ACT Math For Dummies (2011)

Part IV. Visualizing Plane Geometry and Trigonometry

In this part. . .

In the chapters in this part, I cover plane geometry and trigonometry. I help you sharpen your skills on lines and angles, triangles, quadrilaterals, circles, solids, trig ratios, matrices, logarithms, and imaginary numbers. I also include some practice problems to get you up to speed quickly.

Chapter 10. Plain Talk about Plane Geometry

In This Chapter

Answering questions involving angles

Familiarizing yourself with triangles

Making peace with quadrilaterals

Discovering the ins and outs of circles

In this chapter, the topic is geometry. Although a typical geometry course spends a lot of time focusing on geometric proofs, what you need to succeed on the ACT is a review of the basics.

The chapter begins with a refresher on angles, including vertical angles, supplementary angles, angles that involve parallel lines, and angles in a polygon. Then I move on to triangles. I discuss the area formula for a triangle, the Pythagorean theorem, and a variety of common right triangles that you’re sure to see on the ACT.

Next, I move on to quadrilaterals, showing you the formulas for the area and perimeter of squares and rectangles as well as the area of a parallelogram and a trapezoid. I also show you how to find the area and perimeter of a circle and how to answer ACT questions that involve tangent lines, arc length, and chords.

I round out the chapter with a discussion of solid geometry, which is the extension of plane geometry to three dimensional solids. I introduce you to the formulas for the volume and surface area of a cube, a box (rectangular solid), and a sphere. I also give you a few formulas for a variety of other common solids.

Knowing Your Angles

In your geometry class, you probably spent a lot of time proving theorems about angles. This is good training for clear thinking and, of course, for passing your geometry tests. But, luckily, on the ACT, you won’t have to write any geometric proofs. However, you do need to know how to work angles. So, in this section, I give you the most important rules about angles and show you how to work with them to answer a variety of typical ACT questions.

Angles around one point

Here are three important rules for answering ACT questions about angles:

All the angles around a single point add up to 360°.

Supplementary angles — any pair of angles that combine to produce a straight line — add up to 180° degrees. (The angles on both sides of a line add up to the total angles around a single point, because 180° + 180° = 360°.)

Vertical angles — any pair of opposite angles formed when two lines intersect — are equal to each other. (Vertical angles are symmetrical to each other, so they’re equal.)

I show these relationships among angles in Figure 10-1.

In the following figure, what is the value of m?

(A) 10

(B) 12

(C) 15

(D) 18

(E) 20

The figure shows that p° and 4m° are a pair of vertical angles, so

p = 4m

Additionally, p° and (m + p)° are supplementary angles, so

p + (m + p) = 180

Simplify the second equation:

2p + m = 180

Substitute 4m for p and solve:

2(4m) + m = 180

8m + m = 180

9m = 180

m = 20

Therefore, the correct answer is Choice (E).

Angles and parallel lines

A pair of parallel lines extends infinitely in both directions at a constant distance from each other. Geometry includes a bunch of theorems about parallel lines. But don’t fret. For the ACT, you only need to be able to work with the angles that result when a line crosses a pair of parallel lines, as shown in Figure 10-2.

In the following figure, the two horizontal lines are parallel. What is the value of h + j?

(A) 56

(B) 66

(C) 76

(D) 78

(E) 80

Because the two lines are parallel, the preceding figure has only two distinct angles. The smaller angle measures k° and the larger angle measures 4k°. These two angles add up to 180°, so you can create and solve the following equation:

k + 4k = 180

5k = 180

k = 36

The smaller angle measures 36°, so

2j = 36

j = 18

Additionally, the larger angle measures 144° (because 36 + 144 = 180), so

3h = 144

h = 48

Therefore,

h + j = 48 + 18 = 66

So the correct answer is Choice (B).

Interior angles in a polygon

A polygon is a shape made entirely of straight line segments. When you know how many sides a polygon has, you can find the sum of the interior angles for that polygon — that is, the total of all angles inside the polygon — using the following formula:

Total interior angles = 180 (Number of angles – 2)

Table 10-1 gives you the sum of angles for the most common polygons.

Table 10-1 Sum of Interior Angles of Common Polygons

If the measures of four angles in a quadrilateral are 140°, n°, 3n°, and 6n°, what is the value of n?

(A) 10

(B) 20

(C) 22

(D) 24

(E) 36

The total measure of the four angles is 360°, so set up the following equation:

140 + n + 3n + 6n = 360

Now solve for n:

140 + 10n = 360

10n = 220

n = 22

Thus, the correct answer is Choice (C).

