Skip to main content
Code ScholarsCodeScholars
All subjects

Geometry Honors FRQ practice

7 problems · Worked solutions · 9-point rubrics · Mathematics

Code Scholars practice problems. These are not official exam prompts.

Problem 1 · Logic, Axioms, and Proof · 9 points

Prove a property of even integers

Consider the statement: If integers a and b are even, then a + b is even.

  1. A. Prove the statement using the definition of even.
  2. B. Write its converse and decide whether it is true.
  3. C. Write the contrapositive of the original statement and explain its relationship to it.
Show worked solution and rubric
  1. A. Write a=2m,b=2n\displaystyle a=2m,b=2n for integers m,n\displaystyle m,n. Then a+b=2(m+n)\displaystyle a+b=2(m+n), so the sum is even because m+n\displaystyle m+n is an integer.
  2. B. The converse says an even sum implies both addends are even. It is false: 1+3=4\displaystyle 1+3=4 but neither addend is even.
  3. C. If a+b\displaystyle a+b is not even, then at least one of a,b\displaystyle a,b is not even. This is the contrapositive and is logically equivalent to the original implication.

Scoring rubric

  • 3 points: Write a=2m,b=2n\displaystyle a=2m,b=2n for integers m,n\displaystyle m,n
  • 3 points: The converse says an even sum implies both addends are even
  • 3 points: If a+b\displaystyle a+b is not even, then at least one of a,b\displaystyle a,b is not even
Review this unit

Problem 2 · Lines, Angles, and Constructions · 9 points

Bisect a segment by construction

Segment AB has endpoints A=(0,0)\displaystyle A=(0,0) and B=(6,0)\displaystyle B=(6,0). Two equal-radius compass arcs centered at A and B intersect above and below AB.

A (0,0)M (3,0)B (6,0)PQ
Segment AB from (0,0) to (6,0); the perpendicular bisector passes through midpoint M. Not to scale.
  1. A. Describe a valid radius and how to complete the perpendicular bisector.
  2. B. Prove that any arc intersection is equidistant from A and B and lies on the perpendicular bisector.
  3. C. Find the bisector equation and its intersection with y = 2x + 1.
Show worked solution and rubric
  1. A. Choose the same radius greater than 3, such as 5. Draw circles centered at A and B; join the two intersection points with a straightedge.
  2. B. For an intersection (x,y)\displaystyle (x,y), equal radii give x2+y2=(x−6)2+y2\displaystyle x^2+y^2=(x-6)^2+y^2. Hence 12x=36\displaystyle 12x=36, so x=3\displaystyle x=3, the line through midpoint (3,0)\displaystyle (3,0) perpendicular to AB.
  3. C. The bisector is x=3\displaystyle x=3. Substitution gives y=7\displaystyle y=7, so the intersection is (3,7)\displaystyle (3,7).
A (0,0)M (3,0)B (6,0)PQ
Segment AB from (0,0) to (6,0); the perpendicular bisector passes through midpoint M. Not to scale.

Scoring rubric

  • 3 points: Choose the same radius greater than 3, such as 5
  • 3 points: For an intersection (x,y)\displaystyle (x,y), equal radii give x2+y2=(x−6)2+y2\displaystyle x^2+y^2=(x-6)^2+y^2
  • 3 points: The bisector is x=3\displaystyle x=3
Review this unit

Problem 3 · Triangles, Congruence, and Similarity · 9 points

Use a parallel segment

In triangle ABC, D lies on AB and E lies on AC, with DE parallel to BC. Given AD = 4, DB = 2, AE = 6, and BC = 9.

