (x,y) which satisfy the inequality: $ |x| + \frac{|y|}{2} \leq 1$
a) 2 b) 3 c) 4 d) 8 e) 10
How would you go about solving this?
What is the area of the polygon fromed by the points (x,y) which satisfy the inequality: $ |x| + \frac{|y|}{2} \leq 1$
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$\begingroup$
absolute-value
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0Try drawing the region described by $\lvert x \rvert + \lvert y \rvert /2 \le 1$. It should break up nicely into smaller sub-regions of which you can easily calculate the area. – 2017-01-10
3 Answers
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Hint
Divide your graph into quadrants. In each one, draw the region formed. e.g., in the top right hand quadrant you have the half plane $x+y/2\le1$, in the lower left quadrant you have $-x -y/2 \le 1$ and so on.
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The shape above will have twice the area of the points that satisfy $|x|+|y| \le 1$ which is four times the area of the points that satisfy $x+y \le 1$ and $x,y \ge 0$. The latter is easily seen to be ${ 1\over 2}$ hence the answer is $4$.
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$$|x| + \frac{|y|}{2} \leq 1$$
This is a taxicab geometry ellipse or, better yet, a rhombus. The coordinates of the vertices are $(0,\pm 2), (\pm 1, 0)$.
The lengths of its diagonals are $D = 2-(-2) = 4$ and $d=1-(-1) = 2$
Notice that the diagonals break up the rhombus into four congruent right triangles with sides of $\frac 12D$ and $\frac 12d$ and area of $\frac 18dD$.
So the area of the rhombus is $4(\frac 18dD) = \frac 12 dD = 4$