Pythagorean Geometry – Tutorial and Practice Questions
Pythagorus – A Quick Review and Tutorial
If we have a right triangle ABC, where its sides (legs) are a and b and c is a hypotenuse (the side opposite the right angle), then we can establish a relationship between these sides using the following formula:
This formula is proven in the Pythagorean Theorem. There are many proofs of this theorem, but we’ll look at just one geometrical proof:
If we draw squares on the right triangle’s sides, then the area of the square upon the hypotenuse is equal to the sum of the areas of the squares that are upon other two sides of the triangle. Since the areas of these squares are a2, b2 and c2, that is how we got the formula above.
One of the famous right triangles is one with sides 3, 4 and 5. And we can see here that:
32 + 42 = 52
9 + 16 = 25
25 = 25
Example:
We write the information we have about triangle ABC and we draw a picture of it for better understanding of the relation between
its elements:
P = 18 cm
a – b = 3 cm
h=?
We use the formula for the perimeter of the isosceles triangle, since that is what is given to us:
P = a + 2b = 18 cm
Notice that we have 2 equations with 2 variables, so we can solve it as a system of equations:
a + 2b = 18
a – b = 3 / multiply by 2
a + 2b = 18
2a – 2b = 6
a + 2b + 2a – 2b = 18 + 6
3a = 24
a = 24:3 = 8 cm
Now we go back to find b:
a – b = 3
8 – b = 3
b = 8 – 3
b = 5cm
Another Example
The isosceles triangle ABC has a perimeter of 18 centimeters, and the difference between its base and legs is 3 centimeters. Find the height of this triangle.
We write the information we have about triangle ABC and we draw a picture of it for better understanding of the relation between
its elements:
P = 18 cm
a – b = 3 cm
h = ?
We use the formula for the perimeter of the
isosceles triangle, since that is what is given to us:
P = a + 2b = 18 cm
Notice that we have 2 equations with 2 variables, so we can solve it as a system of equations:
a + 2b = 18
a – b = 3 / multiply by 2
a + 2b = 18
2a – 2b = 6
a + 2b + 2a – 2b = 18 + 6
3a = 24
a = 24:3 = 8 cm
Now we go back to find b:
a – b = 3
8 – b = 3
b = 8 – 3
b = 5cm
Using Pythagorean Theorem, we can find the height using a and b, because the height falls on the side a at the right angle. Notice that height cuts side a exactly in half, and that’s why we use in the formula a/2. In this case, b is our hypotenuse, so we have:
Pythagorean Geometry – Practice Questions
1. What is the length of each side of the indicated square above?
a. 10
b. 15
c. 20
d. 5
2. What is the length of the sides in the triangle above?
a. 10
b. 20
c. 100
d. 40
3. For the given diameter, what is the square of the indicated measurement?
a. 6
b. 72
c. 36
d. 64
4. Every day starting from his home Peter travels due east 3 kilometers to the school. After school he travels due north 4 kilometers to the library. What is the distance between Peter’s home and the library?
a. 15 km
b. 10 km
c. 5 km
d. 12 ½ km
5. What is the length of the missing side in the triangle above?
a. 6
b. 4
c. 8
d. 5
Answer Key
4. C
Pythagorean Theorem:
(Hypotenuse)2 = (Perpendicular)2 + (Base)2
h2 = a2 + b2
Given: 32 + 42 = h2
h2 = 9 + 16
h = √25
h = 5
4 Comments
I’m confused by question 4. In the answer you say a squared + b squared = 25. Isn’t 25 just the sum of a and b, not those squared?
The Pythagorean formula is the SUM of a and b – a2 + b2 = c2
You’re right, there’s a big issue with this problem.
It says that when Peter walks to the library, he walks a total of 25 miles. The two paths he takes are like the legs of the triangle, so we could write that as a + b = 25. We wouldn’t square “a” or “b”, since linear distance (home to school, school to library) is not a squared function.
What IS true is that the two legs, when squared, should add to the hypotenuse squared. So, we can write that as:
a^2 + b^2 = x^2
The issue at this point though would be that you only have two functions, yet there are three unknowns (a, b, x).
So, I tried plugging possible solutions into “x” in order to solve for “a” and “b”.
Even if you plug in the largest value (15) for “x”, it’s still not solvable, and here’s why:
If we assume that “x” is 15, then:
a^2 + b^2 = 15^2 Which would simplify to…
a^2 + b^2 = 225
Then, we can use the a + b = 25 equation that the original problem gave us and isolate a variable (I chose “b”).
b = 25 – a
Now, I can plug this value of “b” into the a^2 + b^2 = 225 equation, which results in:
a^2 + (25 – a)^2 = 225
Next, by distributing the exponent into the parenthesis, I would get:
a^2 + 625 – 50a + a^2 = 225
After some cleaning up and combining of like terms, I get:
2a^2 – 50a + 400 = 0
I divided through by 2 to make things a bit easier:
a^2 – 25a + 200 = 0
And now, all you have to do is find two numbers that multiply to 200, and add to 25!!!
…
…
Did you find it yet? No? Well, you shouldn’t. Because there aren’t any two numbers that would solve this…
Yes – an old version of the question – corrected! thanks!