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Try This

Creating Right Triangles

Right triangle geometry can be traced back to ancient Egypt. Using only a piece of rope, the ancient Egyptians solved the problem of how to lay out a \(90^{\circ}\) corner for the base of a pyramid.

The following steps demonstrate how they did this:

Step 1: Tie \(13\) equally-spaced knots on a rope.
 

Step 2: Pin the \(1^{st}\) and \(13^{th}\) knot where the \(90^\circ\) angle is desired.

Step 3: Pull the rope at the \(5^{th}\) knot to form a line segment. Pin the \(5^{th}\) knot.

Step 4: Pull the \(10^{th}\) knot to form a triangle.
Pin the \(10^{th}\) knot.

There will be a \(90^{\circ}\) angle at the \(1^{st}\) and last knots. It's worth noting that Egyptians mostly studied specific examples of right triangles and that they understood right triangles in terms of ratios.

It wasn't until much later that the Greek mathematician, Pythagoras, discovered that there was an equation that described the relationship between the sides of a right triangle.

Lesson Goals

• Investigate the relationship between the side lengths of a right triangle.
• Develop the Pythagorean Theorem.
• Solve for the missing side length of a right triangle.

Try This!

In the Try This problem, we are going to explore a neat property of right triangles.

Step 1: Draw a right triangle.

Step 2: Draw a regular polygon on the largest side of this triangle, so that the side of the triangle is also one side of your polygon.

Now, your regular polygon can be an equilateral triangle, a square, a regular pentagon, or like me, you can draw a regular hexagon.

Step 3: Draw similar regular polygons on the other sides of the right triangle.

Since I drew a hexagon in Step 2, I'm going to draw hexagons in this step as well. But you would draw the same type of shape that you drew in Step 2.

Now that you have three similar regular polygons on all three sides of your right triangle, calculate the area of the three polygons you created. What do you notice?

Think about this problem, then move on to the next part of the lesson.

The Pythagorean Relationship

Right Triangle Terminology

A triangle that contains a right angle is called a right triangle.

Recall that the sum of the three angles in any triangle must be \(180^\circ\).

Take a moment and try this problem on your own.

Solution

Since the sum of the three angles must be \(180^\circ\), and since \(\angle A\) is equal to \(90^\circ\), we know that the sum of the remaining two angles must be \(90^\circ\). So, 

\(\angle B + \angle C = 90^\circ\)

We conclude that both \(\angle B\) and \(\angle C\) must always be less than \(90^\circ\). 

What this tells us is that in a right triangle, the \(90^\circ\) angle is the largest angle.

You might remember from an earlier geometry lesson that the largest side of a triangle is always opposite to the largest angle, and so we call the largest side of a right triangle the hypotenuse.

The other two sides of the triangle are called legs. The two legs meet at the \(90^\circ\) angle.

Squares on the Sides of Right Triangles

Many thousands of years ago, a Greek mathematician named Pythagoras drew squares on the sides of a right triangle to examine the relationship between the side lengths of the triangle.

He drew these squares so that one side of the largest square is the hypotenuse of the right triangle.

One side of the smallest square is the smallest side of the right triangle.

And one side of the remaining square is the remaining side of the right triangle.

Pythagoras discovered that the areas of these three squares were related.

Explore This

Description

Use the following investigation to explore the relationship between the areas of the squares drawn on the three sides of a right triangle.

Based on the investigation, what can you conclude about the squares drawn on the sides of a right triangle?

Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0. https://ggbm.at/tarGNuVX

Online Version

https://ggbm.at/tarGNuVX

The Pythagorean Relationship

In the investigation, there are five pieces, one piece is the entire square of the smallest leg and the other four pieces are created from slicing the square on the other leg. These pieces fit perfectly inside the larger square.

Created with GeoGebra. Author: University of Waterloo. CC BY-NC-SA 4.0. https://ggbm.at/tarGNuVX

This tells us that the area of the large square on the hypotenuse must be equal to the sum of the areas of the two smaller squares on the two legs. This important property of right triangles is called the Pythagorean Relationship.

Example 1

Solution

Let's start by letting \(A\) represent the area of the square on the hypotenuse of the right triangle. 

By the Pythagorean Relationship, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the two legs. The result is

\(\begin{align*} A &=8 + 12 \\ &= 20 \end{align*}\)

Therefore, the area of the square on the hypotenuse is \(20\) cm\(^2\).

Example 2

Solution

Let's start by letting \(A\) represent the area of the square on the remaining leg of the right triangle.

