On the Math test in Praxis Core, you’ll be asked a handful of geometry questions. While the subset of geometry questions is not the largest category on the exam, there will still be a significant number of geometry questions on test day. You can expect geometry answers to be about 15% of your final Core Math score.
So obviously you’re going to want to do well on the types of geometry questions seen in Praxis Core math. Geometry problems on this exam fall into three broad categories: coordinate plane geometry, geometry with two dimensional shapes (squares, cubes, triangles, circles, etc..), and geometry with three dimensional shapes (cylinders, cones, angular blocks, and so on). Below, you can test your Praxis Core Math geometry skills with one practice question from each category. An answer key is available at the very bottom of this post.
Practice coordinate plane geometry question
Note: I almost considered including this question type in the practice set for numbers and quantity. This is because coordinate plane geometry on the Praxis Core always boils down to addition or subtraction. However, doing these problems efficiently and correctly does require a keen sense of geometric space. So I’m going to count these as true geometry questions.
If rhombus ABCD in the xy-plane is shifted 3 units down and 5 units to the left, what would be the coordinates of point D after the shift?
A) (-3, 10)
B) (-3, 0)
C) (10, 3)
D) (0, -3)
E) (5, -3)
Practice two-dimensional geometry question
The rectangle above is four centimeters high and 8 centimeters wide. What is the approximate length of each diagonal (diagonals AC and BD)?
Practice three-dimensional geometry question
The figure above shows a three dimensional object comprised of a cylinder with a radius of 4 feet and a height of 5 feet, topped by a right cylinder cone with a radius of 4 feet and a height of 3 feet. What is the approximate total volume of this object? (The volume of a right cylinder cone with a base radius r and height h is 1/3 pi(r^2)h.)
A) 200 cubic feet
B) 300 cubic feet
C) 335 cubic feet
D) 415 cubic feet
E) 610 cubic feet
Notes about the answers:
The answer to question 1 is a simple matter of correctly subtracting 3 from the y coordinate, and 5 from the x coordinate. Answer two requires you to know and correctly apply the Pythagorean theorem, which states that in a right triangle where a and b are the lengths of the sides of the right angle, and c is the length of the side opposite the right angle, a^2 + b^2 = c^2. The answer to the third question gives you the formula for the volume of a right circular cone, but it also requires you to know and use the formula for cylinder volume: V =pi(r^2)h.
The answer given for the “Practice two-dimensional geometry question” is incorrect.
Using the Pythagorean theorem:
4 squared + 8 squared = x squared
16 + 64 = 80
The correct answer should be the square root of 80: 8.944
(This is the second practice question I have completed today in which an incorrect answer was credited with being correct.)
In this case, the answer in the post actually is correct Ted. It looks like you missed one important keyword in the problem– you’re asked to find the approximate length of each diagonal, not the exact length. 8.944 would be the correct answer (rounded to the nearest thousandth), but 9 is an acceptable whole number approximation of the square root of 80.
(But great catch on the incorrect numeric entry problem over in the Practice Algebra Questions post, Ted!)
The answer key says “B” which is 3???
Thanks for bringing that to my attention! The answer to the second question should actually be “C,” which is 9. I’ve just edited the blog post to get rid of the error.
Can you explain the third problem? I come up with 80 pi plus 16 pi which gives me an approximate value of 301
NOTE: Edited to correct a math mistake!
To David G, and Fee, the commenter below— You are absolutely correct, and my original answer was wrong! The formula for the right cone is 1/3 π(r^2)h . Height is 3 and radius is 4. Plug in those numbers, and you get (1/3)*π*(4^2)*3. (1/3)*3 is 1, so the 1/3 and the 3 cancel each other out, for just 16π. 16π is roughly 50, and 80π is roughly 251. So your calculation of 301 is right, and the approximate answer should actually be B.
wouldn’t you use the height of the right cone which is 3ft instead of the height of the cylinder when finding the total volume of the cone?
You’re right, Fee! Thanks for bringing this to my attention! I’ve edited my original response to David G, and edited the post to reflect the correct answer, which is actually (B) and not (C).
where did 80n come from? I’m lost. I had trouble with geometry on the test.
I’m happy to help 🙂 First, just to make sure, the value is 80pi (not 80n). This value is the volume of the right cylinder, while 16pi is the volume of the cone.
The volume of a cylinder with a base radius r and height h is pi*(r^2)h
where r is the radius of the circular base and h is the height of the cylinder.
In the figure, we see that
r = 4
h = 5
Plugging those values into the formula above, we get
pi*(4^2)5 = pi*16*5 = 80pi
I hope this clears up your doubts! If not, let us know 🙂
Hmmm-this is a little scary. You have alot of mistakes after going through all the blog links, please don’t get it twisted, I am having a terrible time passing the core math so don’t take it as judging, but before I sign up is the paid service better than the blog problems as far as the error rates?