## Basic Rules

There are some rules that are true for all the different kinds of triangles. Below are the four most important rules. Note: The triangles in the figures represent *any* triangle.

**Rule 1**:

The sum of the interior angles of a triangle is 180°.

*x*° + *y*° + *z*° = 180°

**Rule 2**:

The length of the longest side of a triangle must be less than the sum of the lengths of the other two sides of the triangle, but greater than the difference of the lengths.

(*AB* – *BC*) < *AC* < (*AB* + *BC*)

**Rule 3**:

The longest side of a triangle is opposite the largest angle and the shortest side is opposite the smallest angle.

*∠B* is the largest angle so *AC* is the longest side.

*AB* is the shortest side so ∠*C* is the smallest angle.

**Rule 4**:

If you extend one side of the triangle, the *exterior angle* formed will be equal to the sum of the other two interior angles in the triangle.

*x*° + *z*° = *n*°

## Example

Find the value of

y.

### Solution

Start with the triangle on the right. Using exterior angles, *x*° + 120° = 150°, so *x*° = 30°.

The arcs in the top of the triangle mean that the angles are congruent. Since *x*° = 30°, its congruent angle is also 30°.

From here, there are two methods for finding the value of *y*.

One method is to use exterior angles. *y*° + 30° = 120°, so *y*° = 90°.

The other method is to first find the other unknown angle. Since the unknown angle forms a straight angle with the 120° angle, its measure is 60°.

The sum of the interior angles of a triangle is 180°.

So 60° + 30° + *y*° = 180°.

Again, *y*° = 90°.

## Perimeter

The ** perimeter** of any figure is the distance around the outside of the figure, or the sum of the sides of the figure. Many SAT questions about triangle perimeters involve Rule 2 from above.

(a)

## Example

Triangles

AandBshare a side. The lengths of two sides of triangleAare 4 inches and 6 inches. The lengths of two sides of triangleBare 3 inches and 8 inches. What is the range of possible lengths of the third, shared side?

### Solution

Drawing a sketch can make any question easier to visualize.

The lengths of two sides of triangle *A* are 4 inches and 6 inches.

So the range of values for the third side of triangle *A* is

(6 – 4) < *x* < (6 + 4), or 2 < *x* < 10.

The lengths of two sides of triangle *B* are 3 inches and 8 inches.

So the range of values for the third side of triangle *B* is

(8 – 3) < *x* < (8 + 3), or 5 < *x* < 11.

Combine the two inequalities.

2 < *x* < 10 and 5 < *x* < 11

5 < *x* < 10

So the third side must be longer than 5 inches and shorter than 10 inches.

## Area

The** area** of any figure is the amount of surface that is covered by the figure. Each type of figure has a different formula for finding its area.

To find the area of a triangle, you need two measurements: the ** base** and the

**.**

*height*The formula for the area of a triangle is *A* = \dfrac{1}{2} (base × height) = \dfrac{1}{2} *bh*

The base of a triangle can be any of its sides. The height of the triangle is the perpendicular distance from the base to the opposite angle. Here are some examples of triangles with different bases and heights. Every triangle has three bases and three corresponding heights. Notice that a height can be inside the triangle, outside the triangle, or one of the sides of the triangle.

## Types of Triangles

There are three important types of triangle on the SAT.

An ** isosceles triangle** has two equal sides and two equal angles. In triangle

*ABC*, ∠

*B*= ∠

*C*and

*AB*=

*AC*. This is an application of Rule 3 above. Two sides are equal, so their opposite angles are equal as well.

The other side, *BC* can be longer, shorter, or equal to the length of *AB* = *AC*.

The height from side *BC* bisects the opposite angle ∠*A*. *BD* = *DC* and ∠*DAB* = ∠*DAC*. This creates two congruent right triangles.

## Example

Triangle

RSTis an isosceles triangle with ∠R= 110°,RT= 12 andST= 20. The height fromSTto ∠Ris 7. Find ∠S, ∠T,RS, the perimeter and the area of triangleRST.

### Solution

Drawing a sketch is the best way to solve the problem.

