## SAT Test Prep

**CHAPTER 10**

ESSENTIAL GEOMETRY SKILLS

ESSENTIAL GEOMETRY SKILLS

**Lesson 2: Triangles**

**Angles in Polygons**

Remembering what you learned about parallel lines in the last lesson, consider this diagram:

We drew line ℓ so that it is parallel to the opposite side of the triangle. Do you see the two Z’s? The angles marked *a* are equal, and so are the angles marked *c*. We also know that angles that make up a straight line have a sum of 180°, so . The angles inside the triangle are also *a, b*, and *c*.

Therefore, *the sum of angles in a triangle is always 180°*.

Every polygon with *n* sides can be divided into *n* – 2 triangles that share their vertices (corners) with the polygon:

Therefore, *the sum of the angles in any polygon with n sides is 180(n – 2) °*.

**Angle-Side Relationships in Triangles**

A triangle is like an alligator mouth with a stick in it: The wider the mouth, the bigger the stick, right?

Therefore, the largest angle of a triangle is always across from the longest side, and vice versa. Likewise, the smallest angle is always across from the shortest side.

**Example:**

In the figure below, , so

An isosceles triangle is a triangle with two equal sides. If two sides in a triangle are equal, then the angles across from those sides are equal, too, and vice versa.

**The Triangle Inequality**

Look closely at the figure below. The shortest path from point *A* to point *B* is the line segment connecting them. Therefore, unless point *C* is “on the way” from *A* to *B*, that is, unless it’s on , the distance from *A* to *B* through *C* must be longer than the direct route. In other words:

The sum of any two sides of a triangle is always greater than the third side. This means that the length of any side of a triangle must be between the sum and the difference of the other two sides.

**The External Angle Theorem**

The extended side of a triangle forms an external angle with the adjacent side. The external angle of a triangle is equal to the sum of the two “remote interior” angles. Notice that this follows from our angle theorems:

and ; therefore,

**Concept Review 2: Triangles**

__1.__ The sum of the measures of the angles in a quadrilateral is __________.

__2.__ The sum of the measures of the interior angles in an octagon is __________.

__3.__ In Δ*ABC*, if the measure of ∠*A* is 65° and the measure of ∠*B* is 60°, then which side is longest? __________.

__4.__ The angles in an equilateral triangle must have a measure of __________.

__5.__ Can an isosceles triangle include angles of 35° and 60°? Why or why not?

__6.__ Draw a diagram to illustrate the external angle theorem.

__7.__ If a triangle has sides of lengths 20 and 15, then the third side must be less than __________ but greater than __________.

__8.__ Is it possible for a triangle to have sides of lengths 5, 8, and 14? Why or why not?

__9.__ If an isosceles triangle includes an angle of 25°, the other two angles could have measures of __________ and __________ or __________ and __________.

__10.__ In the figure above, and are diameters of the circle and *P* is *not* the center. Complete the statement below with <, >, or=.

**SAT Practice 2: Triangles**

**1**__.__ In the figure above, if , then *x* =

(A) 25

(B) 30

(C) 35

(D) 50

(E) 65

**2**__.__ In the figure above,

**3**__.__ The three sides of a triangle have lengths of 9, 16, and *k*. Which of the following could equal *k*?

I. 6

II. 16

III. 25

(A) I only

(B) II only

(C) I and II only

(D) I and III only

(E) I, II, and III

__Note__: Figure not drawn to scale.

**4**__.__ Which of the following statements about *a* and *b* in the figure above must be true?

(A) I only

(B) II only

(C) I and II only

(D) I and III only

(E) I, II, and III

**5**__.__ In the figure above, if , then

(A) 70

(B) 80

(C) 90

(D) 100

(E) 120

**6**__.__ In the figure above, which of the following expresses *a* in terms of *b* and *c*?

__Note__: Figure not drawn to scale.

**7**__.__ Which of the following represents the correct ordering of the lengths of the five segments in the figure above?

**8**__.__ A triangle has two sides of lengths 4 centimeters and 6 centimeters. Its area is *n* square centimeters, where *n* is a prime number. What is the greatest possible value of *n*?

(A) 11

(B) 12

(C) 19

(D) 23

(E) 24

**Answer Key 2: Triangles**

**Concept Review 2**

__1.__ 360°

__2.__ 1,080°

__3.__ Draw a diagram. If the measure of ∠*A* is 65° and the measure of ∠*B* is 60°, then the measure of ∠*C* must be 55°, because the angles must have a sum of 180°. Since ∠*A* is the largest angle, the side opposite it, , must be the longest side.

__4.__ 60°. Since all the sides are equal, all the angles are, too.

__5.__ No, because an isosceles triangle must have two equal angles, and the sum of all three must be 180°. Since , and , the triangle is impossible.

__6.__ Your diagram should look something like this:

__7.__ If a triangle has sides of lengths 20 and 15, then the third side must be less than 35 (their sum) but greater than 5 (their difference).

__8.__ No. The sum of the two shorter sides of a triangle is always greater than the third side, but is not greater than 14. So the triangle is impossible.

__9.__ 25° and 130° or 77.5° and 77.5°

__10.__ Draw in the line segments *PQ, PR, PS*, and *PT*. Notice that this forms two triangles, Δ*PQS* and Δ*PRT*. Since any two sides of a triangle must have a sum greater than the third side, , and Therefore, .

**SAT Practice 2**

__1.__ **A** If , then, by the Isosceles Triangle theorem, ∠*BAD* and ∠*BDA* must be equal. To find their measure, subtract 50° from 180° and divide by 2. This gives 65°. Mark up the diagram with this information. Since the angles in the big triangle have a sum of 180°, , so .

__2.__ **500** Drawing two diagonals shows that the figure can be divided into three triangles. (Remember that an *n*-sided figure can be divided into *n* – 2 triangles.) Therefore, the sum of all the angles is . Subtracting 40° leaves 500°.

__3.__ **B** The third side of any triangle must have a length that is between the sum and the difference of the other two sides. Since and , the third side must be between (but not including) 7 and 25.

__4.__ **A** Since the big triangle is a right triangle, must equal 90. The two small triangles are also right triangles, so is also 90. Therefore, and statement I is true. In one “solution” of this triangle, *a* and *b* are 65 and *x* is 25. (Put the values into the diagram and check that everything “fits.”) This solution proves that statements II and III are not necessarily true.

__5.__ **C** If , then, by the Isosceles Triangle theorem, the angles opposite those sides must be equal. You should mark the other angle with an *x* also, as shown here. Similarly, if then the angles opposite those sides must be equal also, and they should both be marked *y*. Now consider the big triangle. Since its angles must have a sum of 180, . Dividing both sides by 2 gives . (Notice that the fact that *ADB* measures 100° doesn’t make any difference!)

__6.__ **E** Label the two angles that are “vertical” to those marked *b* and *c*. Notice that the angle marked *a* is an “external” angle. By the External Angle theorem, .

__7.__ **D** Write in the missing angle measures by using the fact that the sum of angles in a triangle is 180°. Then use the fact that the biggest side of a triangle is always across from the biggest angle to order the sides of each triangle.

__8.__ **A** Consider the side of length 4 to be the base, and “attach” the side of length 6. Notice that the triangle has the greatest possible height when the two sides form a right angle. Therefore, the greatest possible area of such a triangle is , and the minimum possible area is 0. The greatest prime number less than 12 is 11.