8.1 Geometry Essentials

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Learning Objectives

By the end of this section, you will be able to:

identify essential features of, as well as compute area and perimeter of, basic geometric shapes: circles, triangles, polygons
identify and use right triangles and related facts such as the Pythagorean Theorem
identify essential features of, as well as compute volume and surface area of, basic 3D solids: cylinders, prisms, spheres

This section is going to cover the essentials of geometry that everybody should have on instant recall before going into Calculus I. Many application problems use physical situations involving cylindrical or conical tanks, or spherical balloons, or spreading circles of oil spills... Calc I is basically devoted to the concept of a derivative, which is code for "rate of change" physically, so related rates problems involve things like water leaking from a container or the volume of a balloon expanding. In optimization problems, you need to know basic geometry of boxes or cylinders, so that you can find the dimensions that will result in the maximum volume or minimum surface area under certain constraints, for example. A professor might provide you with more obscure formulas, like the surface area of a cone, say, but they will certainly expect you to know how to find the area of a circle or triangle right off the top of your head. In other words, this section is again a big survey of "fun facts" about shapes that you need to memorize if you haven't already!

Area and Perimeter

My brother just bought a house. In taking up the rattling old tiles, he discovered that there is terrazzo flooring underneath. Instead of covering up that nice surface, he is going to have it restored/refinished. Such services are priced depending on the size of the job. If the refinishers will charge $2 per square foot, then what he needs to know is the area of his house's floors! Area comes with units that are "square" such as square feet (ft$^2$) or square miles (mi$^2$).

Perimeter is like measuring how many yards of fencing material you would need to enclose a whole entire yard. It's the sum of the lengths of all the sides of a shape (or in a circle, the length of a string if you laid it all the way around the edge of the circle, and then stretched it out and measured it). As such, the units of a perimeter are distance measures like centimeters, feet, miles, etc.

Here are the basic shapes for which you must know how to find area and perimeter.

Rectangles

Rectangles have four sides, with opposite pairs having equal length, and all their internal angles are right angles ($90^\circ$). Squares are just special rectangles where all the sides are the same length.

$rectangles.png$

The area of a rectangle is "length times width," $A_{\text{rect}} = lw$ square units. In the case of a square, length and width are the same, so if the side length is $x$, then $A_{\text{sq}} = x^2$ square units. The perimeter of a rectangle adds up the two sides of length $l$ and the two sides of length $w$, so $P_{\text{rect}} = 2l + 2w$ units. What is the perimeter of a square with side length $x$?

Exercise $\PageIndex{1}$

Find the area and perimeter for each rectangle. Figures may not be drawn to scale!

$rect1.png$	$rect2.png$	$rect3.png$
1.	2.	3.

Answer

$A = 200$ square units. $P = 60$ units.
$A = 72$ square units. $P = 34$ units.
$A = 6x^2 $ square units. $P = 2x+2(6x) = 14x$ units.

Triangles

Triangles are three-sided shapes. There are a few types of triangles to know about:

Right triangles have a right ($90^\circ$) internal angle. The sides on either side of the right angle are the legs and the third (longest) side is the hypotenuse. For these triangles, the height is the same as the length of a vertical leg.
Equilateral triangles have all three sides of the same length. All three of their internal angles are also the same, $60^\circ$. The height of these triangles is NOT the same as their side length.
Isosceles triangles have two sides of equal length. The angles opposite those sides will also be equal. The height of these triangles is NOT the same as their side length. (Technically, an equilateral triangle is a special isosceles triangle, just like a square is a special rectangle...)
Scalene triangles have all different length sides. The height of these triangles is NOT the same as their side length.

$triangles.png$

The area of a triangle is always "one-half base times height," $A_{\text{tri}} = \frac{1}{2}bh $. The perimeter of a triangle is the sum of its side lengths. If you know those lengths, you just add them up. If you don't know them, it can be slightly tricky to find them... But if it's a right triangle, there is a super important math fact that makes life easier.

Pythagorean Theorem

In a right triangle with leg lengths $a$ and $b$, and hypotenuse length $c$, the following is true:

\[ a^2 + b^2 = c^2 \notag \]

This can be used to find the length of any side of a right triangle if you know the other two. Simply plug in the known values and solve for the third.

