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}

In the right triangle $A B C,$ there is an inscribed square.

a right triangle with a square inscribed

What is the area of the square?

To calculate the area of the square we need to know its side. We will label this as x.

In the diagram we can identify two right triangles. These triangles are similar because they are both right and they share an angle at A. This is enough information to claim similarity by the Angle-Angle Similarity Theorem.

Since the ratio between corresponding sides in similar triangles is the same, we can write the following equation. x/x+4 = 2/x+2 Let's solve for x.

We want to find the area of the square which is the square of its side. A=x^2 ⇓ A=( sqrt(8) )^2=8 cm^2

Zain is solving a puzzle in a role-playing game that requires figuring out the measurements of a triangle similar to another triangle with the sides $7,$ $5,$ and $3$ centimeters. The triangle he is solving has the side lengths of $14,$ $x,$ and $y$ centimeters. What could $x$ and $y$ be? He has shared his notes with you.

Did Zain get it right?

Zain has made correct calculations, but we still cannot say for sure that the solution is correct. It seems Zain has assumed which sides are corresponding. Howeer, there is nothing in the exercise indicating which sides are corresponding. We could be equally correct to draw any of the following triangles.

The side that is 14 centimeters in the greater triangle could also correspond to the side that is 5 centimeters or the side that is 3 centimeters. This would have given us the following ratios. 14/5=2.8 or 14/3 ≈ 4.67 By multiplying the rest of the sides in the smaller triangle with these ratios, we get three possible triangles.

To solve the exercise correctly, Zain would need to report all of these possible solutions to the game master.

Given the information in the diagram, determine the distance between the swimmer and the shoreline.

a swimmer that is at a distance from the shoreline

Let's label the distance between the swimmer and the shoreline x. We will also mark a few angles.

Examining the diagram, we can identify two right triangles. Since they also share an angle, we know they are similar by the Angle-Angle Similarity Theorem.

Using that the triangles are similar, we can write the following equation. x/x+30=150/175 Let's solve this equation for x.

The distance between the ship and the shoreline is 180 meters.

For a triangle, the altitude can be drawn against any of its sides.

Two triangles with sides of a an b. One heights is drawn in each triangle.

Use similarity to find the ratio of

h_{a}

h_{b} .

Each of the altitudes divides the triangle into two right triangles. Let's highlight those that have the angle opposite side c as one of its acute angles.

Since these two right triangles share one angle we know they are similar by the Angle-Angle Similarity Theorem.

Using the triangles' similarity, we can express the ratio of h_a to h_b. h_a/h_b = b/a

Kevin and Ali are sailing on a nice summer's day. As they are approaching a bridge with a clearance of $14$ meters, Kevin is worried that the mast will not clear the bridge.

They know that the height from the surface of the water to the beginning of the mast is $1.5$ meters. However, they do not know the height of the mast. Before it is too late, they make some quick measurements. Notice that the two lengths between the mast and backstay are parallel.

Will they clear the bridge? Support your reasoning with calculations.

The measured distance between the backstay and the mast creates a pair of triangles. These are similar triangles because they are both right triangles and share an angle. This means we can claim similarity by the Angle-Angle Similarity Theorem.

Let's separate the triangles. We will label the vertical leg of the smaller triangle as x.

Using the similarity between the triangles, we can write and solve an equation for x.

The rest of the mast has a length of 11.20 meters. If we add this to the 0.8 meter section of the mast, we can determine the total height of the sailboat. We already measured the 1.5 meters between the water surface and the mast. 11.20+1.5+0.8=13.5 m Since 13.5 meters is less than 14 meters, the sailboat will clear the bridge.

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