Since triangle *RST* is an isosceles triangle, it has two equal angles.

Since ∠*R* = 110°, it cannot be one of the congruent angles.

*∠R* + ∠*S* + ∠*T* = 180°

so ∠*S* + ∠*T* = 180° – 110° = 70°.

*∠S* = ∠*T* so they both equal

70° ÷ 2 = 35°.

Since triangle *RST* is an isosceles triangle, it has two equal sides.

*∠S* = ∠*T* so the sides opposite those angles are congruent.

*RT* = *RS* and *RT* = 12, so *RS* = 12.

The perimeter is 12 + 12 + 20 = 44 units.

Since the height is from *ST* to ∠*R*, *ST* is the base and *ST* = 20.

The area is (1/2)(20)(7) = 70 square units.

An ** equilateral triangle** has three equal sides and three equal angles. Because a triangle’s angles add up to 180°, each angle of an equilateral triangle is 60°.

An equilateral triangle is a type of isosceles triangle, because it has two equal sides and two equal angles.

The height of an equilateral triangle bisects the triangle into two congruent right triangles. The angles of the right triangle are 30° : 60° : 90°.

A ** right triangle** is a triangle with a 90° angle. The two perpendicular sides are called

**and the side opposite the right angle is called the**

*legs***.**

*hypotenuse*To find the area, the legs can be used as the height and base of the right triangle.

The hypotenuse is always the longest side of a right triangle. This is an application of Rule 3 above: the right angle is the largest angle, so the hypotenuse is the longest side.

## Right Triangles

For right triangles, the relationship between the legs and the hypotenuse is defined by the ** Pythagorean Theorem**. The Pythagorean Theorem states that the square of the hypotenuse will equal the sum of the squares of the legs.

*a*^{2} + *b*^{2} = *c*^{2}* or*

(leg)^{2} + (leg)^{2} = (hypotenuse)^{2}

The Pythagorean Theorem can be used to find the length of any side of a right triangle.

**Pythagorean Theorem**

Video Courtesy of **Kaplan SAT prep.**

ExampleFind the lengths of the unknown sides.

(b)

(c)

### Solution

(a)

3^{2} + 4^{2} = *a*^{2}

9 + 16 = *a*^{2}

25 = *a*^{2}

5 = *a*

(b)

12^{2} + *b*^{2} = 13^{2}

144 + *b*^{2} = 169

*b*^{2} = 25

*b* = 5

(c)

4^{2} + 8^{2} = *c*^{2}

16 + 64 = *c*^{2}

80 = *c*^{2}

80 = 4 *×* 4 *×* 5

*c* =\sqrt{80} = 4\sqrt{5}

## Special Right Triangles based on Side Lengths

There are several special right triangles where the lengths of their sides form ratios of whole numbers. These triangles appear commonly on the SAT, so being able to quickly recognize them will help you solve problems more quickly and accurately.

Common Pythagorean Triples and Some of Their Multiples

**3, 4, 5**

6, 8, 10

9, 12, 15

30, 40, 50

3*x*, 4*x*, 5*x*

**5, 12, 13**

10, 24, 26

15, 36, 39

50, 120, 130

5*x*, 12*x*, 13*x*

**8, 15, 17**

16, 30, 34

24, 45, 51

80, 150, 170

8*x*, 15*x*, 17*x*

**7, 24, 25**

14, 48, 50

21, 72, 75

70, 240, 250

7*x*, 24*x*, 25*x*

It is important to remember that the longest side is always the hypotenuse. If two legs of a right triangle are 3 and 5, the hypotenuse is not 4.

Also note that multiples of these triangles are also special right triangles. The 3 : 4 : 5 triangle is also the 6 : 8 : 10 and the 9 : 12 : 15.

The 5 : 12 : 13 is also the 10 : 24 : 26.

## Example

The sides of an isosceles triangle are 10, 10 and 16 feet. Find the area of the triangle.

### Solution

To find the area, draw the height of the triangle.

Since it is an isosceles triangle, the height bisects the base and creates two congruent right triangles.

