OMPT drill — Geometry

OMPT geometry basics practice: lesson and 10 problems

Q: How do I find the centre and radius of a circle on the OMPT?

Get the equation into (x-a)^2 + (y-b)^2 = r^2, by completing the square in x and in y if it arrives expanded. The centre is (a, b) and the radius is r, the square root of the right-hand side. Two marks go missing here every time: the bracket (y+2)^2 means b = -2, not +2, and the answer to "radius" is sqrt(25) = 5, not 25.

Q: How do I find the distance from a point to a line on the OMPT?

Write the line as ax + by + c = 0, then use d = frac|ax_0 + by_0 + c|sqrt(a^2+b^2). The rearranging is half the work, because the question usually hands you the line as y = mx + q. The absolute value bars are not decoration: without them a point on the far side of the line returns a negative distance.

Angles, triangles, Pythagoras, perimeter and area, then the same shapes on a grid: distances, lines and circles. The chapter three variants test and almost no OMPT prep material covers. Draw every one of these before you calculate; almost every error in this strand is visible in a sketch and invisible in the algebra.

Tested inOMPT-C OMPT-D OMPT-G

OMPT geometry basics: the lesson

What the OMPT tests in geometry

Geometry on the OMPT comes in two halves that look unrelated and share every formula. The first half is plane geometry: angles on a straight line and around a point, angles made by a transversal crossing parallel lines, the angle sum of a triangle and of an $n$ -gon, the Pythagorean theorem, and the perimeter and area of the standard shapes. The second half is the same geometry on a coordinate grid: the distance between two points, the equation of a line, perpendicular and parallel slopes, the distance from a point to a line, and circles with their tangents and intersections. The bridge between the halves is Pythagoras, which is where the distance formula comes from, so if the theorem is automatic then half of the coordinate work is too.

Three of the seven variants test this, at three depths. The OMPT-C has four syllabus topics (notions on lines, angles, triangles, Pythagoras) for 15% of a 30-question paper, all plane, no coordinates. The OMPT-G splits it into two chapters, "Geometry, basics" with 8 topics and "Geometry, continued" with 9, together 10% of the paper: the basics chapter adds polygons, perimeter and area, and the continued chapter is entirely lines and circles on the grid. The OMPT-D is the deep one, 20 topics for 20% of the paper, adding parametric curves and vectors on top of lines and circles. The A, B, E and F have no geometry chapter at all.

Notation and answer format the OMPT expects

Angles in degrees with the unit implied by the question. This chapter is not trigonometry, so radians almost never appear here.
Lengths and areas exact: $20$ becomes $25$ , and an area of $\frac{15}{2}$ stays a fraction unless the question asks for a decimal.
A circle answer usually wants two things, the centre as a coordinate pair and the radius as a single number. Give $r$ , not $r^{2}$ .
A line can be typed as $y = m x + q$ or as $a x + b y + c = 0$ ; the Maxima grader accepts equivalent forms, but the distance formula needs the second one, so rearrange before substituting.
On the OMPT-D the note field is compulsory, and a geometry answer with no construction written down is where partial credit disappears fastest. Sketch, even roughly, and say which theorem you used.

Worked example 1: the angles of a regular polygon

Find one interior angle of a regular twelve-sided polygon.
Split the polygon into triangles from one vertex: an $n$ -gon gives $n - 2$ of them, so the interior angles add to $(n - 2) \cdot 18 0^{\circ}$ .
For $n = 12$ : $10 \cdot 18 0^{\circ} = 180 0^{\circ}$ in total.
Regular means all twelve angles are equal, so each is $1800 \div 12 = 15 0^{\circ}$ .
Check with exterior angles, which always sum to $36 0^{\circ}$ : each is $360 \div 12 = 3 0^{\circ}$ , and $180 - 30 = 150$ . Final answer: $15 0^{\circ}$ .