By the Pythagorean Relationship, the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the two legs. The result is

\(\begin{align*} 34 &= A + 16 \end{align*}\)

Notice that this time, the unknown area, \(A\), is on the right side of the equation, and so finding the missing area is less straightforward than in our previous example. But for us this is not a problem, because we know how to solve for \(A\). We can subtract \(16\) from both sides of the equation. We get that

\(\begin{align*} 34 &= A + 16 \\ 34 \class{hl2}{-16} &\;= A + 16 \class{hl2}{-16} \\ 18 &\;= A \end{align*}\)

Therefore, the area of the square on the remaining leg is \(18\) cm\(^2\).

Check Your Understanding 1

Question

The square on the hypotenuse of a right triangle has an area of \(90\) mm\(^2\). The square on one leg has an area of \(24\) mm\(^2\). What is the area of the square on the remaining leg of the right triangle?

Answer

\(66\) mm\(^2\)

Feedback

We start by letting \(A\) represent the area of the square on the remaining leg. 

The Pythagorean Relationship tells us the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the two legs. So, 

\(\begin{align*} 90 &=A+24 \\ 90 \class{hl2}{-24} &= A + 24 \class{hl2}{-24} \\ 66 &=A \end{align*}\)

Therefore, the area of the square on the remaining leg is \(66\) mm\(^2\).

Try This Problem Revisited

Let's take another look at the Try This problem.

Solution

The Try This problem is an interesting extension of the Pythagorean Relationship and explores the fact that if any similar shapes, not just squares, are drawn on the three sides of a right triangle, then the area of the largest shape is equal to the sum of the areas of the smaller two shapes.

Side Lengths of Right Triangles

Example 3

Solution

Step 1: Area of Square on Hypotenuse

Step 2: Length of Hypotenuse

Then \( h^2 = 25 \), since squaring the side length of the square produces the area. 

We need to find a side length that squares to \(25\), and we get that

\(\begin{align*} h^2 &= 25 \\ h &= 5 \end{align*}\)

Therefore, the length of the hypotenuse is \(5\) cm.

Example 4

Solution

If we know the length of two sides of a right triangle, then we can use our knowledge of squares and the Pythagorean Relationship to find the length of the third side.

Step 1: Area of Square on Remaining Leg

Notice that in my diagram, I have drawn a square on each side of the triangle.

I have done this because using the Pythagorean Relationship, the area of these squares will help us find the missing side length. The length of the square on the hypotenuse is \(10\) cm\(^2\), which means the area of the square is equal to \(100\) cm\(^2\). The length of the other side is \(6\) cm\(^2\), which means the area of the square is equal to \(36\) cm\(^2\).

Let \(A\) represent the area of the square on the remaining leg.

Then by the Pythagorean Relationship, the area of the larger square is equal to the sum of the areas of the smaller two squares, which gives us that

\( 100=A + 36 \)

Again, notice that the unknown area \(A\) is on the right side of the equation, and determining its value is not as straightforward as our previous example. But we can still determine the value of \(A\) by solving the equation. And so we subtract \(36\) from both sides, and we get that

\(\begin{align*} 100&=A + 36 \\ 100 \class{hl2}{-36} &= A + 36 \class{hl2}{-36} \\ 64 &= A \\ \end{align*}\)

The area is \(64\) cm\(^2\).

Step 2: Length of Remaining Leg

Let's now move our focus on to this square. Notice that the unknown side length of the triangle is equal to the side length of this square.

Let \(a\) represent the length of the remaining leg.

Then we have that \(a^2=64\). We need to find a side length that squares to \(64\).

\(\begin{align*} a^2 &= 64 \\ a &= 8 \end{align*}\)

Therefore, the length of the remaining leg is \(8\) cm.

Check Your Understanding 2

Question

The length of one leg of a right triangle is \(24\) m. The length of the hypotenuse is \(25\) m. What is the length of the remaining leg of the right triangle? 

Answer

\(7\) m

Feedback

The Pythagorean Theorem

Example 5

Solution

We start by assigning a variable \(h\) to represent the length of the hypotenuse. 

You can imagine a square being drawn on each side of the right triangle, and we know because of the Pythagorean Relationship that there is a relationship between the areas of these three squares.

First, note that the area of the square on the hypotenuse would be \(h^2\). The area of the squares on the two legs would be \(6^2\) and \(9^2\).