The sides of each right triangle are *h* : 8 : 10. This is a multiple of a 3 : 4 : 5 right triangle, so *h* = 6.

*A* = \Big(\,\dfrac{\,1\,}{2}\,\Big) *bh* = \Big(\,\dfrac{\,1\,}{2}\,\Big)(16)(6) …Remember that the base of the entire triangle is 16.

= (8)(6) = 48 …The area of the triangle is 48 square feet.

## Example

The area of a right triangle is 60 cm

^{2}and the height is 15 cm. Find the perimeter of the triangle.

### Solution

Substitute the values into the area formula.

*A* = \Big(\,\dfrac{\,1\,}{2}\,\Big)* bh* so 60 = \Big(\,\dfrac{\,1\,}{2}\,\Big)(*b*)(15) …Solve for *b*.

120 = 15*b*

*b* = \dfrac{120}{15} …The base of the triangle is 8.

So, the base is 8 and the height is 15, which are the legs of a right triangle.

Rather than calculate the hypotenuse, notice that these are the legs of an 8 : 15 : 17 right triangle.

The perimeter is 8 + 15 + 17 = 40 cm.

## Special Right Triangles based on Angles

There are two special “angle-based” right triangles on the SAT. The first is the 45° : 45° : 90° triangle.

A 45° : 45° : 90° triangle is an isosceles triangle, so it has two equal sides and two 45° angles.

There is a constant relationship between the lengths of the legs and the hypotenuse. The hypotenuse is always \sqrt{2} times the length of one leg.

Therefore, you only need the length of one side to be able to find the area.

## Example

The longest side of an isosceles right triangle is 10 units. Find the area of the triangle.

### Solution

An isosceles right triangle is a

45° : 45° : 90° triangle.

The longest side is the hypotenuse, which is equal to 10.

So each leg is

\dfrac{10}{\sqrt{2}}= \dfrac{10}{\sqrt{2}} \times \dfrac{\sqrt{2}}{\sqrt{2}} = \dfrac{10\sqrt{2}}{2} = 5\sqrt{2}

The legs are height and base , so the area is

A = \Big(\dfrac{1}{2}\Big)\textit{bh}

= \Big(\dfrac{1}{2}\Big)(5\sqrt{2})^{\displaystyle{2}} = \Big(\dfrac{1}{2}\Big)(25 × 2) = 25

The area of the triangle is 25 square units.

A 30° : 60° : 90° right triangle works in the same way as the 45° : 45° : 90° triangle, but with different dimensions.

The dimensions of a 30° : 60° : 90° triangle are *x* : *x*\sqrt{3} : 2*x*.

Again, you only need the length of one side to be able to find the area.

One way to remember this is to apply Rule 3 from above – the smallest side is opposite the smallest angle and vice versa.

Since *x* is the smallest side, it is always opposite the 30° angle, and 2*x, *which is the largest, must always be opposite the 90° angle.

This means 2*x* is the hypotenuse. Since *x*\sqrt{3} is between *x* and 2*x* (*x*\sqrt{3} is approximately 1.7), it is always opposite the 60° angle.

An equilateral triangle can be bisected into two 30° : 60° : 90° right triangles.

The height bisects the base and one angle of the triangle, creating two triangles

that are congruent 30° : 60° : 90° right triangles.

Therefore, you only need the length of one side or the height to be able to find the area of an equilateral triangle.

## Example

An isosceles triangle has two 30° angles. The side between the 30° angles has a length of 4. How long are the equal sides?

### Solution

Use the description to draw a sketch.

Let *y* be the unknown lengths. The 30° angles give you a hint to look for right triangles. You can then use these right triangles to solve for the sides of the isosceles triangle.

Draw the height to divide the isosceles triangle into two

30° : 60° : 90° triangles.

The length of the longest side of the original triangle is 4, so the length of the longer legs of the 30° : 60° : 90° triangles is 2.

The side you are looking for is the hypotenuse using 2 as the length of the longer leg. Use the ratio *x* : *x*\sqrt{3} : 2*x* with *x*\sqrt{3} = 2 to find *x*, which is the base of the 30° : 60° : 90° triangles.

