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
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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!