$pyth.png$

Example $\PageIndex{1}$

Find the area and perimeter of the triangles below. Note that diagrams may not be to scale, so you must use math.

$tri.png$	$tri1.png$
1.	2.

Solution

1. This is a right triangle and we're given two side lengths: one leg and the hypotenuse. To find the area, we need to find the height, the length of that other leg. We use the Pythagorean Theorem, plugging in the sides we know and using a variable name for the other leg:

\[ 1^2 + b^2 = 2^2 \quad \rightarrow \quad b^2 = 4 - 1 = 3, \quad \quad \rightarrow \quad b = \sqrt{3} \notag \]

Note that we only use the positive square root because a negative side length doesn't make sense. Now that we know all three sides (and thus the height), we compute:

\[ A = \frac{1}{2}bh = \frac{1}{2}(1)(\sqrt{3}) = \frac{\sqrt{3}}{2} \quad \quad P = 1 + 2 + \sqrt{3} = 3 + \sqrt{3} \notag \]

2. (Tick marks in sides of shapes mean that the lengths are equal, if you didn't know.) This is an isoceles triangle, and we're given the base and the height. We can get the area straightaway: $A = \frac{1}{2}bh = \frac{1}{2}(12)(10) = 60 $. To find the perimeter, we need to know length of the other sides (which are the same). To find this, notice that cutting the triangle in half down the given height line creates two right triangles! The base of each right triangle is just half of the whole triangle's base, so it must be $5$. The unknown side we want is the hypotenuse of the right triangle. We can find it using the Pythagorean Theorem.

\[ 5^2 + 12^2 = c^2 \quad \rightarrow \quad 25+144 = c^2 \quad \rightarrow \quad 169 = c^2 \quad \rightarrow \quad c = 13 \notag \]

Now we can compute the perimeter, knowing the base is 10 and each of the other sides is 13: $P = 10+13+13 = 36$.

Exercise $\PageIndex{2}$

Find the area and perimeter of the triangles. Leave answers exact (don't use a calculator, just write square roots).

$1.png$	$3.png$	$2.png$
1.	2.	3.

Answer

$A = 12$ square units, $P = 12$ units.
$A = 24$ square units. $P = 2\sqrt{52}+12 $ units.
$ A = 8$ square units. $P = \sqrt{5} + \sqrt{13} + 4 $ units.

Special Triangles

Here are some special triangle types that, if you learn to recognize them, will make your life wayyyy easier. I explain them below the graphic.

$special right.png$

45-45-90 Right Triangles: These triangles have a $90^\circ$ angle so they're right triangles, and the other two angles are both $45^\circ$. This means that the legs have equal length. It also turns out that the hypotenuse is $ \sqrt{2}$ times the length of the legs (use the Pythagorean Theorem to check this).

30-60-90 Right Triangles: These triangles have a $90^\circ$ angle so they're right triangles, and the other two angles are $30^\circ$ and $60^\circ$. Whatever the shortest leg (across from the $30^\circ$) is, the hypotenuse is twice that value, and the other leg is $\sqrt{3}$ times that value (again, comes from Pythagorus).

Pythagorean Triples: These are right triangles whose side lengths are all integers. For example, a triangle with side lengths $(3, 4, 5)$ is always a right triangle. The other easiest one is $(5, 12, 13)$, and you might see $(7, 24, 25)$. Note that any MULTIPLE of these side lengths is also guaranteed to be a right triangle. Stuff like $ (6, 8, 10)$.

Circles

Circles have one essentially defining dimension: the radius, $r$. This is the uniform distance from any point on the edge of the circle to the center. The total maximum distance across a circle passing through the center is called the diameter, and it is simply double the radius. The area of a circle is $ A_c = \pi r^2 $. The perimeter of a circle, or the distance all the way around its edge, is called the circumference, and computed $C = 2\pi r$.

Exercise $\PageIndex{3}$

Give the radius, diameter, circumference, and area of the circles.

$a.png$	$b.png$
1.	2.