ADEBC
D lies on AB and E on AC; DE is parallel to BC. Diagram not to scale.
  1. A. Prove triangles ADE and ABC are similar.
  2. B. Find AC, EC, and DE.
  3. C. Find the ratio of their areas and justify why it is not the side ratio.
Show worked solution and rubric
  1. A. Angle A is shared and corresponding angles at D and B are equal by parallel lines. The triangles are similar by AA.
  2. B. AD/AB=4/6=2/3\displaystyle AD/AB=4/6=2/3. Thus AC=6÷(2/3)=9\displaystyle AC=6\div(2/3)=9, EC=9−6=3\displaystyle EC=9-6=3, and DE=9(2/3)=6\displaystyle DE=9(2/3)=6.
  3. C. The area ratio is (2/3)2=4/9\displaystyle (2/3)^2=4/9 because both base and corresponding height scale by 2/3\displaystyle 2/3.
ADEBC
D lies on AB and E on AC; DE is parallel to BC. Diagram not to scale.

Scoring rubric

  • 3 points: Angle A is shared and corresponding angles at D and B are equal by parallel lines
  • 3 points: AD/AB=4/6=2/3\displaystyle AD/AB=4/6=2/3
  • 3 points: The area ratio is (2/3)2=4/9\displaystyle (2/3)^2=4/9 because both base and corresponding height scale by 2/3\displaystyle 2/3.
Review this unit

Problem 4 · Polygons and Quadrilaterals · 9 points

Prove a quadrilateral is a parallelogram

Quadrilateral ABCD has A=(0,0),B=(4,0),C=(6,3),D=(2,3)\displaystyle A=(0,0), B=(4,0), C=(6,3), D=(2,3) in that order.

A (0,0)B (4,0)C (6,3)D (2,3)
Parallelogram ABCD with horizontal bases; the height is 3 units. Not to scale.
  1. A. Prove it is a parallelogram using slopes.
  2. B. Find the diagonal midpoints as a second verification.
  3. C. Decide if it is a rectangle or rhombus, and find its area.
Show worked solution and rubric
  1. A. AB\displaystyle AB and CD\displaystyle CD have slope 0; BC\displaystyle BC and AD\displaystyle AD have slope 3/2\displaystyle 3/2. Both opposite pairs are parallel.
  2. B. The midpoints of AC and BD are both (3,3/2)\displaystyle (3,3/2), so the diagonals bisect each other.
  3. C. Adjacent side lengths are 4\displaystyle 4 and 13\displaystyle \sqrt{13}, so it is not a rhombus. Adjacent sides are not perpendicular, so it is not a rectangle. Area is base times height, 4⋅3=12\displaystyle 4\cdot3=12.
A (0,0)B (4,0)C (6,3)D (2,3)
Parallelogram ABCD with horizontal bases; the height is 3 units. Not to scale.

Scoring rubric

  • 3 points: AB\displaystyle AB and CD\displaystyle CD have slope 0; BC\displaystyle BC and AD\displaystyle AD have slope 3/2\displaystyle 3/2
  • 3 points: The midpoints of AC and BD are both (3,3/2)\displaystyle (3,3/2), so the diagonals bisect each other.
  • 3 points: Adjacent side lengths are 4\displaystyle 4 and 13\displaystyle \sqrt{13}, so it is not a rhombus
Review this unit

Problem 5 · Circles and Conic Sections · 9 points

Find a tangent to a circle

A circle has equation x2+y2−4x+6y−12=0\displaystyle x^2+y^2-4x+6y-12=0.

Problem diagram
-6-4-2246810-8-6-4-22xycenterP
  1. A. Complete the square to find its center and radius.
  2. B. Show that P = (5, 1) is on the circle.
  3. C. Find the tangent line at P and justify perpendicularity.
Show worked solution and rubric
  1. A. (x−2)2+(y+3)2=25\displaystyle (x-2)^2+(y+3)^2=25, so the center is (2,−3)\displaystyle (2,-3) and radius is 5.
  2. B. The displacement from the center to P is (3,4)\displaystyle (3,4), with squared length 9+16=25\displaystyle 9+16=25.
  3. C. The radius slope is 4/3\displaystyle 4/3, so the tangent slope is −3/4\displaystyle -3/4. Its equation is y−1=−34(x−5)\displaystyle y-1=-\frac34(x-5). The product of the slopes is −1\displaystyle -1.
Solution diagram
-6-4-2246810-8-6-4-22xycenterP