Using the Pythagorean Relationship, we know that the area of the square on the hypotenuse is equal to the sum of the area of the squares on the other two sides. So we write

\(\begin{align*} h^2 &= 6^2 + 9^2 \\ h^2 &= 36 + 81 \\ h^2 &= 117 \end{align*}\)

We need to find a side length that squares to \(117\). Since \(117\) is not a perfect square, then \(h\) is not an integer. So what are some perfect squares that are near \(117\)? 

Perhaps you'll recall that \(10^2\) is equal to \(100\) is a perfect square, and \(11^2\) equals \(121\) is also a perfect square. We have that

\(100 \lt 117 \lt 121\)

Therefore, 

\(10 \lt h \lt 11\)

Using our knowledge of square roots, we can determine that 

\(\begin{align*} h^2 &= 6^2 + 9^2 \\ h^2 &= 36 + 81 \\ h^2 &= 117 \\ h &\approx10.8 \end{align*}\)

Therefore, the length of the hypotenuse is approximately \(10.8\) cm.

Notice that in this example, we represented the area of each square in terms of its side length. The result is an equation that represents a relationship between the three side lengths of a right triangle.

The Pythagorean Theorem

We can restate the Pythagorean Relationship using algebra, and we're now going to call it the Pythagorean Theorem. 

I want you to remember that this statement is not really any different from the Pythagorean Relationship that we have been working with throughout the lesson.

To convince yourself of this, start by drawing the right triangle, where \(c\) represents the length of the hypotenuse and \(a\) and \(b\) represent the lengths of the two legs. 

You can draw a square on each side of the triangle.

The area of the square on the hypotenuse is \(c^2\), since the side length is \(c\).

Similarly, the areas of the squares on the two legs are \(a^2\) and \(b^2\) because the side lengths are \(a\) and \(b\) respectively.

Notice that the equation \(c^2 = a^2 + b^2 \) is simply saying that the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Example 6

Take a moment and try this problem on your own.

Solution

Let's start by letting \(a\) represent the length of the unknown leg.  

By the Pythagorean Theorem,

\(\begin{align*} 17^2 &= a^2 + 3^2 \\ 289 &= a^2 + 9 \end{align*}\)

To solve for \(a^2\), our first step is to subtract \(9\) from both sides. We get that

\(\begin{align*} 17^2 &= a^2 + 3^2 \\ 289 &= a^2 + 9 \\ 289 \class{hl2}{-9} &= a^2 + 9 \class{hl2}{-9} \\ 280 &= a^2 \end{align*}\)

We need to find a side length that squares to \(280\). Since \(280\) is not a perfect square, then \(a\) is not going to be an integer. But we can find perfect squares that are close to \(280\). Specifically, \(16^2=256\) and \(17^2 = 289\) are both perfect squares.

We have that 

\(256 \lt 280 \lt 289\)

Therefore

\(16 \lt a \lt 17\)

Using our knowledge of square roots, we can determine that 

\(\begin{align*} 17^2 &= a^2 + 3^2 \\ 289 &\;= a^2 + 9 \\ 289 \class{hl2}{-9} &\;= a^2 + 9 \class{hl2}{-9} \\ 280 &\;= a^2 \\ 16.7 &\;\approx a \end{align*}\)

Therefore, the length of the missing leg is approximately \(16.7\) cm.

Check Your Understanding 3

Question

The length of one leg of a right triangle is \(10\) mm. The length of the hypotenuse is \(12\) mm. What is the length of the remaining leg of the right triangle? Round your answer to 1 decimal place.

Answer

\(6.6\) mm

Feedback

Let \(a\) represent the length of the remaining leg. 

By the Pythagorean Theorem, 

\(\begin{align*} 12^2 &= a^2 + 10^2 \\ 144 &= a^2 + 100 \\ 144 \class{hl2}{-100} &= a^2 +100 \class{hl2}{-100} \\ 44 &=a^2\end{align*}\)

We need to find a number that squares to \(44\). Since \(44\) is not a perfect square, we know that \(a\) is not going to be an integer. 

We can start by finding perfect squares that are close to \(44\). 

\(\begin{align*} 36 \lt 44 &\lt 49 \\ 6^2 \lt a^2 &\lt 7^2 \\ 6 \lt a &\lt 7 \end{align*}\)

We know the value of \(a\) is going to be between \(6\) and \(7\). Using our knowledge of square roots, we can determine that the value of \(a\) is approximately \(6.6\). 

Therefore, the length of the remaining leg is approximately \(6.6\) mm.

Take It With You

The side lengths of a triangle are \(20\) cm, \(25\) cm, and \(35\) cm.

Can you determine whether or not this is a right triangle? Explain.