Geometry Quiz: Constructing Inscribed And Circumscribed Circles
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Constructing Inscribed And Circumscribed CirclesQuestion 1 of 10

Triangle PQRPQR is inscribed in a circle with center OO and radius 10. If arc PQPQ measures 120°120° and arc QRQR measures 100°100°, what is the measure of QPR\angle QPR?

70°70°
50°50°
60°60°
40°40°
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Geometry Quiz

Geometry Quiz: Constructing Inscribed And Circumscribed Circles

Practice Constructing Inscribed And Circumscribed Circles in Geometry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Constructing Inscribed And Circumscribed Circles, giving you a quick way to practice the rules, question types, and explanations that matter most for Geometry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Triangle PQRPQR is inscribed in a circle with center OO and radius 10. If arc PQPQ measures 120°120° and arc QRQR measures 100°100°, what is the measure of QPR\angle QPR?

  1. 70°70°
  2. 50°50° (correct answer)
  3. 60°60°
  4. 40°40°
Explanation: An inscribed angle measures half the central angle that subtends the same arc. QPR\angle QPR is an inscribed angle that intercepts arc QRQR. Since arc QRQR measures 100°100°, QPR=100°2=50°\angle QPR = \frac{100°}{2} = 50°. Choice A incorrectly uses arc PRPR (which measures 360°120°100°=140°360° - 120° - 100° = 140°, giving 70°70°). Choice C incorrectly uses arc PQPQ (120°/2=60°120°/2 = 60°). Choice D uses an incorrect calculation or wrong arc measurement.

Question 2

Refer to the figure below. Quadrilateral ABCDABCD is inscribed in a circle. If mA=85m\angle A = 85^\circ and mB=74m\angle B = 74^\circ, what is mDm\angle D?

  1. 9595^\circ
  2. 106106^\circ (correct answer)
  3. 116116^\circ
  4. 7474^\circ
Explanation: In a cyclic quadrilateral, opposite angles are supplementary. Angles BB and DD are opposite, so mD=18074=106m\angle D = 180^\circ - 74^\circ = 106^\circ. Choice A would be mCm\angle C (opposite to A\angle A). Choice C incorrectly adds instead of subtracting. Choice D is the value of B\angle B, chosen by mistakenly making opposite angles equal.

Question 3

In the figure, triangle ABCABC is inscribed in circle OO, and BAC=35¬\angle BAC = 35¬∞. Point DD is on the circle such that BDC\angle BDC and BAC\angle BAC intercept the same arc. What is the relationship between BDC\angle BDC and BAC\angle BAC?

  1. BDC=BAC=35¬\angle BDC = \angle BAC = 35¬∞ because inscribed angles intercepting the same arc are equal (correct answer)
  2. BDC=2BAC=70¬\angle BDC = 2\angle BAC = 70¬∞ because DD is on the opposite side of the arc
  3. BDC=180¬BAC=145¬\angle BDC = 180¬∞ - \angle BAC = 145¬∞ because they form a cyclic quadrilateral
  4. BDC=90¬BAC=55¬\angle BDC = 90¬∞ - \angle BAC = 55¬∞ because of the inscribed angle theorem
Explanation: The inscribed angle theorem states that inscribed angles intercepting the same arc are equal, regardless of where the vertex lies on the circle (as long as it's on the same side of the chord). Since both BAC\angle BAC and BDC\angle BDC intercept arc BCBC, they are equal: BDC=BAC=35¬\angle BDC = \angle BAC = 35¬∞. Choice B incorrectly applies a central angle relationship. Choice C confuses this with supplementary angles in a cyclic quadrilateral (which would apply if ABDCABDC formed a quadrilateral). Choice D incorrectly applies a right triangle relationship.

Question 4

In the figure, triangle ABCABC has its incircle touching side BCBC at point DD, side ACAC at point EE, and side ABAB at point FF. If AB=13AB = 13, BC=14BC = 14, and AC=15AC = 15, what is the length of BDBD?

