What is the area of the polygon fromed by the points (x,y) which satisfy the inequality: $ |x| + \frac{|y|}{2} \leq 1$

Question

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?

Answer 1

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.

Answer 2

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 $4$.

Answer 3

$$|x| + \frac{|y|}{2} \leq 1$$

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 $4(\frac 18dD) = \frac 12 dD = 4$