Worked example 2: a circle hiding in an expanded equation

Find the centre and radius of $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ .
Complete the square in $x$ : $x^{2} - 6 x = (x - 3)^{2} - 9$ .
Complete the square in $y$ : $y^{2} + 4 y = (y + 2)^{2} - 4$ .
Substitute both and move the constants right: $(x - 3)^{2} + (y + 2)^{2} = 12 + 9 + 4 = 25$ .
Read off against $(x - a)^{2} + (y - b)^{2} = r^{2}$ : centre $(3, - 2)$ and radius $25 = 5$ . The $+ 2$ in the bracket means $b = - 2$ , and the answer to "radius" is 5, not 25.

How each OMPT variant tests geometry basics

The chapter weights are the official ones from the omptest.org learning-assessment pages. The question counts are my estimates of how much of that chapter this one topic takes up. The 21-question papers (OMPT-D and OMPT-F) ask long multi-part questions, so a topic there usually shows up as a step inside a bigger problem.

OMPT-A
52 questions
Not tested
The eight OMPT-A chapters run from numbers to differentiation with no geometry. A right triangle there would be scenery in a word problem.
OMPT-B
64 questions
Not tested
OMPT-B has no geometry chapter. Pythagoras shows up inside trigonometry, where it belongs to the unit circle rather than to shapes.
OMPT-C
30 questions
Testedroughly 4 to 5 of 30 questions (15% of the paper)
Official chapter: Geometry, 15% of the paper.
The shortest geometry chapter on any variant: four syllabus topics, covering notions on lines, angles, triangles and the Pythagorean theorem. Short numeric answers, no coordinates, no circles. Angle chases between parallel lines and a missing side in a right triangle are the two shapes.
OMPT-D
21 questions
Tested3 to 4 of the 21 long questions (20% of the paper)
Official chapter: Geometry, 20% of the paper.
The largest geometry chapter, 20 syllabus topics. Lines by several descriptions, angles between lines, distance from a point to a line, circles, their intersections and tangents, then parametric curves and vectors. Multi-step, with the written note required.
OMPT-E
40 questions
Not tested
OMPT-E swaps geometry for statistics and probability. Nothing in its twelve chapters is geometric.
OMPT-F
21 questions
Not tested
OMPT-F is algebra, functions and differentiation over six chapters. No geometry, no trigonometry.
OMPT-G
26 questions
Testedroughly 2 to 3 of 26 questions (10% across both geometry chapters)
Official chapter: Geometry and coordinates, 10% of the paper.
Split in two: "Geometry, basics" (8 topics: lines, parallel and perpendicular lines, angles, triangles, polygons, Pythagoras, perimeter, area) and "Geometry, continued" (9 topics: descriptions of a line, angles between lines, perpendicular lines, distance from a point to a line, and circles with their intersections, tangents and distances). Final answers only.

OMPT geometry basics practice problems

10 original problems in the formats the OMPT uses: typed numeric answers, multiple choice, and spot-the-error. The last few are harder than the exam average on purpose. Solutions and the common mistake sit behind each problem.

Problem 1Numeric answer

A right-angled triangle has legs of length 9 and 12. What is its perimeter?

Show the worked solution

Answer: 36

The hypotenuse comes from Pythagoras: $c^{2} = 9^{2} + 1 2^{2} = 81 + 144 = 225$ , so $c = 15$ . The perimeter is the sum of all three sides, $9 + 12 + 15 = 36$ . The 9-12-15 triangle is the 3-4-5 triangle scaled by 3, and recognising the scaled version saves the arithmetic.

Problem 2Numeric answer

What is the size, in degrees, of one interior angle of a regular twelve-sided polygon?

Show the worked solution

Answer: 150

The interior angles of an $n$ -gon add to $(n - 2) \cdot 18 0^{\circ}$ , because the polygon splits into $n - 2$ triangles. For $n = 12$ that is $10 \cdot 18 0^{\circ} = 180 0^{\circ}$ , and in a regular polygon the twelve angles are equal, so each is $1800 \div 12 = 15 0^{\circ}$ . The exterior angle check: $360 \div 12 = 30$ , and $180 - 30 = 150$ .

Problem 3Multiple choice

A triangle has vertices $A (0, 0)$ , $B (6, 0)$ and $C (4, 5)$ . What is its area?