Working with Triangles

A triangle is a shape with three sides that are line segments. These three-sided shapes are tremendously important in geometry and are the basis of trigonometry (the study of triangles), which I discuss in Chapter 11. Here are three important types of triangles:

An equilateral triangle has three equal sides and three equal angles all measuring 60°.

An isosceles triangle has two equal sides and two corresponding angles that are equal to each other.

A right triangle has one right angle — that is, one 90° angle.

In this section, you get a solid basis for answering ACT questions that include triangles. First, I introduce the formula for the area of a triangle. Then, I discuss a set of special triangles with distinct properties. Next, I move on to the Pythagorean theorem, which allows you to calculate the length of one side of a right triangle when you know the lengths of the other two sides.

Finding the area of a triangle

To find the area of any triangle, you need to know the base (b), which can be any side, and its height (h), which is the distance from the base to the opposite corner. I illustrate this concept in Figure 10-3. Here’s the formula for the area of a triangle:

In the following figure, if the area of ΔPQT is 14, what is the area of ΔPQR?

(A) 10

(B) 20

(C) 24

(D) 28

(E) 48

You know the area and base of ΔPQT, so use the formula for the area of a triangle to find the height:

The height of ΔPQT is 4, so the height of ΔPQR is also 4. The base of ΔPQR is 12, so

So the correct answer is Choice (C).

Answering questions containing right triangles

A right triangle is a triangle that has one right angle — an angle that measures 90°. Right triangles appear often on the ACT, so you need to know how to work with them and recognize them. In this section, I show you how to calculate a missing length of a right triangle using the Pythagorean theorem. I also provide some info on recognizing some commonly used right triangles.

Taking advantage of the Pythagorean theorem

The Pythagorean theorem is a formula that allows you to calculate the length of a side of a right triangle using the lengths of the other two sides. This theorem is based on the lengths of the two legs (a and b) of the triangle — that is, the two short sides that form the right angle — and the length of the hypotenuse (c), which is the long side directly across from the right angle. Here’s what the theorem looks like:

a² + b² = c²

What is the area of the following right triangle?

(A) 2

(B) 4

(C) 5

(D) 8

(E) 10

You know the length of one leg (5) and the hypotenuse (), so plug these values into the formula for the Pythagorean theorem:

As you can see, the remaining leg of the triangle has a length of 2. In a right triangle, the two legs are also the base and height, so plug these two values into the formula for the area of a triangle:

Therefore, the correct answer is Choice (C).

Recognizing some common right triangles

Two of the most common right triangles on the ACT are identified by the measurements of their angles: the 45-45-90 triangle and the 30-60-90 triangle. You can see examples of these two triangles in Figure 10-4.

The 45-45-90 triangle has three sides whose ratios are of these lengths: . The 30-60-90 triangle has three sides whose ratios are of these lengths: .

You can put these two triangles to use when trying to answer a wide variety of ACT questions that seem unrelated to right triangles. For example, notice that when you bisect a square diagonally, the result is two 45-45-90 triangles. Similarly, when you bisect an equilateral triangle vertically, the result is two 30-60-90 triangles. Check out Figure 10-5 to see these bisections and resulting triangles.

In the following figure, what is the ratio of FN to GH?

(A) 1 to 2

(B) 1 to

(C) 1 to

(D) 2 to

(E) to

Angle F measures 45°, so ΔGNF is a 45-45-90 triangle. Similarly, ∠H measures 30°, so ΔHNG is a 30-60-90 triangle. Let FN = x. Then, GN = x and GH = 2x, and you can calculate the ratio of FN to GH like this:

Therefore, the correct answer is Choice (A).

You can identify three more common right triangles by the lengths of their sides. I show examples of them in Figure 10-6.

Look for right triangles whose sides are multiples of the 3-4-5 triangle: 6-8-10, 9-12-15, and so on.

In the following figure, what is the value of x?

(A) 9

(B) 10

(C) 12

(D) 13

(E) 15

The figure shows two right triangles. The triangle on the right-hand side of the figure has a leg that is the length of 8 and a hypotenuse that is the length of 17, so it’s an 8-15-17 triangle; therefore, the length of its remaining leg is 15. The triangle on the left has a leg with the length of 9 and a hypotenuse the length of 15, so it’s the 9-12-15 variation of the 3-4-5 triangle. Therefore, x = 12, making the correct answer Choice (C).

Breezing through Your Work with Quadrilaterals

A quadrilateral is a shape with four sides that are line segments. For the ACT, you need to be able to work with the following four basic quadrilaterals:

Squares: A square has four right angles and four sides that are equal in length.