Answer

$r = 1$ units, $d = 2$ units, $C = 2\pi$ units, $A = \pi$ sq. units
$d = 8$ units, $r = 4$ units, $C = 8\pi$ units, $A = 16\pi$ sq. units

Pieces of Pizzas

You can also use your common sense about the area of portions of circles. This is very easy when you see semicircles or quarter-circles: just take the appropriate fraction of the area of an imaginary full circle!

Example $\PageIndex{2}$

This shape is a semicircle, which is half of a full circle that I've shown with a dashed line. Find the area and perimeter of the semicircle.

$semicircle.png$

Solution

The radius of the full circle would be $r$, so the area of the full circle would be $A = \pi r^2$. The area of the semicircle is half of that, so we report $\frac{1}{2}\pi r^2$. Now, be careful with the perimeter... Part of the perimeter is that round boundary of the semicircle, which would be half of the full circle's circumference. That piece must be $ \frac{1}{2}(2\pi r) = \pi r $. But there is also that bottom boundary, which is a straight line. It would actually be the diameter of the full circle, so it has length $2r$. The total perimeter of the shape is then $2r + \pi r $ or $(2+\pi)r$, if you prefer.

Decomposing Shapes

For other 2D shapes, the key to finding area boils down to just cutting them into pieces whose area you can find easily. I personally don't usually have the area formula of a trapezoid on rote recall, because it's easy for me to re-derive it if I really need it, using slice-and-dice reasoning. Let's do some examples.

Example $\PageIndex{3}$

Find the area. (In 1. you may assume that this is an isosceles trapezoid, meaning the slanty sides are equal.)

$1.1.png$	$1.4.png$	$1.7.png$
1.	2.	3.

Solution

1. This is a symmetrical trapezoid, and rather than memorizing the formulas for those, I just imagine slicing it into pieces. One option is to cut two congruent right triangles off the ends, like this:

$1.2.png$

This breaks the shape down into three simple shapes: two identical right triangles with base 1 and height 3, and a rectangle with length 4 and width 3. You can get those areas and add them up, or another shortcut is to imagine cutting off one triangle, rotating it, and sliding it over to the other side... This shows that the area of the trapezoid is exactly the same as the area of a rectangle with length 5 and width 3!

$1.3.png$

Either way, the area of the shape must be $A = 15$ square units. By the way, if you do want the formula for the area of a trapezoid with base $b$ and top side $a$ and height $h$, it would be $A = \frac{a+b}{2}h $.

2. This is a parallelogram, and we've been given the base and the height. To find the area, once again break down the shape into triangles and rectangles, or imagine cutting off the triangular portion on the left and sliding it over to fit against the right side of the shape... The area of a parallelogram is the same as the area of a rectangle with the same base and height!

$1.5.png$

Thus the area is $A = 15$.

3. This shape can be divided into two pieces: a rectangle with length 4 and width 2, and a semicircle of some size. We observe that the diameter of the circle is 4, so the radius must be 2. The area of a circle with radius 2 is $\pi r^2 = \pi (2^2) = 4\pi$, so the area of a semicircle is half of that. Thus the total area of the shape is $A = 2(4) + \frac{1}{2} 4\pi = 8+2\pi$ square units.

Exercise $\PageIndex{4}$

Find the area. Figures may not be drawn to scale!

$shape.png$	$L.png$	$corners.png$
1.	2.	3.

Answer

$A = \frac{3}{4} (4\pi) = 3\pi $ square units. (This is three-fourths of a circle with radius 2.)
$A = 16$ square units.
$A = 104$ square units.

Volume and Surface Area

Let's move from 2-dimensional shapes in the plane on to 3-dimensional shapes like cubes, spheres, cylinders, rectangular prisms (boxes), etc. Imagine filling a cylindrical tin can with a liquid. The volume of the can is how much liquid it can hold, reported in "cubic" units like cubic centimeters (cm$^3$) or cubic feet (ft$^3$), or other volume units like liters (1 L = 0.001 m$^3$). Now imagine that you take the tin can, cut off the circular top and bottom, and then slice down the side and flatten out the tin sheet it was made from:

$sa.png$

The total area of those three 2D shapes is called the surface area of the cylinder. There are handful of important standard 3D shapes you should know.

Boxes

A box is a rectangular prism (this includes cubes, as they're just special boxes where all sides have the same length). You can think of them as having a top, bottom, and four sides. There are three defining dimensions for a box: the length, width, and height.

