What is the area of the shape contained by the points A (0, 6), B (-2, 4), C (0, 2), and D (2, 4)?
The first thing we should do is draw this out. That'll give us a better sense of what this shape is other than the extremely unhelpful "quadrilateral."
Rather than using the distance formula to check whether the sides are of equal length, we can take a look at the diagonals. Each of the diagonals is 6 – 2 = 2 – (-2) = 4 units long, and is bisected by the other. At the very least, we have a rhombus.
We can use the area formula for a rhombus using the diagonals, setting d1 = d2 = 4 units.
A = 8 units2
What is the area of the shape contained by the points A (-1, 0), B (-2, -2), C (0, -5), D (3, -3), and E (3, 0)?
There are several ways to go about this, but they all start by drawing out the figure.
We could find the area of the square contained by (-2, 0), (-2, -5), (3, -5) and (3, 0) and then subtract the areas of the triangles on the corners. There are other ways too, but whichever method you choose, you should end up with the same answer. It's just a matter of what you find easier.
We prefer subtracting the areas of the triangles from the square. Why use the distance formula if we don't have to?
The area of the square is 3 – (-2) = 5 by 0 – (-5) = 5, or A = s2 = 52 = 25 units2. The triangles we have to subtract are all right (as in right angles, not feelin' alright), so finding their areas shouldn't be too difficult.
They have areas of ½(1)(2) = 1 unit2, ½(2)(3) = 3 units2, and ½(3)(2) = 3 units2. Subtracting these areas from the area of the big square should give us the area of the pentagon we're interested in. (Not that way.)
First, let's start by splitting it up into pieces we can identify. The easiest to spot is the semicircle on the lower left side (going from A to B to C). The remaining pentagon (ACDEF) can be cut from A to D to form a triangle (ΔACD) and a quadrilateral (ADEF). If we add these individual areas, we'll get the area of the entire figure.
Let's start by calculating the area of the semicircle, AC. Since its diameter goes from A to C, it's clear that the radius of the semicircle is 3. We can multiply the area formula for a circle by ½ to find the area of the semicircle.
AC = 4.5π ≈ 14.1 units2
It's clear that ΔACD is a right triangle, with legs of lengths 3 – 0 = 3 and 2 – (-4) = 6. We can plug those numbers in as the base and height and find the area.
AT = ½bh AT = ½(3)(6) AT = 9 units2
Now for that quadrilateral. Before we can do anything, let's find out what type of quadrilateral it really is. Calculating side lengths is a good way to do this. Go, go, distance formula!
We have two pairs of congruent sides. What we have here is a parallelogram, so we know to use A = bh to find the area. Only one slight problem: what's the base and what's the height?
We know that the height has to be perpendicular to the base, so let's start there. We'll set our base to DE, which has a slope of 4. Our height should have a slope of (since negative reciprocal slopes mean perpendicular lines).
Calculating the height of this thing might be trickier than you'd expect, so we'd suggest setting up a system of linear equations. For instance, the intersection of lines and y = 4x – 16 will give us the point on a line perpendicular to DE that goes through (0, 2). Using the distance formula between that point, , and (0, 2) should give us the height.
Now we can use A = bh to find the area of the parallelogram. Rather than plugging in , we can estimate it at 16.49.
AP = bh AP = 16.49 × 4.37 AP ≈ 72.1 units2
We're still not done. Adding the pieces together will solve the last piece of the puzzle.
A = AP + AT + AC A = 72.1 units2 + 9 units2 + 14.1 units2 A = 95.2 unis2