Rectangles: A rectangle has four right angles and two pairs of opposite sides that are equal in length.

Parallelograms: A parallelogram has two pairs of opposite sides that are parallel to each other and equal in length.

Trapezoids: A trapezoid has one pair of opposite sides that are parallel to each other.

Figure 10-7 shows you these four basic quadrilaterals.

In this section, I get you up to speed on a variety of questions that cover these four types of quadrilaterals.

Squares

A square is a quadrilateral with four equal sides and four right angles. The area and perimeter of a square are both based on the length of one side (s):

Area: A = s²

Perimeter: P = 4s

If the perimeter of a square is 8n and its area is 20n, what is the value of n?

(A) 2

(B) 4

(C) 5

(D) 8

(E) 10

The perimeter of the square is 8n, so you can use the perimeter formula to express the length of a side (s) in terms of n:

P = 4s

8n = 4s

2n = s

Now plug this value of s into the area formula:

A = s² = (2n)² = 4n²

The question tells you that the area is also 20n, so:

4n² = 20n

n² = 5n

n = 5

Therefore, the correct answer is Choice (C).

Rectangles

A rectangle is a quadrilateral with four right angles and two pairs of equal sides. The area and perimeter of a rectangle are both based on its length (l), which is its longer side, and its width (w), which is its shorter side:

Area: A = lw

Perimeter: P = 2l + 2w

A rectangular field has a length and width that are in a ratio of 4:3. If the diagonal distance from one corner of the field to the opposite corner is 200 feet, what is the perimeter of the field?

(A) 360 feet

(B) 480 feet

(C) 540 feet

(D) 560 feet

(E) 640 feet

Begin by drawing a picture like the following to get a sense of what the question asks you for:

This right triangle has two legs that are 3x and 4x, so it’s a 3-4-5 triangle. If you notice this fact, you can see that the hypotenuse is 5x. As a result, you can set up the equation 5x = 200, which solves as x = 40.

But if you don’t notice this fact, you can still see that the right triangle created by the diagonal has two legs that are labeled 3x and 4x (using the ratio) with the diagonal being 200. So it’s easy to set up for the Pythagorean theorem:

(3x)² + (4x)² = 200²

9x² + 16x² = 40,000

25x² = 40,000

x² = 1,600

x = 40

Thus, the length of the field is 4 × 40 = 160 and the width is 3 × 40 = 120. Plug these values into the formula for the perimeter of a rectangle:

A = 2l + 2w = 2(160) + 2(120) = 320 + 240 = 560

So the correct answer is Choice (D).

Parallelograms

A parallelogram is a quadrilateral with two pairs of parallel sides. No special formula exists for finding the perimeter of a parallelogram. But, to find the area of one, you need to know the base (b), which can be any side, and itsheight (h), which is the shortest distance from the base to the opposite side:

A = bh

The height of a parallelogram is n and the length of its base is 2.5n. What is the value of n if the area of the parallelogram is 20?

(A) 1

(B) 2

(C)

(D)

(E)

Plug the height, base, and area into the formula for the area of a parallelogram:

Thus, the correct answer is Choice (D).

Trapezoids

A trapezoid is a quadrilateral with one pair of parallel sides. No special formula exists for finding the perimeter of a trapezoid. You simply add up the side lengths. The area of this quadrilateral depends upon the bases (b₁ and b₂), which are the two parallel sides, and its height (h), which is the distance between the two bases. Here’s what the formula for the area looks like:

If you need help remembering this formula, notice that the fractional part is the average of the two bases. This average multiplied by the height of the trapezoid gives you the area.

In the following figure, the trapezoid GHJK has an area of 25. Which of the following is the approximate whole-number distance from H to K?

(A) 7

(B) 8

(C) 9

(D) 10

(E) 11

The figure shows that the lengths of the two bases are 4 and 6, and the area of the trapezoid is 25, so use the area formula for a trapezoid to find the height:

Thus, HG = 5. The distance from H to K is the hypotenuse of ΔGHK. The lengths of the two sides of this triangle are 5 and 6, so use the Pythagorean theorem to find the hypotenuse:

c² = a² + b² = 5² + 6² = 25 + 36 = 61

Because , the correct answer is Choice (B).

Wheeling and Dealing with Circles

A circle is the set of points that are all an equal distance from a single center point. ACT questions often require you to find values related to a circle, such as its diameter, area, or circumference. Knowing how to work with a line tangent to a circle is also helpful. Finally, you need to know a bit about arc length. In this section, I help you roll through all these concepts.