A $12$
B $15$
C $20$
D $30$

Show the worked solution

Answer: B — $15$

Take $A B$ as the base: it lies along the $x$ -axis and has length 6. The height is the perpendicular distance from $C$ up to that axis, which is simply its $y$ -coordinate, 5. So the area is $\frac{1}{2} \cdot 6 \cdot 5 = 15$ . The $x$ -coordinate of $C$ never enters the calculation: sliding $C$ left or right along the line $y = 5$ does not change the area.

Problem 4Numeric answer

Find the distance between the points $(- 2, 3)$ and $(4, - 5)$ .

Show the worked solution

Answer: 10

The horizontal gap is $4 - (- 2) = 6$ and the vertical gap is $- 5 - 3 = - 8$ . The distance formula is Pythagoras on those two gaps: $d = 6^{2} + (- 8)^{2} = 36 + 64 = 100 = 10$ . Signs cannot hurt you here because both differences get squared, but you still have to subtract in a consistent order.

Problem 5Spot the error

A student is asked for the centre and radius of the circle $(x - 1)^{2} + (y + 2)^{2} = 9$ . Step 1: recall that $(x - a)^{2} + (y - b)^{2} = r^{2}$ has centre $(a, b)$ and radius $r$ . Step 2: read off the centre as $(- 1, 2)$ . Step 3: read off the radius as 3. Which step contains the error?

Show the worked solution

Answer: Step 2

Step 1 states the standard form correctly and Step 3 is right, since $r^{2} = 9$ gives $r = 3$ . Step 2 flips both signs that should have stayed and keeps the one that should have flipped. Matching $(x - 1)^{2}$ against $(x - a)^{2}$ gives $a = 1$ ; matching $(y + 2)^{2} = (y - (- 2))^{2}$ gives $b = - 2$ . The centre is $(1, - 2)$ .

Problem 6Numeric answer

What is the radius of the circle $x^{2} + y^{2} - 6 x + 4 y - 12 = 0$ ?

Show the worked solution

Answer: 5

Complete the square in each variable. For $x$ : $x^{2} - 6 x = (x - 3)^{2} - 9$ . For $y$ : $y^{2} + 4 y = (y + 2)^{2} - 4$ . Substituting gives $(x - 3)^{2} - 9 + (y + 2)^{2} - 4 - 12 = 0$ , so $(x - 3)^{2} + (y + 2)^{2} = 25$ . The centre is $(3, - 2)$ and the radius is $25 = 5$ .

Problem 7Numeric answer

Find the distance from the point $(4, - 1)$ to the line $3 x + 4 y - 12 = 0$ .

Show the worked solution

Answer: 0.8

With the line written as $a x + b y + c = 0$ , the distance is $\frac{∣ a x _{0} + b y _{0} + c ∣}{a ^{2} + b ^{2}}$ . Substituting: the numerator is $∣3 \cdot 4 + 4 \cdot (- 1) - 12∣ = ∣12 - 4 - 12∣ = ∣ - 4∣ = 4$ , and the denominator is $9 + 16 = 5$ . So the distance is $\frac{4}{5} = 0.8$ .

Problem 8Multiple choice

Which line passes through $(2, 5)$ and is perpendicular to $y = - \frac{2}{3} x + 1$ ?

A $y = - \frac{2}{3} x + \frac{19}{3}$
B $y = \frac{2}{3} x + \frac{11}{3}$
C $y = \frac{3}{2} x + 2$
D $y = - \frac{3}{2} x + 8$

Show the worked solution

Answer: C — $y = \frac{3}{2} x + 2$

Perpendicular slopes multiply to $- 1$ , so from $m = - \frac{2}{3}$ the perpendicular slope is $\frac{3}{2}$ . That rules out every option but the third. Checking the point: $\frac{3}{2} \cdot 2 + 2 = 3 + 2 = 5$ , which matches. The first option is parallel, not perpendicular, and the fourth has the negative reciprocal with an extra sign change.

Problem 9Spot the error

A right-angled triangle has a hypotenuse of 13 and one leg of 5. A student finds the other leg. Step 1: write $a^{2} + b^{2} = c^{2}$ . Step 2: substitute $5^{2} + 1 3^{2} = b^{2}$ . Step 3: compute $b = 194 \approx 13.9$ . Which step contains the error?

