Printable · GCSE Higher · ages 14-16
Circle theorems worksheet — GCSE Higher
Fifteen questions on "circle theorems" — DfE statement G10. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
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Answer key: Circle theorems worksheet — GCSE Higher
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- (b) 35° — Method: angle CBE and angle ABC lie on a straight line at B, so they are supplementary; angle ABC then equals angle ADC because B and D are both on the major arc AC (angles in the same segment). Working: angle ABC = 180 − 145 = 35 degrees, using angles on a straight line. Since B and D are both on the major arc AC, angle ADC = angle ABC = 35°. Answer: 35°. Find angle ABC FIRST from the straight line at B before applying the circle theorem: using the given 145° directly, doubling or halving it, or subtracting it from 180° a second time all give the wrong angle.
- (a) 71° — Method: first use the straight line through T to find the tangent-chord angle RTC, then apply the alternate segment theorem, which states that this angle equals the angle subtended by the chord in the alternate segment. Working: the two rails form a straight line through T, so angle RTC = 180 − 109 = 71 degrees. D lies in the alternate segment of the chord TC, so angle TDC = angle RTC = 71°. Answer: 71°. Find angle RTC FIRST from the straight line before applying the theorem: using the given 109° directly, doubling the tangent-chord angle, or taking its complement from 90° all give the wrong angle at D.
- (b) 160° — B is on the minor arc AC, so the angle at the centre theorem applies to the reflex angle AOC: reflex angle AOC = 2 × 100° = 200°. The non-reflex angle AOC is the rest of the full turn: 360° − 200° = 160°. Reporting the reflex angle itself, without subtracting it from 360°, gives 200°. Subtracting angle ABC from 180° instead, as if this were a cyclic quadrilateral, gives 80°. Halving angle ABC instead of doubling it gives 50°. Double first, then take the angle away from a full turn, and 160° is what's left.
- (a) 104° — Method: find each inscribed angle separately using the angle-at-the-centre theorem: for a point on the arc NOT cut off by the given central angle, halve that central angle; for a point on the OTHER arc, halve the REFLEX central angle instead; then subtract the smaller from the larger. Working: for C on the major arc, angle ACB = 76 ÷ 2 = 38 degrees. For D on the minor arc, D sees the reflex angle at the centre, 360 − 76 = 284 degrees, so angle ADB = 284 ÷ 2 = 142 degrees. The difference is 142 − 38 = 104 degrees. Answer: 104°. Both inscribed angles need the theorem applied separately, using the correct arc's central angle each time (the reflex angle for D), and the question asks for the DIFFERENCE between the two, not either angle on its own and not their sum, 180°, which is simply the opposite-angle total for the cyclic quadrilateral ACBD.
- (a) 24 cm — The line from the centre to the midpoint of a chord is perpendicular to the chord, so triangle OMA has a right angle at M. By Pythagoras' theorem, AM² = OA² − OM² = 169 − 25 = 144, so AM = 12 cm. AB is twice AM, since M is the midpoint: AB = 2 × 12 = 24 cm. Finding AM = 12 cm correctly but forgetting to double it for the full chord gives 12 cm. Working out 169 − 25 = 144 and forgetting to take the square root gives 144 cm. Subtracting first and then doubling the wrong way, (13 − 5) × 2, gives 16 cm. Doubling AM after finding it correctly is the step that's missing from all three — do it, and you get 24 cm.
- (b) 2x + 2y = 180° — Method: use the fact that OA, OB and OC are all radii, so triangles OAC and OBC are each isosceles, and set up the angle sum of triangle ABC as a whole. Working: since OA = OC, triangle OAC is isosceles, so angle OAC = angle OCA = x. Since OB = OC, triangle OBC is isosceles, so angle OBC = angle OCB = y. The angle at A in triangle ABC is x, the angle at B is y, and the angle at C is angle OCA plus angle OCB, which is x + y. The three angles of triangle ABC sum to 180°, giving x + y + (x + y) = 180°, which is 2x + 2y = 180°. Answer: 2x + 2y = 180°. Dividing this by 2 gives x + y = 90°, and since angle ACB = x + y, this proves angle ACB = 90°: the point C was never restricted to any particular position on the circle, so the proof holds for every choice of C. Include ALL THREE angles of the triangle in the sum, do not use a quadrilateral's 360° total, and remember angle ACB is the SUM x + y, not x alone.
- (b) 30 cm — The perpendicular from the centre of a circle to a chord bisects the chord, so this line, half the chord and the radius form a right-angled triangle. Using Pythagoras' theorem, half the chord = √(17² − 8²) = √(289 − 64) = √225 = 15 cm. The full chord AB is twice this length: AB = 2 × 15 = 30 cm. Stopping after finding the half-chord, without doubling it for the whole chord, gives 15 cm. Adding the radius and the perpendicular distance directly, 17 + 8 = 25 cm, ignores that these two lengths are the two shorter sides of a right-angled triangle, not parts of a straight line. Subtracting instead, 17 − 8 = 9 cm, makes the same mistake in the other direction.