Scoring rubric

  • 3 points: (x−2)2+(y+3)2=25\displaystyle (x-2)^2+(y+3)^2=25, so the center is (2,−3)\displaystyle (2,-3) and radius is 5.
  • 3 points: The displacement from the center to P is (3,4)\displaystyle (3,4), with squared length 9+16=25\displaystyle 9+16=25.
  • 3 points: The radius slope is 4/3\displaystyle 4/3, so the tangent slope is −3/4\displaystyle -3/4
Review this unit

Problem 6 · Transformations and Coordinate Geometry · 9 points

Compose two rigid motions

Triangle ABC has A=(1,1),B=(4,1),C=(1,3)\displaystyle A=(1,1),B=(4,1),C=(1,3). Reflect across the y-axis, then translate by vector (2, -1).

  1. A. Write the combined coordinate rule.
  2. B. Find the image of each vertex and compare side lengths.
  3. C. Explain what happens to orientation and area.
Show worked solution and rubric
  1. A. The rule is (x,y)↦(−x+2,y−1)\displaystyle (x,y)\mapsto(-x+2,y-1). The reflection happens before the translation.
  2. B. A′=(1,0)\displaystyle A\prime=(1,0), B′=(−2,0)\displaystyle B\prime=(-2,0), C′=(1,2)\displaystyle C\prime=(1,2). The original legs have lengths 3 and 2; their images also do, and the hypotenuse remains 13\displaystyle \sqrt{13}.
  3. C. One reflection reverses orientation; translation preserves it. Both are rigid motions, so area remains 12(3)(2)=3\displaystyle \frac12(3)(2)=3 square units.

Scoring rubric

  • 3 points: The rule is (x,y)↦(−x+2,y−1)\displaystyle (x,y)\mapsto(-x+2,y-1)
  • 3 points: A′=(1,0)\displaystyle A\prime=(1,0), B′=(−2,0)\displaystyle B\prime=(-2,0), C′=(1,2)\displaystyle C\prime=(1,2)
  • 3 points: One reflection reverses orientation; translation preserves it
Review this unit

Problem 7 · Solid Geometry and Geometric Systems · 9 points

Compare a cylinder and cone

A cylinder and a cone both have radius 3 cm and height 8 cm.

  1. A. Find both volumes and their ratio.
  2. B. Find the total exterior surface area of the cone, including its base.
  3. C. If every length is doubled, explain how volume and surface area change.
Show worked solution and rubric
  1. A. Vcyl=πr2h=72π\displaystyle V_{cyl}=\pi r^2h=72\pi cubic cm and Vcone=13πr2h=24π\displaystyle V_{cone}=\frac13\pi r^2h=24\pi cubic cm. The cylinder-to-cone volume ratio is 3 to 1.
  2. B. The slant height is ℓ=32+82=73\displaystyle \ell=\sqrt{3^2+8^2}=\sqrt{73}. Total cone area is πrℓ+πr2=3π73+9π\displaystyle \pi r\ell+\pi r^2=3\pi\sqrt{73}+9\pi square cm.
  3. C. Doubling all dimensions multiplies volumes by 23=8\displaystyle 2^3=8 and surface areas by 22=4\displaystyle 2^2=4. The cone-to-cylinder volume ratio remains unchanged.

Scoring rubric

  • 3 points: Vcyl=πr2h=72π\displaystyle V_{cyl}=\pi r^2h=72\pi cubic cm and Vcone=13πr2h=24π\displaystyle V_{cone}=\frac13\pi r^2h=24\pi cubic cm
  • 3 points: The slant height is ℓ=32+82=73\displaystyle \ell=\sqrt{3^2+8^2}=\sqrt{73}
  • 3 points: Doubling all dimensions multiplies volumes by 23=8\displaystyle 2^3=8 and surface areas by 22=4\displaystyle 2^2=4
Review this unit

Need help with this topic?

Work through questions with a tutor and get help with the steps you find difficult.

Get help with this topic