Show the worked solution

Answer: Step 2

Step 1 is the theorem and Step 3 evaluates its own equation correctly. Step 2 puts the hypotenuse on the wrong side: $c$ is always the side opposite the right angle, the longest one, so the equation is $5^{2} + b^{2} = 1 3^{2}$ , giving $b^{2} = 169 - 25 = 144$ and $b = 12$ . The student's answer fails the sanity test on sight, since no leg can be longer than the hypotenuse.

Problem 10Numeric answer

How many points does the line $y = 2 x + 5$ have in common with the circle $x^{2} + y^{2} = 5$ ?

Show the worked solution

Answer: 1

Substitute the line into the circle: $x^{2} + (2 x + 5)^{2} = 5$ , which expands to $x^{2} + 4 x^{2} + 20 x + 25 = 5$ , so $5 x^{2} + 20 x + 20 = 0$ and, dividing by 5, $x^{2} + 4 x + 4 = 0$ . The discriminant is $16 - 16 = 0$ , so there is exactly one solution, $x = - 2$ , giving the single point $(- 2, 1)$ . The line is a tangent.

OMPT geometry basics: questions students ask

Which OMPT variants have geometry?

Three of the seven: OMPT-C (4 topics, 15% of the paper), OMPT-D (20 topics, 20%) and OMPT-G (two chapters of 8 and 9 topics, 10% between them). The A, B, E and F have no geometry chapter at all. The three that do test very different depths, so check which variant your offer names before deciding how much of this page applies to you.

What geometry is on the OMPT-C?

Four syllabus topics: some notions on lines, angles, triangles, and the Pythagorean theorem. No coordinate geometry, no circles, no area formulas beyond the triangle. On a 30-question paper that is still four or five questions, which is why a C candidate who skips geometry entirely can lose 15% before starting.

Does the OMPT-D geometry chapter include vectors?

Yes. The OMPT-D chapter runs lines, circles, parametric curves and vectors: length of vectors, vectors and parametric equations, and derivatives of parametric curves. That last group is why the D chapter is 20 topics against the G's 17, and it is the part D candidates most often meet for the first time. The C and G geometry chapters have no vectors.

How do I find the centre and radius of a circle on the OMPT?

Get the equation into $(x - a)^{2} + (y - b)^{2} = r^{2}$ , by completing the square in $x$ and in $y$ if it arrives expanded. The centre is $(a, b)$ and the radius is $r$ , the square root of the right-hand side. Two marks go missing here every time: the bracket $(y + 2)^{2}$ means $b = - 2$ , not $+ 2$ , and the answer to "radius" is $25 = 5$ , not 25.

How do I find the distance from a point to a line on the OMPT?

Write the line as $a x + b y + c = 0$ , then use $d = \frac{∣ a x _{0} + b y _{0} + c ∣}{a ^{2} + b ^{2}}$ . The rearranging is half the work, because the question usually hands you the line as $y = m x + q$ . The absolute value bars are not decoration: without them a point on the far side of the line returns a negative distance.

Do I need to memorise area formulas for the OMPT geometry chapter?

For the G, yes: perimeter and area are two of the eight basics topics, so triangle, rectangle, parallelogram, trapezium and circle formulas should be automatic, and the test provides no reference sheet. The C chapter stops before area. The D chapter works with areas through coordinates and integration instead, so a formula sheet matters less there than knowing that a triangle's height must be perpendicular to the base you chose.

Test hubs that cover geometry basics:OMPT-C OMPT-D OMPT-G All 28 topics

OMPT geometry basics practice: lesson and 10 problems

OMPT geometry basics: the lesson

What the OMPT tests in geometry

Notation and answer format the OMPT expects

Worked example 1: the angles of a regular polygon

Worked example 2: a circle hiding in an expanded equation

How each OMPT variant tests geometry basics

OMPT geometry basics practice problems

OMPT geometry basics: questions students ask

Related OMPT topics and test hubs