  1. 66 (correct answer)
  2. 77
  3. 88
  4. 55
Explanation: When an incircle touches the sides of a triangle, the tangent segments from each vertex to the points of tangency are equal. Let AF=AE=xAF = AE = x, BF=BD=yBF = BD = y, and CD=CE=zCD = CE = z. Then AB=x+y=13AB = x + y = 13, BC=y+z=14BC = y + z = 14, and AC=x+z=15AC = x + z = 15. Solving this system: Adding all three equations gives 2(x+y+z)=422(x + y + z) = 42, so x+y+z=21x + y + z = 21. Therefore z=2113=8z = 21 - 13 = 8, x=2114=7x = 21 - 14 = 7, and y=2115=6y = 21 - 15 = 6. So BD=y=6BD = y = 6. Choice B gives the value of x. Choice C gives the value of z. Choice D is an arithmetic error.

Question 5

The incircle of triangle MNPMNP has radius 4 and center II. If the triangle's semiperimeter is 15, what is the area of triangle MNPMNP?

  1. 4545
  2. 3030
  3. 6060 (correct answer)
  4. 7575
Explanation: When you encounter problems involving an incircle (the circle inscribed within a triangle), you're dealing with one of geometry's most elegant relationships. The key insight is that the incircle's radius connects directly to the triangle's area and perimeter. The fundamental formula for any triangle with an incircle is: Area = radius × semiperimeter. This relationship exists because the incircle's center connects to all three sides, creating three smaller triangles whose combined area equals the original triangle. Given an inradius of 4 and semiperimeter of 15, we can calculate: Area = 4×15=604 \times 15 = 60. This confirms answer choice C is correct. Let's examine why the other answers represent common mistakes. Answer A (45) likely comes from incorrectly using the formula Area = 12×base×height\frac{1}{2} \times \text{base} \times \text{height} and assuming the radius somehow represents half the height. Answer B (30) suggests someone may have confused the semiperimeter with the full perimeter, calculating 4×7.5=304 \times 7.5 = 30. Answer D (75) might result from adding the radius and semiperimeter instead of multiplying: 4+15=194 + 15 = 19, though this doesn't directly yield 75, it represents the type of operational error students sometimes make under pressure. Remember this powerful relationship: whenever you see an incircle problem with given radius and semiperimeter, immediately think "Area = r × s." This formula works for any triangle and often provides the most direct path to the solution, bypassing more complex approaches involving side lengths or angles.

Question 6

A regular hexagon is inscribed in a circle of radius 8. What is the area of the region between the circle and the hexagon?

  1. 64π96364\pi - 96\sqrt{3} (correct answer)
  2. 64π48364\pi - 48\sqrt{3}
  3. 32π96332\pi - 96\sqrt{3}
  4. 64π144364\pi - 144\sqrt{3}
Explanation: The area between the circle and hexagon equals the circle's area minus the hexagon's area. Circle area = πr2=64π\pi r^2 = 64\pi. A regular hexagon inscribed in a circle of radius rr has area 332r2\frac{3\sqrt{3}}{2}r^2. With r=8r = 8, hexagon area = 33264=963\frac{3\sqrt{3}}{2} \cdot 64 = 96\sqrt{3}. Therefore, the area between = 64π96364\pi - 96\sqrt{3}. Choice B uses an incorrect hexagon area formula (missing factor of 2). Choice C uses wrong circle area (missing factor of 2). Choice D uses an incorrect hexagon area calculation (extra factor of 1.5).

Question 7

A quadrilateral PQRSPQRS is inscribed in a circle. If P=3x+20°\angle P = 3x + 20°, Q=2x10°\angle Q = 2x - 10°, and R=x+40°\angle R = x + 40°, what is the measure of S\angle S?

  1. 70°70°
  2. 110°110° (correct answer)
  3. 90°90°
  4. 80°80°
Explanation: In a cyclic quadrilateral, opposite angles are supplementary. From P+R=180°\angle P + \angle R = 180°: (3x+20)+(x+40)=180°(3x + 20) + (x + 40) = 180°, so 4x+60=180°4x + 60 = 180°, giving x=30°x = 30°. Therefore Q=2(30)10=50°\angle Q = 2(30) - 10 = 50°. Since Q+S=180°\angle Q + \angle S = 180°, we have S=180°50°=130°\angle S = 180° - 50° = 130°. Wait, this doesn't match the choices. Let me recalculate: Q=2(30)10=70°\angle Q = 2(30) - 10 = 70°, so S=180°70°=110°\angle S = 180° - 70° = 110°. Choice A gives angle Q's measure. Choice C assumes a right angle. Choice D uses an arithmetic error.

Question 8

The circumcenter of triangle ABCABC is equidistant from all three vertices. If the circumcenter lies outside the triangle, what can be concluded about triangle ABCABC?