- (b) 80° — Method: opposite angles of a cyclic quadrilateral sum to 180°, so form and solve an equation for x, then substitute back to find angle DAB. Working: (2x + 10) + (3x − 5) = 180, so 5x + 5 = 180, x = 35, and angle DAB = 2(35) + 10 = 80°. Using 360° instead of 180° as the total (as if the two angles were the whole circle rather than a pair of opposite angles) gives x = 71 and angle DAB = 152°; solving correctly for x but reporting x itself instead of substituting it back into 2x + 10 gives 35°; and mishandling the signs when combining the two expressions, using (2x + 10) + (3x + 5) = 180 instead of (2x + 10) + (3x − 5) = 180, gives x = 33 and angle DAB = 76°. Opposite angles of a cyclic quadrilateral — not adjacent ones — are the pair that sums to 180°.
- (b) 58° — Method: the angle at the centre is TWICE the angle at the circumference when both stand on the same arc. Working: angle AOC is the angle at the centre standing on arc AC, and angle ABC is the angle at the circumference standing on the same arc AC, with B on the major arc, so angle AOC = 2 × angle ABC. Rearranging, angle ABC = 116 ÷ 2 = 58 degrees. Answer: 58°. Halve the centre angle, do not double it, and use the angle AOC exactly as given, 116°: you do not need its reflex angle here, because B sits on the major arc, which is precisely the arrangement the theorem is stated for.
- (c) Tangent-chord angle = angle in the alternate segment. — This diagram has a diameter AC, a centre O, and a cyclic quadrilateral ABCD, but no tangent anywhere in it. 'Angle in a semicircle = 90°' applies directly, because AC is a diameter. 'Opposite angles of a cyclic quadrilateral sum to 180°' applies directly, because ABCD is a cyclic quadrilateral. 'Angle at the centre = twice angle at circumference' applies directly, because O is given as the centre of the circle. The tangent-chord fact relates the angle between a tangent and a chord to an angle elsewhere in the circle, and since this diagram has no tangent, there is nothing in it for that fact to describe — so it is the one theorem that does not apply here.
- (b) 63° — Method: a tangent meets the radius drawn to the point of contact at a right angle, so triangle OPT has a 90° angle at T; the three angles of the triangle then sum to 180°. Working: angle OTP = 90°, angle OPT = 27°, so angle POT = 180 − 90 − 27 = 63 degrees. Answer: 63°. The tangent-radius angle is a fixed 90°, not something to assume equal to another angle in the triangle, and the three angles of ANY triangle sum to 180°, never 360°: that total belongs to a quadrilateral, not a triangle.
- (a) 90° — Method: the angle in a semicircle theorem states that the angle subtended at the circumference by a diameter is always a right angle, whatever the circle's size. Working: since AC is a diameter, the angle ABC subtended at the circumference by AC is 90 degrees, regardless of the circle's radius. Answer: 90°. This result holds for every position of B on the circle and for every size of circle, so no length or radius is ever needed: do not confuse angle ABC with the straight angle along the diameter itself, which is 180°, and do not halve the right angle again.
- (a) 65° — OT and OC are both radii, so triangle OTC is isosceles with OT = OC, and its base angles are equal: angle OTC = angle OCT = (180° − 130°) ÷ 2 = 25°. A tangent is perpendicular to the radius at the point of contact, so angle OTP = 90°. Since C lies inside angle OTP, the chord TC splits this right angle into angle OTC and angle PTC, so angle PTC = angle OTP − angle OTC = 90° − 25° = 65°. Stopping after finding the base angle of the isosceles triangle, without subtracting it from the right angle at T, leaves 25° instead of the angle actually asked for. Adding the base angle to the right angle instead of subtracting it, 90° + 25° = 115°, reverses the direction the two angles combine in. Finding the sum of the two base angles of the isosceles triangle, 180° − 130° = 50°, and stopping there, gives another wrong value entirely.
- (d) Opposite angles in a cyclic quadrilateral sum to 180° — ABCD is a cyclic quadrilateral, and angle ABC and angle ADC are a pair of opposite angles in it — they don't sit next to each other around the quadrilateral. The theorem that fixes the sum of a cyclic quadrilateral's opposite angles at 180° is exactly the one needed here. The other three theorems don't apply to this figure at all: the angle-in-a-semicircle theorem gives a 90° angle only when one side is a diameter, and no diameter is mentioned; the angle-at-the-centre theorem needs a centre point and an inscribed angle on the same arc, and no centre is given here; the tangent–radius theorem gives a 90° angle only where a tangent line touches the circle, and there is no tangent in this figure. Only the cyclic-quadrilateral theorem fits what is actually drawn.
- (a) 46° — Method: use the angle at the centre, then the cyclic quadrilateral, then the angle sum of a triangle. Working: angle ABC = 152° ÷ 2 = 76° (angle at the centre is twice the angle at the circumference); angle ADC = 180° − 76° = 104° (opposite angles of a cyclic quadrilateral sum to 180°); angle DAC = 180° − 104° − 30° = 46° (angle sum of triangle ACD). Skipping the halving step and taking angle ABC = 152° gives angle ADC = 180° − 152° = 28° and then angle DAC = 122°; using angle ABC = 76° directly as angle ADC, skipping the cyclic quadrilateral step, gives 74°; and finding angle DAC as 180° − 104° without subtracting the given 30° gives 76°. Each of the three steps must be carried through in order — halve, then subtract from 180°, then subtract from 180° again with the extra angle.
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