## Triangles

#### The SAT is absolutely obsessed with triangle-based problems, and as a result, the SAT Math section includes many questions on the SAT Math section. The following set of formulas will help eradicate your trouble once and for all.

## Angles of a Triangle

#### The angles in any given triangle have two key properties:

- The sum of the three angles of a triangle equals 180° ∆.
- Angles correspond to their opposite sides.

This means that the largest angle is opposite the longest side, while the smallest angle is opposite the shortest side.

## Sides of a Triangle

#### Consider the following “impossible” triangle ∆ABC and what it reveals about the relationship between the three sides of any triangle.

The triangle can never be drawn with the given measurements. Why? Consider that the shortest distance between any two points is a straight line.

According to the triangle shown, the direct straight line distance between point C and point B is 14; however, the indirect path from point C to B (the path that goes from C to A to B) is 10 + 3, or 13, which is shorter than the direct path! This is clearly impossible.

The leads to the following *Triangle Inequality Law:** **The sum of any two sides of a triangle must be great than the third side.*

Therefore, the maximum integer distance for side BC in the triangle above is 12. If the length of side BC is not restricted to integers, then this length has to be less than 13.

Note that the length cannot be as small as we wish, either. It must be greater than the difference between the lengths of the other two sides. In this case, side BC must be longer than 10 - 3 = 7. This is a consequence of the same idea.

## The Right Angle Theorem

#### A right triangle is a triangle with one right angle (90°). Every right triangle is composed of two legs and a hypotenuse. The **HYPOTENUSE** is the side opposite the right angle. Given the lengths of two sides of a right triangle, you determine the length of the third side using the *Pythagorean Theorem*.

**Pythagorean Theorem: \(A^2 + B^2 = C^2\)**

### Special Right Angle Triangles

Certain right triangles appear over and over on the SAT: the 30-60-90 right triangle and the 45-45-90 right triangle. It pays to memorize these common combinations in order to save time on the test. Instead of using the Pythagorean Theorem to solve for the lengths of the sides, you should know the following Pythagorean triples from memory.

*1. THE 45-45-90 TRIANGLE:* By 45-45-90, we mean the three angles of a right triangle are 45°, 45°and 90°. The lengths of the legs of every 45-45-90 triangle have a specific ratio.

*1. THE 45-45-90 TRIANGLE:*

#### Given that the length of side AB is 5, what are the lengths of sides BC and AC?

*Solution:** *Since AB is 5, use the ratio 1:1:√2 for sides AB : BC : AC to determine that the multiplier x is 5. You then find that the sides of the triangle have lengths 5:5:5√2.

*2. THE 30-60-90 TRIANGLE*. The most popular of all right triangles is the 30-60-90 triangles. The lengths of the legs of every 30-60-90 triangle will have the following ratio, which you must memorize:

*2. THE 30-60-90 TRIANGLE*

*Solved Example 1*

Find the lengths of sides a and c of the triangle shown?

*Solved Example 1*

** A.** a = 7.5√3 cm; c = 7√3 cm

**a = 3.5√3 cm; c = 7√3 cm**

*B.***a = 7.5√3 cm; c = 7 cm**

*C.***a = 3.5√3 cm; c = 7 cm**

*D.*** Solution:** We know that one side is 10.5 cm long and that that side is across from the 600 angle.

In the side ratio, that side is x√3.

Therefore, the value of x is \(\frac{10.5}{\sqrt{3}}=\frac{10.5}{\sqrt{3}}\) x \(\frac{\sqrt{3}}{\sqrt{3}}=\frac{10.5\sqrt{3}}{3}=3.5\sqrt{3}\: cm\)

The length of the side marked ‘a’ is the length of the side across from the 30° angle. That side is x in the side ratio and is equal to 3.5√3. Similarly, The side ‘c’ is the 2x in the proportion. Therefore its value is 7√3.

**The correct answer is B.**

### Exterior Angles of a Triangle

#### An EXTERIOR ANGLE of a triangle is equal in measure to the sum of the two non-adjacent (opposite) interior angles of the triangle.

This property is frequently tested on the SAT! In particular, look for exterior angles within more complicated diagrams.

*Solved Example 2*

*Solved Example 2*

Which of the following expresses Z in terms of x and Y? 2x + 3y − 180

A.

