Angle-Angle Similarity Theorem & Side-Side-Side Similarity Theorem Explained

Ratio	Expression	Simplification
$\frac{A C}{A B}$	$\frac{4 5 9 + 1 2 7 5}{4 5 9} = \frac{1 7 3 4}{4 5 9}$	$\frac{3 4}{9}$
$\frac{A E}{A D}$	$\frac{4 0 5 + 1 1 2 5}{4 0 5} = \frac{1 5 3 0}{4 0 5}$	$\frac{3 4}{9}$
$\frac{E C}{D B}$	$\frac{1 3 6 0}{3 6 0}$	$\frac{3 4}{9}$

Ratio

Expression

Simplification

\frac{A C}{A B}

\frac{4 5 9 + 1 2 7 5}{4 5 9} = \frac{1 7 3 4}{4 5 9}

\frac{3 4}{9}

\frac{A E}{A D}

\frac{4 0 5 + 1 1 2 5}{4 0 5} = \frac{1 5 3 0}{4 0 5}

\frac{3 4}{9}

\frac{E C}{D B}

\frac{1 3 6 0}{3 6 0}

\frac{3 4}{9}

Ratio	Expression	Simplified Form
$\frac{A B}{A E}$	$\frac{7}{5 + 9} = \frac{7}{1 4}$	$\frac{1}{2}$
$\frac{A C}{A D}$	$\frac{5}{7 + 3} = \frac{5}{1 0}$	$\frac{1}{2}$

Ratio

Expression

Simplified Form

\frac{A B}{A E}

\frac{7}{5 + 9} = \frac{7}{1 4}

\frac{1}{2}

\frac{A C}{A D}

\frac{5}{7 + 3} = \frac{5}{1 0}

\frac{1}{2}

You have a white rectangular shaped sheet of paper with a gray backside. The paper is folded such that the lower right corner meets the upper horizontal side and the crease goes through the lower left corner.

a sheet of paper which is about to be folded

Calculate the area of the backside that becomes visible after the paper has been folded. The unfolded paper is

15

12

centimeters. Round the answer to the nearest square centimeter.

Let's first draw a diagram showing how the paper will look like when folded. At the same time, we will introduce some labels to make it easier to refer to the diagram.

In order to calculate the area of △ DEF, we need to know two side lengths. That means we need the length of EF. From the diagram, we see that x makes the leg of a right triangle where the hypotenuse and second leg are known. Therefore, we can determine x by using the Pythagorean Theorem.

When we know the length of the side labeled x, we can determine the length of EB by subtracting this from 15. 15-9 = 6 Examining the diagram, we can identify two right triangles, △ AED and △ EBF. We need to prove that these triangles are similar. Let's label the acute angles of these triangles.

At point E we can see that the sum of u, v and the right angle makes a straight angle. Therefore, we can write the following equation. m∠ u +m∠ v+90^(∘)= 180^(∘) ⇓ m∠ u = 90^(∘)- m∠ v Next, we will write an equation for the sum of the angles in △ AED. Then we can use the relationship just found to show that ∠ v and ∠ w are congruent.

As we have shown, ∠ w and ∠ v are congruent. Furthermore, since △ ADE and △ BFE have two pairs of congruent angles, we can claim similarity by the Angle-Angle Similarity Theorem.

Using this similarity, we can write an equation preparing to solve for EF.. EF/15=6/12 Let's go for it!

When we know the length of EF, we can calculate the gray triangle's area.

The visible gray area is about 56 square centimeters.

The city council is planning to build a school on a plot of land. Below we see a diagram of the school as seen from above. The school's two shorter buildings each have an entrance facing the courtyard. These are labeled $E_{1}$ and $E_{2} .$

diagram showing the bird's eye view of a school

The city planner wants to place a third entrance,

E_{3},

somewhere along the longer building's side facing the courtyard such that the distance between

E_{1},

E_{2},

and

E_{3}

is minimized. What is the minimum distance between the three entrances so that can be achieved? The council has asked for the city planner to give their answer in exact form.

Let's start by drawing an outline of the school and adding some lengths to the buildings. We will also label the buildings as A, B, and C.

To determine the minimal distance, we remember that the shortest distance between two points is a straight line. Therefore, we need to place E_3 somewhere along the side of the longest building such that the distance from E_1 to E_3 to E_2 is as short as if it was a straight line. To find where we must place E_3, we can reflect E_2 across the longer side of B.

The distance from E'_2 to E_1 is minimized by a straight segment between them. Where this segment intersects the longer side of B, is where we should place the third entrance to minimize the distance. If we draw this segment we can identify two similar triangles by the Angle-Angle Similarity Theorem.

Using the similarity between the triangles, we can write an equation. x/250-x=100/150 Let's solve this equation for x.

The third door should be placed 100 feet to the left of where B and C meet on the courtyard.

Now we can determine the minimum distance by using the Pythagorean Theorem. 150^2+150^2=c^2 &⇔ c =150sqrt(2) 100^2+100^2=d^2 &⇔ d=100sqrt(2) Finally, we will add the distance of c and d to obtain the minimal distance between the three entrances. c+d=150sqrt(2)+100sqrt(2) ⇓ c+d=250sqrt(2)feet

The Conditions for Triangle Similarity

Catch-Up and Review

Using Similarity of Triangles to Solve Problems

Investigating Triangles With Two Pairs of Congruent Angles

Angle-Angle Similarity Theorem

Proof

Solving Problems Using Angle-Angle-Similarity Theorem

Hint

Solution

Practice Solving Problems Using Similar Triangles

Side-Side-Side Similarity Theorem

Proof

Solving Problems Using Similar Triangles

Answer

Hint

Solution

Step 1

Step 2

Step 3

Answering the Question

Side-Angle-Side Similarity Theorem

Proof

Proving Similarity Between Triangles Given Sides

Hint

Solution

Applying Triangle Similarity Theorems to Solve Problems

Hint

Solution

Extra

The Conditions for Triangle Similarity

Recommended exercises

	11 Theory slides
	11 Exercises - Grade E - A
	Each lesson is meant to take 1-2 classroom sessions