Rounding up the basic circle formulas

The distance from the center point to any point on a circle is called the radius. When you know the radius (r) of a circle, you can find three other values related to that circle using the following formulas:

Diameter: D = 2r. The diameter of a circle is the length of a line drawn from any point on the circle through the center to the opposite point on the circle.

Area: A = πr².

Circumference: C = 2πr. The circumference of a circle is the distance around its edge — essentially, its perimeter.

Two of these three formulas use the value π, an irrational number whose value is approximately 3.14. However, an ACT question won’t ask you to substitute a numerical value for π without specifying what value to use.

The area of a circle is n and its circumference is also n. What is its diameter?

(A) 2

(B) 2n

(C) 2πn

(D) 4

(E) 4πn

The area of the circle is n, so

n = πr²

The circumference is also n, so

n = 2πr

Because both values equal n, you can set them equal to each other and solve for r:

Now use the radius to find the diameter:

D = 2r = 2 × 2 = 4

So the correct answer is Choice (D).

Understanding tangent lines

A tangent line is a line that touches a circle at a single point, as shown in Figure 10-8. The most important feature of a tangent line is that it forms a right angle with the radius (or diameter) that also crosses that point. Knowing this fact is essential to answering questions on the ACT.

Don’t confuse a tangent line to a circle with the concept of a tangent in trigonometry (which I discuss in Chapter 11). You don’t need trig to answer an ACT question about tangent lines.

When an ACT question includes a circle with a tangent line, look for a right triangle, and then see whether it’s one of the three common right triangles I discuss in the earlier section “Recognizing some common right triangles.” Even if it isn’t, you still may be able to use the Pythagorean theorem to answer the question.

In the following figure, the circle has an area of 3π, and is a segment of a tangent line that has a length of 1. What is the length of ?

(A) 2

(B) 2π

(C)

(D) 3

(E)

The key to this question is to add to the figure like this:

Now you can see that you’re working with a right triangle. One leg has a length of 1, and the other leg is a radius of the circle. To find this radius, use the formula for the area of a circle:

Because , the triangle is a 30-60-90 triangle. Therefore, the hypotenuse OB = 2, making the correct answer Choice (A). (If you don’t recognize that this is a 30-60-90 triangle, you can still find the solution using the Pythagorean theorem. However, keep in mind that it’s a more complex calculation.)

Making sense of arc length

Arc length is a partial distance around the edge of a circle. Arc length usually is measured in degrees. For example, Figure 10-9 shows a circle with a 45° arc highlighted.

Arc length makes sense in relation to circumference of a circle, C = 2πr. In fact, the circumference of a circle is simply a 360° arc length. With that understanding, the arc length formula becomes easier to handle:

What is the length of a 60° arc of a circle with a radius of 12?

(A) π

(B) 2π

(C) 3π

(D) 4π

(E) 6π

Plug the degrees of the arc and the radius into the arc length formula:

Now simplify:

Thus, the correct answer is Choice (D).

If a 40° arc of a circle has a length of , what is the area of the circle?

(F) π

(G) 9π

(H) 25π

(J) 36π

(K) 81π

Begin by plugging the degrees and arc length into the formula:

Next, simplify and solve for r:

Now plug r into the area formula for a circle:

So the correct answer is Choice (H).

Striking a few chords

A chord of a circle is a line segment drawn between two points on that circle. See Figure 10-10 for an example of a chord.

A line drawn from the center of a circle to the nearest point on the chord bisects the chord at a right angle. This fact can be useful for answering ACT questions, as the following example illustrates.

P and Q are two points on a circle with a center named O, such that is a chord of the circle and PQ = 6 centimeters. If the shortest distance from to O is 1 centimeter, what is the area of the circle in square centimeters?

(A) 5π

(B) 9π

(C) 10π

(D) 37π

(E) 100π

Begin by sketching out the information from the question like this:

Notice that the line segment from O to meets at a right angle. Furthermore, it bisects . After the radius is drawn ( or ), the result is a right triangle whose legs are 1 and 3 in length, so you can use the Pythagorean theorem to find the hypotenuse:

The hypotenuse of this triangle is also the radius of the circle, so use the formula for the area of a circle:

The correct answer is Choice (C).

Examining Solid Geometry

Solid geometry is geometry in three dimensions (3-D). It includes everything you know from plane geometry, but it also adds solids such as cubes, boxes, prisms, cylinders, pyramids, cones, and spheres.

Generally speaking, an ACT question probably won’t require you to remember an obscure formula to solve a problem. However, the question may provide a useful formula and ask you to apply it to get an answer. So to feel completely comfortable applying them, make sure you familiarize yourself with a few formulas from solid geometry.