**x + 2y − 180**

*B.***. 180 − x − y**

*C***360 − 2x − 3y**

*D.*** Solution:** Each of the three triangles we are focusing on would all add up to 180°. We have three different triangles and three missing angles, so their equations would look like:

180 - x − y

180 − x − y

180 - y - z

We know that all of those equations will find us one of the three unmarked angles. We also know that those three angles add up to 180° So when we add the equations together and set them equal to 180° we get:

(180−x−y) + (180−x−y) + (180−y−z) = 180

540 − 2x − 3y - z = 180

−2x − 3y - z = −360

−z = −360 + 2x + 3y

z = 360 − 2x − 3y

**The correct answer is D.**

## Area of a Triangle

**TRIANGLE AREA** is another important concept in the SAT exam. It is defined as the half of the product of base and height i.e.

Area of a Triangle \(A=\frac{1}{2}\bullet base\bullet height\: OR\: A=\frac{1}{2}BH\)

#### Note that it is not necessary that always the bottom side be a base. You can consider any side as the base and define height accordingly. See the figures below.

#### Another important formula that can save you significant time is the **area of an equilateral triangle** with a side of length S is \(\frac{S^{2}\sqrt{3}}{4}\)

The formula is derived because an equilateral triangle can be split into two 30-60-90 triangles, If the side length of the equilateral triangle is S, then S is also the hypotenuse of each of the 30-60-90 triangles, so their sides are as shown in the diagram.

The equilateral triangle has a base of length S and a height of length (S√3)/2. Therefore its area is \(\frac{1}{2}\) x (S) x \(\frac{S\sqrt{3}}{2}=\frac{S^{2}\sqrt{3}}{4}\)

## Similar Triangles

One final tool that you can use for SAT triangle problems is the similar triangle strategy. Often, looking for similar triangles can help you solve complex problems.

The two triangles are said to be similar if their corresponding angles are equal. Once you find that two triangles have two pairs of equal angles, you know that the triangles are similar. If two sets of angles are congruent, then the third angles must be congruent, since the sum of the angles in any triangle is 180°. It is not necessary that the lengths of these sides also be equal. The sides of the similar triangles can be expressed in terms of a ratio. For example, if the triangles ABC and PQR are similar, then the ratio of the side AB to the side BC is equal to the ratio of the side PQ to the side QR.

For example: \(\frac{AB}{BC}=\frac{PQ}{QR}\)

## Ratio of Areas

If two triangles are similar, the ratio of their areas is square of the ratio of their sides. **If two similar triangles have corresponding side lengths in ratio **a:b, **then their areas will be in ratio** \(a^{2}:b^{2}\)

The lengths being compared do not have to be sides - they can represent heights or perimeters. In fact, the figures do not have even have to be triangles. The principle holds true for any similar figures: quadrilaterals, pentagons, etc.

**Example:** In the figure, AE || CD and segment AD intersects segment CE at B. What is the length of segment CE?

** Solution:** Angles ABE and DBC are vertically opposite angles, and, therefore, they have the same measurement. Since segment AE is parallel to segment CD, angles A and D are equal to each other. Since all angles are equal in triangles ABE and DBC, triangle ABE is similar to triangle DBC, with vertices A, B, and E matching up to vertices D, B, and C, respectively.

Solving this you get CB = 4, and so CE = CB + BE = 4 + 8 = 12.

*Solved Example 3*

*Solved Example 3*

A summer camp counselor wants to find a length, x, in feet, across a lake as represented in the sketch above. The lengths represented by AB, EB, BD, and CD on the sketch were determined to be 1800 feet, 1400 feet, 700 feet, and 800 feet, respectively. Segments AC and DE intersect at B, and ∠AEB and ∠CDB have the same measure. What is the value of x? 1200

A.

**1600**

*B.***2000**

*C.***2400**

*D.*** Solution:** The question tells you that ∠AEB and ∠CDB have the same measure. Since ∠ABE and ∠CBD are vertical angles, they measure the same. Triangle EAB is similar to triangle DCB because the triangles have two pairs of corresponding congruent angles. Since the triangles are similar, the corresponding sides are in the same proportion: \(\frac{CD}{x}=\frac{BD}{EB}\)

Substituting the given values of 800 for CD, 700 for BD, and 1400 for EB, gives: \(\frac{800}{x}=\frac{700}{1400}\)

Therefore, x = 1600. **The correct answer is B.**