So the base of the 30° : 60° : 90° triangles is *x* = \dfrac{2\sqrt{3}}{3}

The hypotenuse of the 30°- 60°- 90° triangles will be 2*x*, which is the value for *y*.

*y* = 2*x* = 2\Big(\dfrac{2\sqrt{3}}{3}\Big) = \dfrac{4\sqrt{3}}{3}

## Example

In an isosceles triangle, ∠

A= 60° and sideAB= 12 units. Find the perimeter and area of triangleABC.

### Solution

This trick to this question is realizing that an isosceles triangle with one angle equal to 60° is an equilateral triangle.

Think it through: An isosceles triangle must have two equal angles. So either ∠*B* or ∠*C* is equal to ∠*A* so is 60°. Combined, these two angles are 120°, and because the sum of all three angles must equal 180°, the third angle must equal 60°.

Since one side of triangle *ABC* is equal to 12, all sides are 12.

3 *×* 12 = 36

One side of triangle *ABC* is equal to 12, which is the hypotenuse of the 30° : 60° : 90° right triangles created by the height.

So 12 = 2*x*, *x* = 6, and the height

= 6\sqrt{3}.

The area is A = \Big(\dfrac{1}{2}\Big)\textit{bh}

= \Big(\dfrac{1}{2}\Big)(12)(6\sqrt{3}) = 36\sqrt{3}.

The perimeter is 36 units. The area of the triangle is 36\sqrt{3} square units.

**Note:** Since the sides of an equilateral triangle are always equal, there is a special formula for the area. If the length of a side of the triangle is *s*, the formula is

*A* = \dfrac{\textit{s}^{\displaystyle{2}}\sqrt{3}}{4}

In the Example above, the area would be A = \dfrac{12^{\displaystyle{2}}\sqrt{3}}{4} = \dfrac{144\sqrt{3}}{4} = 36\sqrt{3}

## Example

A triangle has angles of 45° and 75°. The side opposite the 45° angle has a length of 6. What is the length of the side opposite the 75° angle?

### Solution

Sketch the triangle.

The third angle is 180° – (75° + 45°) = 60°.

Again, see if you can solve the problem by creating right triangles.

Draw the height of the triangle and form two right triangles.

You now have a 30° : 60° : 90° triangle and a 45° : 45° : 90° triangle.

Label the unknown lengths *x*, *y*, and *z*.

In the 30° : 60° : 90° triangle, the hypotenuse is equal to 6, so the shorter leg *x* = 3 and the longer leg *y* = 3\sqrt{3} .

In the 45° : 45° : 90° triangle, *x* = *z* so *z* = 3.

The length of the side opposite the 75° angle is *y* + *z* = 3 + 3\sqrt{3} or 3 (1 + \sqrt{3} )

Video Courtesy of __Kaplan SAT prep__.

## Similar Triangles

Triangles that have the same shape are called ** similar triangles**.

There are two ways to know that two triangles are similar.

If the *corresponding angles* are equal, the triangles are similar.

If the *corresponding sides* have the same ratio, the triangles are similar.

The ratio of the lengths of the corresponding sides is called the** scale factor**. The scale factor for congruent triangles is 1.

Triangle *ABC* is similar to triangle *DEF*.

∠*A* = ∠*D*

∠*B* = ∠*E*

∠*C* = ∠*F*

\dfrac{\textit{AB}}{\textit{DE}} = \dfrac{\textit{BC}}{\textit{EF}} = \dfrac{\textit{AC}}{\textit{DF}} This ratio is the scale factor.

There are three common presentations of similar triangles on the SAT.

#### Two triangles, same angles.

You can see these are similar triangles because they have the same angles or because the sides are in proportion. You only need angles or sides to know the triangles are similar.

The scale factor is

\dfrac{small}{large} = \dfrac{4}{6} = \dfrac{2}{3}\, \textit{ or } \,\dfrac{small}{large} = \dfrac{6}{4} = 1.5

#### One triangle, with a line parallel to the base of the triangle.

This figure is composed of two triangles.

They share an angle of *y*°.

