MCA High School / MCA EOC Geometry. Practice it free.

Uses congruence, similarity, right-triangle trigonometry, and circle theorems to model and solve problems. This High School / MCA EOC reporting domain maps to 9 practice skills and 8 representative questions from the playable bank.

High School / MCA EOC9 mapped skills
What the test measures

Geometry skills

  1. Experiment with transformations in the plane

  2. Prove geometric theorems

  3. Define trigonometric ratios and solve problems involving right triangles

  4. Apply similarity and use theorems about circles

  5. Translate between geometric description and the equation of a conic section

Standards basis

Minnesota Academic Standards in Mathematics (2007; assessed by MCA-III in grades 3-8 and 11 through 2026-27. 2022 standards phase in with MCA-IV beginning 2027-28) — Minnesota

How the MCA reports it

Minnesota’s MCA math is a state assessment tied to Minnesota Academic Standards in Mathematics rather than a separate national framework. The official public resources available here point to Minnesota-specific standards implementation, and absent a test blueprint, the coverage map should be organized by the state’s grade-level math domains that closely parallel CCSS-M. Note: the current MCA-III assesses the 2007 Minnesota standards in grades 3-8 and 11; Minnesota administers a single comprehensive high-school (grade 11) math test, NOT Algebra I / Geometry / Algebra II end-of-course exams — the 'High School / MCA EOC' band below is the grade-11 HS administration. (MCA-IV on the 2022 standards begins 2027-28.)

Blueprint & weighting

MCA uses the Independent / state-specific framework (by grade domains structure).

Try it now

8 free questions · 0/0 correct

Practice playeasy

In parallelogram PQRSPQRS, m∠P=(4x+6)°m\angle P=(4x+6)° and m∠Q=(2x+24)°m\angle Q=(2x+24)°. What is the measure of ∠P\angle P?

Consecutive angles are supplementary: (4x+6)+(2x+24)=180(4x+6)+(2x+24)=180, so 6x+30=1806x+30=180 and x=25x=25. Then m∠P=4(25)+6=106°m\angle P=4(25)+6=106°.

Practice playeasy

A circle in the coordinate plane has a diameter with endpoints (2.5,−1)(2.5, -1) and (2.5,7).(2.5, 7)\textsf{.} What is the radius of the circle?

The endpoints share the same xx-coordinate, so the diameter is vertical with length ∣7−(−1)∣=8.|7 - (-1)| = 8\textsf{.} The radius is half the diameter, so r=82=4.r = \dfrac{8}{2} = 4\textsf{.}

Practice playmedium

The equation (x−1)2+(y+4)2=49(x - 1)^2 + (y + 4)^2 = 49 represents Circle A. Circle B is obtained by shifting Circle A up 66 units in the coordinate plane. Which of the following equations represents Circle B?

Circle A has center (1,−4)(1, -4) and radius 77. Shifting up 66 moves the center to (1,2)(1, 2), so Circle B is (x−1)2+(y−2)2=49(x - 1)^2 + (y - 2)^2 = 49.

Practice playeasy

Identify the conic: x29+y24=1\dfrac{x^{2}}{9} + \dfrac{y^{2}}{4} = 1.

Both variables are squared with positive coefficients and different denominators: this is an ellipse.

Practice playeasy

A circle in the coordinate plane has equation (x+1)2+(y−7)2=4.(x + 1)^{2} + (y - 7)^{2} = 4\textsf{.} What is the yy-coordinate of the center of the circle?

(y−7)2(y - 7)^{2} means k=7,k = 7\textsf{,} so the center is (−1,7).(-1, 7)\textsf{.}

Practice playeasy

In a right triangle, cos⁡θ=1213\cos\theta = \dfrac{12}{13} and tan⁡θ=512\tan\theta = \dfrac{5}{12}. What is sin⁡θ\sin\theta?

sin⁡θ=tan⁡θ⋅cos⁡θ=512⋅1213=513\sin\theta = \tan\theta \cdot \cos\theta = \dfrac{5}{12} \cdot \dfrac{12}{13} = \dfrac{5}{13}.

Practice playeasy

Which transformation creates an image that is congruent to its preimage, but is located at a different angle around a center point?

A rotation is a rigid transformation that turns a figure around a fixed point, called a center of rotation, by a specific angle in a specific direction. All points remain equidistant from the center of rotation, keeping the image congruent to the preimage. In this example, the image is congruent to the preimage and is located at a different angle around that center point.

Practice playeasy

In a 30∘30^{\circ}-60∘60^{\circ}-90∘90^{\circ} triangle, the hypotenuse is 14.14\textsf{.} What is the short leg?

In a 30∘30^{\circ}-60∘60^{\circ}-90∘90^{\circ} triangle, the short leg is half the hypotenuse: 142=7.\dfrac{14}{2} = 7\textsf{.}

Keep practicing

Turn geometry into game time.

The MCA placement starts with this test's real coverage map and finds the right difficulty.