$box.png$

The volume of a box is "length times width times height," $V = lwh$, or in the case of a cube where all sides have length $x$, $V = x^3$. The surface area can be found by imagining cutting apart a cardboard box like in the figure above. There are pairs of sides with different areas, $hw$ or $hl$ or $lw$. The total surface area of the box is $S=2hw+2hl+2lw$. What is the surface area of a cube then?

Exercise $\PageIndex{5}$

Find the volume and surface area.

$box1.png$	$box2.png$
1.	2.

Answer

$V = 400$ cubic units. $S = 340$ square units.
$V = 27$ cubic units. $S = 54$ square units.

Cylinders (Extruded Solids)

As a child, you might have played with Play-Doh and certain accoutrements called extruders, which have some kind of shaped opening, through which you force a blob of dough. The result is a long snake with consistent cross section, such as a long cylinder, or a long triangular prism. It's easy to calculate the volume of any such shape with consistent cross section area and height $h$. You just multiply the area of the base times the height. Here are some examples:

$extruded solids.png$

Thus the volume of the cylinder is $V_{\text{cyl}} = \pi r^2 h $. The volume of the box is consistent with the formula we just learned: $V_{\text{box}} = (lw)(h)$. The volume of the triangular prism shape is $V_{tr} = \frac{1}{2}abh $.

Finding surface area of such shapes is all about imagining cutting apart the shape as if it were made out of cardboard. Just like the tin can example we used before:

$sa.png$

The surface area of a cylinder consists of two circles, both with area $\pi r^2$, and a rectangular sheet when flattened out. The height of that rectangle is the height of the cylinder, but what is the length? Well, it was wrapped all the way around the circular base, so the length is exactly the circumference of the circle! Thus the area of the rectangle is $2\pi r h$. The surface area of a cylinder as a whole is thus $S_{\text{cyl}} = 2 \pi r^2 + 2\pi r h $.

Exercise $\PageIndex{6}$

Find the volume (you can use a calculator for the numbers) and the surface area (break down into the pieces shown).

$tris.png$

Answer: $V = \frac{1}{2}(6)(4.9)(15) =220.5$ cubic units. $S = 2 \left( \frac{1}{2}(6)(4.9) \right) + 5(15) + 6(15) + 7(15) = 299.4$ square units.

Spheres

A sphere, just like a circle, has just one defining dimension: its radius $r$. A sphere is all the points in 3D space that are exactly a distance of $r$ away from the center.

$sphere.png$

Once you know $r$, you can compute the volume and surface area:

\[ V_{\text{sphere}} = \frac{4}{3} \pi r^3, \quad \quad \quad S_{\text{sphere}} = 4 \pi r^2 \notag \]

Unfortunately, I don't have a super quick and easy trick for you to see why these formulas are true, but if you want to do a bit of a deep dive, this is a cool video! For now, you'll probably just want to memorize these formulas. At least, you can remember that since we're doing something circlish, there better be a $\pi$ in there somewhere! And then since volume comes in cubic units, it makes sense that the $r$ should be cubed in that formula, while the surface area has $r$ squared and area comes in square units. Maybe those tricks will get you started. Meanwhile, practice using the formulas in the exercise below.

Exercise $\PageIndex{7}$

Find the volume and surface area of the sphere

with radius $1$ in.
with radius $ 4$ cm.
with radius $x$ units.

Answer

$V = \frac{4}{3}\pi$ cubic inches. $S = 4\pi$ square inches.
$V = \frac{256}{3} \pi $ cm$^3$. $S = 64\pi $ cm$^2$.
$V = \frac{4}{3} \pi x^3 $ cubic units. $S = 4 \pi x^2$ square units.

Okay, those are the basic formulas that you will need going forward. There are, of course, other shapes that professors may or may not expect you to know off the top of your head. For example, the volume of a right circular cone is one-third the volume of a cylinder with the same base and height:

$cone.png$

\[ V_{\text{cyl}} = \pi r^2 h \quad \rightarrow \quad V_{\text{cone}} = \frac{1}{3} \pi r^2 h \notag \]

When in doubt, just ask your professor what kind of geometric shapes and solids they want you to have on instant recall.

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