What are the center and radius of the circle 2x^2 + 2y^2 − 12x + 8y − 6 = 0?

Center (6, −4); radius 2 sqrt(5)

To prove that the midpoint of the hypotenuse of a right triangle is equidistant from all three vertices, a student places right triangle OAB with O(0, 0), A(2a, 0), and B(0, 2b), where a > 0 and b > 0. Point M is the midpoint of hypotenuse AB. Which computation correctly completes the proof?

M = (a, b), and MO, MA, and MB each equal sqrt(a^2 + b^2)

M = (2a, 2b), and MO, MA, and MB each equal 2 sqrt(a^2 + b^2)

M = (a, b), and MO = a + b while MA and MB each equal sqrt(a^2 + b^2)

M = (a, b), and MO, MA, and MB each equal sqrt(2a^2 + 2b^2)

Praxis Geometry (5163)

Free Praxis Geometry (5163) Practice Questions

10 sample questions for the Praxis Geometry (5163), with the correct answer and an explanation for each. No account needed. Try each one before you open the answer.

Sections covered on the exam: Plane Figures, Lines, and Angles, Transformations, Congruence, Proof, and Constructions, Similarity and Trigonometry, Circles.

Question 1 · Plane Figures, Lines, and Angles
Parallel lines l and m are cut by transversal t. Angles 1 and 2 are same-side interior angles. m∠1 = (3x + 15)° and m∠2 = (5x - 11)°.
What is m∠2?
- A.126°
- B.99°
- C.54°
- D.81°
Show answer and explanation
Answer: B. 99°
Same-side interior angles formed by parallel lines are supplementary, so 8x + 4 = 180 and x = 22. Then m∠2 = 5(22) - 11 = 99. Setting the two angles equal, as if they were alternate interior angles, gives x = 13 and 54°; the value 81° is m∠1.
Question 2 · Transformations, Congruence, Proof, and Constructions
Segments AD and BE intersect at point C, forming triangles ABC and DEC. It is given that AC = DC.
Which additional fact, together with AC = DC, is enough to prove that triangle ABC is congruent to triangle DEC?
- A.∠A ≅ ∠E
- B.AB ≅ DE
- C.BC ≅ DC
- D.AB ∥ DE
Show answer and explanation
Answer: D. AB ∥ DE
If AB ∥ DE, then ∠A ≅ ∠D as alternate interior angles along transversal AD, and ∠ACB ≅ ∠DCE as vertical angles, so the triangles are congruent by ASA with included side AC ≅ DC. Adding AB ≅ DE gives two sides and a non-included angle (SSA), which is not a valid criterion. ∠A ≅ ∠E pairs angles that do not correspond under this naming, so no criterion applies.
Question 3 · Similarity and Trigonometry
For an angle measure x, sin(2x + 10°) = cos(3x - 15°), where both 2x + 10° and 3x - 15° are acute angles. What is the value of x?
- A.19
- B.25
- C.17
- D.37
Show answer and explanation
Answer: A. 19
For acute angles, sin α = cos β exactly when α + β = 90°, so (2x + 10) + (3x - 15) = 90, which gives 5x = 95 and x = 19 (the angles are 48° and 42°). Setting the two angles equal gives 25, which would require sin 60° = cos 60°, and making the sum 180° gives 37.
Question 4 · Transformations, Congruence, Proof, and Constructions
The line through (k, 3) and (2, −1) is perpendicular to the line 3x + 4y = 7. What is the value of k?
- A.−10/3
- B.5
- C.−1
- D.22/3
Show answer and explanation
Answer: B. 5
The line 3x + 4y = 7 has slope −3/4, so a perpendicular line has slope 4/3. Using slope −4/3 (opposite sign but not the reciprocal) leads to k = −1, and using the parallel slope −3/4 leads to k = −10/3. The correct equation (3 − (−1))/(k − 2) = 4/3 gives k − 2 = 3, so k = 5.
Question 5 · Circles
What are the center and radius of the circle 2x^2 + 2y^2 − 12x + 8y − 6 = 0?
- A.Center (3, −2); radius 16
- B.Center (6, −4); radius 2 sqrt(5)
- C.Center (3, −2); radius 4
- D.Center (−3, 2); radius 4
Show answer and explanation