Their bases are parallel, so they both have angles of *x*° and z°.

You can also see that these are similar triangles because the sides are in proportion.

The scale factor is

\dfrac{small}{large} = \dfrac{4.5}{9} = \dfrac{1}{2}\, \textit{ or } \,\dfrac{small}{large} = \dfrac{9}{4.5} = 2

#### Two triangles, connected by vertical angles and having parallel bases.

This figure is composed of two triangles. They have vertical angles of *x*°.

Their bases are parallel, so they both have angles of *y*° and *z*°.

Notice that the angles of *y*° and *z*° are on “opposite sides” from one another.

## Example

Find the measure of ∠

Aand the length ofAD.

### Solution

First confirm that the triangles are similar. The triangles share ∠*A*. The bases of the triangles, *CD* and *BE*, are parallel. Using the properties of angles formed by parallel lines, ∠*AEB* = ∠*ADC* and ∠*ABE* = ∠*ACD*. Since the triangles have 3 equal angles, the triangles are similar.

∠*AEB* = 180° – 135° = 45° = ∠*ADC*

To find the measure of ∠*A*, use the sum of the angles in triangle *ADC*.

∠*A* = 180° – (45° + 30°) = 105°.

Use the scale factor to find the side lengths.

\dfrac{small}{large} = \dfrac{1.5}{3 + 1.5} = \dfrac{1}{\textit{AD}} = \dfrac{1}{1 + \textit{ED}}

\\[3ex] \dfrac{1.5}{4.5} = \dfrac{1}{1 + \textit{ED}} …cross multiply

1.5(1 + *ED*) = 4.5

1 + *ED* = 3 = *AD*

## Example

Triangle

Ris similar to triangleS. Find the ratio of the area of triangleRto triangleS.

### Solution

To find the areas, you need the base and height for both triangles. Use the only pair of corresponding sides that have values to find the scale factor.

\dfrac{side \,\textit{R}}{side \,\textit{S}} = \dfrac{34}{51} = \dfrac{2}{3}\\[3ex]Use corresponding sides and the scale factor to find the values of *x* and *w*.

\dfrac{2}{3} = \dfrac{\textit{x}}{30}, so *x* = 20.

\\[2ex]\dfrac{2}{3} = \dfrac{\textit{w}}{63}, so *w* = 42.

Remember that the SAT is looking for reasoning, not long calculations. Use what you know about right triangles.

Notice that the sides of the smallest triangle in *R* have side lengths

12 : *y* : 20.

This is a multiple of a 3 : 4 : 5 right triangle, so the height *y* = 4(4) = 16.

Find the height *z* in triangle *S* using the scale factor.

\dfrac{2}{3} = \dfrac{16}{\textit{z}}, so *z* = 24.

Calculate the areas. Remember that the question is asking for a ratio, not a value, so leave the answers in factored form.

area of *R* = \dfrac{1}{2}(42)(16) = (42)(8)

= 6 × 7 × 2 × 4

area of *S* =\dfrac{1}{2}(63)(24) = (63)(12)

= 7 × 9 × 2 × 6

The ratio for the areas is

\dfrac{6 × 7 × 2 × 4}{7 × 9 × 2 × 6} = \dfrac{4}{9}

So, the ratio of the area of triangle *R* to triangle *S* is \dfrac{4}{9}.

Notice that this is the square of the scale function: \Big(\dfrac{2}{3}\Big)^{\displaystyle{2}} = \dfrac{4}{9}.

#### Video Quiz

#### Triangles

Best viewed in landscape mode

2 questions with video explanations

100 seconds per question

# Are you sure you want to refresh the question?

**Before attempting these problems, be sure to review this section on Quantitative Comparison questions.**

https://www.youtube.com/watch?v=45bbhm0Yv4Q&list=PLD0D070C218D8F5A3&index=22

https://www.youtube.com/watch?v=aNm9RNUZF98&list=PLD0D070C218D8F5A3&index=23

https://www.youtube.com/watch?v=Fs5YKMxHkpI&list=PLD0D070C218D8F5A3&index=25