The most important formula for any solid is usually the formula for volume — the amount of space that the solid occupies. Volume is the 3-D equivalent of area and is expressed in cubic units (such as cubic feet, cubic centimeters, and so forth).

For certain solids, however, you may need to work with their surface area, which is (as you may guess) the area of its total surface. You express surface area in square units (such as square inches, meters, and so forth).

In this section, I discuss everything you need to know about solid geometry to do well on the ACT.

Focusing on cubes and boxes

One of the most common solids you’ll find popping up in ACT questions is the cube, which is a box that has six square faces. Both the volume and the surface area of a cube are based on the length of one side (s):

Volume of a cube: V = s³

Surface area of a cube: A = 6s²

The volume of a cube is k cubic meters, and its surface area is square meters. What length is one side of the cube?

(A) 4 meters

(B) 6 meters

(C) 8 meters

(D)

(E)

The volume of this cube is k, so plug this value into the volume formula that I introduce earlier in the section:

k = s³

The surface area of this cube is , so plug this value into the surface-area formula and solve for k in terms of s:

The right-hand side of this last equation now equals k, so, using the previous equation, substitute s³ for k and then solve for s:

So the side of this cube is 8 meters, making the correct answer Choice (C).

The volume and surface area of a box (rectangular solid) is based on its length (l), width (w), and height (h), as shown in these formulas:

Volume of a box: V = lwh

Surface area of a box: A = 2lw + 2lh + 2wh

A rectangular box whose dimensions are n, 10n, and 12n has a volume of 15. What is the value of n?

(F) 2

(G)

(H)

(J)

(K)

Plug the volume and the length, width, and height (in any order) into the formula for the volume of a box and solve for n:

At this point, solve for n by taking the cube root of both sides:

The correct answer is Choice (G).

Incorporating spheres into your geometric repertoire

A sphere is the set of points in space that are all the same distance from a single point. A ball or globe is a perfect visual aid for a sphere. The formulas for both the volume and surface area of a sphere are based on its radius (r):

Volume of a sphere:

Surface area of a sphere:

An artist is incorporating a globe into a large sculpture. She intends to paint the surface of the globe, which is a sphere with a volume of 36π cubic feet. What is the surface area of the globe?

(A) 9π

(B) 16π

(C) 36π

(D) 64π

(E) 81π

First, use the formula for the volume of the globe to find the radius:

Now use the formula for the surface area of a sphere:

So the correct answer is Choice (C).

Figuring the volume of other solids (prisms, cylinders, pyramids, and cones)

Although cubes, boxes, and spheres (which I discuss in the previous section) are the most common geometric solids on the ACT, you may face a question with one of the less-common solids. In this section, I show you how to work with prisms, cylinders, pyramids, and cones. Because of their similar shapes, prisms and pyramids can be lumped together and cylinders and cones can be as well. So I discuss each of these solids in pairs in the following sections.

Prisms and pyramids

The volume of a prism and a pyramid is based on its height (h) and the area of its base (), regardless of the shape of that base. Here are the formulas to use:

Volume of a prism:

Volume of a pyramid:

To find the area of the base, you may need to use one of the other formulas for area from earlier in this chapter. For example, if the base of a solid is a triangle, use the formula for

the area of a triangle: .

The solid in the following figure is a prism with irregular pentagonal base. If the height of the prism is h units and its volume is 7h cubic units, what is the area of the base in square units?

(A) 7

(B) 7h

(C) 7h²

(D) 7h³

(E) Cannot be determined from the information given.

Plug the volume (7h) and height (h) into the formula for a prism and solve for :

Thus, the correct answer is Choice (A).

Cylinders and cones

The volume of a cylinder and the volume of a cone are both based on the radius (r) of its circular base and its height (h). Check out the formulas:

Volume of a cylinder:

Volume of a cone:

Notice that both of these formulas include πr², which is the area formula for a circle. So if you know the area of the base (), you can substitute this value for πr² in the formula.

A cylinder and a cone both have the same volume and height. If the base of the cylinder has an area of 16π square inches, what is the radius of the cone?

(A)

(B)

(C)

(D)

(E)

The area of the cylinder’s base is 16π, so plug this value into the formula for the volume of a cylinder:

The volumes of the cylinder and the cone are the same, so plug in 16πh for V into the equation for the volume of a cone and simplify:

The heights of the cylinder and the cone are equal, so you can cancel out h on both sides of the equation and solve for r:

So the correct answer is Choice (E).