Answer: C. Center (3, −2); radius 4
Divide by 2 to get x^2 − 6x + y^2 + 4y = 3, then complete the square: (x − 3)^2 + (y + 2)^2 = 3 + 9 + 4 = 16. The center is (3, −2) and the radius is sqrt(16) = 4. Reversing the signs gives (−3, 2), and reporting 16 confuses r^2 with r.
Question 6 · Plane Figures, Lines, and Angles
Angles AOB and BOC form a linear pair. Ray OD bisects angle AOB, and ray OE bisects angle BOC. m∠DOB = (2x + 7)° and m∠BOE = (3x - 2)°.
What is m∠AOB?
- A.82°
- B.41°
- C.49°
- D.98°
Show answer and explanation
Answer: A. 82°
The two halves together are half of 180°, so m∠DOB + m∠BOE = 90 and 5x + 5 = 90, giving x = 17. Then m∠DOB = 41 and m∠AOB = 2 × 41 = 82. The value 41° is only the half angle, and 98° is m∠BOC.
Question 7 · Transformations, Congruence, Proof, and Constructions
A vertical pole AB stands on level ground with its base at B. Two ropes run from the top A to stakes C and D on the ground, and the ropes have equal length, AC = AD.
Which congruence criterion proves that the stakes are the same distance from the base of the pole?
- A.ASA
- B.SSS
- C.HL
- D.SAS
Show answer and explanation
Answer: C. HL
Angles ABC and ABD are right angles because the pole is vertical, the ropes AC and AD are the congruent hypotenuses, and leg AB is shared. Triangles ABC and ABD are congruent by HL, so BC ≅ BD by CPCTC. SSS and SAS would require knowing BC ≅ BD, which is the conclusion.
Question 8 · Similarity and Trigonometry
In △ABC, AB = 8 sqrt(2), m∠A = 45°, and m∠B = 30°. The altitude from C meets AB at point H, between A and B.
What is the exact length of AC?
- A.16(sqrt(3) - 1)
- B.4(sqrt(6) - sqrt(2))
- C.8(sqrt(3) + 1)
- D.8(sqrt(3) - 1)
Show answer and explanation
Answer: D. 8(sqrt(3) - 1)
Let CH = h; then AH = h (45-45-90 triangle) and HB = h sqrt(3) (30-60-90 triangle), so h(1 + sqrt(3)) = 8 sqrt(2) and h = 4 sqrt(2)(sqrt(3) - 1). Then AC = h sqrt(2) = 8(sqrt(3) - 1). The value 4(sqrt(6) - sqrt(2)) is the altitude h itself, not the hypotenuse AC of △AHC.
Question 9 · Transformations, Congruence, Proof, and Constructions
To prove that the midpoint of the hypotenuse of a right triangle is equidistant from all three vertices, a student places right triangle OAB with O(0, 0), A(2a, 0), and B(0, 2b), where a > 0 and b > 0. Point M is the midpoint of hypotenuse AB.
Which computation correctly completes the proof?
- A.M = (2a, 2b), and MO, MA, and MB each equal 2 sqrt(a^2 + b^2)
- B.M = (a, b), and MO, MA, and MB each equal sqrt(a^2 + b^2)
- C.M = (a, b), and MO = a + b while MA and MB each equal sqrt(a^2 + b^2)
- D.M = (a, b), and MO, MA, and MB each equal sqrt(2a^2 + 2b^2)
Show answer and explanation
Answer: B. M = (a, b), and MO, MA, and MB each equal sqrt(a^2 + b^2)
The midpoint of AB is ((2a + 0)/2, (0 + 2b)/2) = (a, b). Then MO = sqrt(a^2 + b^2), MA = sqrt((2a − a)^2 + (0 − b)^2) = sqrt(a^2 + b^2), and MB = sqrt(a^2 + (2b − b)^2) = sqrt(a^2 + b^2), so all three are equal. Writing MO = a + b adds the legs instead of using the distance formula, and (2a, 2b) is not the midpoint at all.
Question 10 · Circles
For which values of k does the point (k, 3) lie inside the circle (x − 1)^2 + (y + 1)^2 = 25?
- A.−4 < k < 2
- B.−4 < k < 6
- C.−2 < k < 4
- D.k < −2 or k > 4
Show answer and explanation
Answer: C. −2 < k < 4
The point is inside when (k − 1)^2 + (3 + 1)^2 < 25, so (k − 1)^2 < 9 and −3 < k − 1 < 3, giving −2 < k < 4. Ignoring the y-term gives (k − 1)^2 < 25 and −4 < k < 6, and k < −2 or k > 4 describes points outside the circle.

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