  1. Triangle ABCABC must be an acute triangle with all angles less than 60°60°
  2. Triangle ABCABC must be an isosceles triangle with two equal sides
  3. Triangle ABCABC must be a right triangle with one angle equal to 90°90°
  4. Triangle ABCABC must be an obtuse triangle with one angle greater than 90°90° (correct answer)
Explanation: When you encounter questions about the circumcenter's location, remember that its position relative to the triangle reveals crucial information about the triangle's angles. The circumcenter is the point where the perpendicular bisectors of all three sides intersect, and it's equidistant from all vertices. The key insight is understanding how the circumcenter's location relates to the triangle's largest angle. In an obtuse triangle, the circumcenter always lies outside the triangle, specifically on the opposite side of the longest side from the obtuse angle. Here's why: when one angle exceeds 90°90°, the perpendicular bisectors of the sides forming that angle must intersect outside the triangle to maintain equal distances to all three vertices. This geometric constraint forces the circumcenter outward. Option A is incorrect because in acute triangles (where all angles are less than 90°90°), the circumcenter lies inside the triangle. The specific constraint about angles being less than 60°60° is irrelevant to circumcenter location. Option B is wrong because a triangle's symmetry properties don't determine circumcenter location. Both isosceles and scalene triangles can have circumcenters inside or outside, depending on their angles. Option C is incorrect because in right triangles, the circumcenter lies exactly on the hypotenuse (the longest side), not outside the triangle. The right angle creates a special case where the circumcenter sits on the triangle's boundary. Study tip: Remember the circumcenter location rule: inside for acute triangles, on the hypotenuse for right triangles, and outside for obtuse triangles. This pattern appears frequently in geometry problems involving triangle centers.

Question 9

In quadrilateral WXYZWXYZ inscribed in a circle, W=2Y\angle W = 2\angle Y and X=Z+30°\angle X = \angle Z + 30°. What is the measure of Y\angle Y?

  1. 45°45°
  2. 60°60°
  3. 50°50° (correct answer)
  4. 40°40°
Explanation: When you encounter a cyclic quadrilateral (a quadrilateral inscribed in a circle), remember that opposite angles are supplementary—they add up to 180°180°. This property is the key to solving this problem. Let's set up the relationships systematically. If we call Y=x\angle Y = x, then W=2x\angle W = 2x (given). Since opposite angles are supplementary: W+Y=180°\angle W + \angle Y = 180°, so 2x+x=180°2x + x = 180°, which gives us 3x=180°3x = 180° and x=60°x = 60°. Let's verify this works with the other condition. We found Y=60°\angle Y = 60°, so W=120°\angle W = 120°. For the other pair of opposite angles, let Z=y\angle Z = y. Then X=y+30°\angle X = y + 30° (given). Since X+Z=180°\angle X + \angle Z = 180°: (y+30°)+y=180°(y + 30°) + y = 180°, so 2y=150°2y = 150° and y=75°y = 75°. This means Z=75°\angle Z = 75° and X=105°\angle X = 105°. Checking: 120°+60°=180°120° + 60° = 180° ✓ and 105°+75°=180°105° + 75° = 180° Choice A (45°45°) would make W=90°\angle W = 90°, giving a sum of 135°135° instead of 180°180°. Choice B (60°60°) is our correct answer. Choice D (40°40°) would make W=80°\angle W = 80°, giving a sum of 120°120°. Study tip: For cyclic quadrilaterals, always start by using the opposite angles property (they sum to 180°180°). Set up your equations systematically using this relationship—it's your most powerful tool for these problems.

Question 10

In triangle ABCABC, the circumcenter lies on side BCBC. If AB=5AB = 5 and AC=12AC = 12, what is the radius of the circumscribed circle?

  1. 6.56.5 (correct answer)
  2. 8.58.5
  3. 1313
  4. 66
Explanation: When the circumcenter lies on a side of the triangle, the triangle is a right triangle with that side as the hypotenuse. Since the circumcenter lies on BC, angle A is 90°. By the Pythagorean theorem, BC=52+122=169=13BC = \sqrt{5^2 + 12^2} = \sqrt{169} = 13. The circumradius of a right triangle equals half the hypotenuse, so R=13/2=6.5R = 13/2 = 6.5. Choice B uses an incorrect formula. Choice C gives the hypotenuse length instead of the radius. Choice D incorrectly uses half of one leg.