Geometry Worksheet 1 – Angles and Circle Theorems 1. [Source: SMC] ABCDEFGH is a regular octagon. P is the point inside the octagon such that triangle ABP is equilateral. What is the size of angle APC? A 90° B 112.5° C 117.5° D 120° E 135° 2. If two chords AB and CD intersect at a point X, prove that π΄π × ππ΅ = πΆπ × ππ· (i.e. prove the Intersecting Chord Theorem). 3. [Source: SMC] The size of each exterior angle of a regular polygon is one quarter of the size of an interior angle. How many sides does the polygon have? A 6 B 8 C 9 D 10 E 12 4. [Source: SMC] In the diagram AB, CB, and XY are tangents to the circle with centre O and angle ABC = 48°. What is the size of angle XOY? C X B 48° O Y A A 42° B 69° C 66° D 48° E 84° 5. [Source: SMC] In the figure shown, what is the sum of the interior angles at A, B, C, D, E? A 90° B 135° C 150° D 180° E more information required. E A B D C 6. [Source: SMC] A point π is chosen inside a square π΄π΅πΆπ·. What is the probability that ∠π΄ππ· is acute? (Hint: your expression will involve π) www.drfrostmaths.com/rzc 7. The largest circle which it is possible to draw inside triangle πππ touches the triangle at π, π and π, as shown in the diagram. The size of ∠πππ = 55°. What is the size of ∠πππ ? π π π π 55° π π 8. [Source: SMC] In the diagram, π is the centre of the circle, ∠π΄ππ΅ = πΌ and ∠πΆππ· = π½. What is the size of ∠π΄ππ΅ in terms of πΌ and π½? A D 1 πΌ 2 1 − 2π½ 180° − πΌ − π½ B E 1 1 90° − 2 πΌ − 2 π½ More information needed. C πΌ−π½ 9. [Source: UKMT Mentoring] If a π-sided polygon has exactly 3 obtuse angles (i.e. 90° < π < 180°), then determine the possible values of π (Hint: determine the possible range for the sum of the interior angles, and use these inequalities to solve). www.drfrostmaths.com/rzc Geometry Worksheet 1 - ANSWERS 1. ABCDEFGH is a regular octagon. P is the point inside the octagon such that triangle ABP is equilateral. What is the size of angle APC? C P B A B. ΔAPB is equilateral hence οABP = 60° But BC = AB = PB so ΔPBC is isosceles with οPBC ο½ 135 ο° – 60ο° ο½ 75ο° . 1 ο οBPC ο½ (180 – 75)ο° and οAPC ο½ 52.5ο° ο« 60ο° . 2 2. If two chords AB and CD intersect at a point X, prove that π¨πΏ × πΏπ© = πͺπΏ × πΏπ« (i.e. prove the Intersecting Chord Theorem). Clearly ∠π΄ππ· = ∠πΆππ΅ (opposite angles are equal). Also, ∠π·π΄π΅ = ∠π΅πΆπ· (angles are in π΄π πΆπ same segment are equal). We thus have two similar triangles, and hence ππ· = ππ΅, leading to the desired result. 3. The size of each exterior angle of a regular polygon is one quarter of the size of an interior angle. How many sides does the polygon have? D. Let the exterior angle be x°. Then x + 4x = 180 and therefore x = 36. As is the case in all convex polygons, the sum of the exterior angles = 360° and therefore the number of sides = 360/36 = 10. 4. C. Let the points of contacts of the tangents be P Q and R as shown and let οXOQ ο½ x and οQOY ο½ y . Then since OX is the axis of symmetry of the tangent kite OPXQ, it bisects οPOQ so οXOP ο½ x . P C X 48° O x y A R B Q Y Similarly οROY ο½ y . Thus, in the quadrilateral OPBR we have 2x + 2y + 2 × 90 o + 48 o = 360 o i.e. 2 (x + y) = 132 o i.e. x + y = 66 o www.drfrostmaths.com/rzc 5. A E X e y a x Y b B D d c C Let a, b, c, d, e, x and y represent the sizes in degrees of certain angles in the figure, as shown and let the points of intersection of AD with EB and EC be X and Y respectively. Angle EXY is an exterior angle of triangle XBD so x = b + d. Similarly, angle EYX is an exterior angle of triangle YAC so y = a + c. In triangle EXY, e + x + y = 180, so a + b + c + d + e = 180. 6. A point π· is chosen inside a square π¨π©πͺπ«. What is the probability that ∠π¨π·π« is acute? Observe the below diagram: A D B C If the chosen point was on the semi-circular arc, then ∠π΄ππ· = 90° (angle on circumference from diameter of a circle is 90). We see can see that if the point is inside the semicircle, the angle is obtuse, and acute outside it. If we let the radius of the semicircle be 1, then its area is π/2 and the area of the square 4. The proportion of the square that is inside the semicircle is therefore π/8, and thus the proportion outside it (where we’ll have an acute π angle) 1 − 8 . 7. The largest circle which it is possible to draw inside triangle π·πΈπΉ touches the triangle at πΊ, π» and πΌ, as shown in the diagram. The size of ∠πΊπ»πΌ = ππ°. What is the size of ∠π·πΈπΉ? By the Alternate Segment Theorem, ∠πππ = 55°, as is ∠πππ. Thus ∠πππ = 180° − (2 × 55°) = 70°. 1 8. ∠π΄πΆπ΅ = 2 πΌ (angle subtended by an arc at the centre of a circle is twice the angle 1 1 1 subtended at the circumference) and, similarly, ∠πΆπ΄π· = 2 π½. Therefore ∠π΄ππ΅ = 2 πΌ − 2 π½ (the exterior angle of a triangle is equal to the sum of the two interior opposite angles). 9. [Source: UKMT Mentoring] If a π-sided polygon has exactly 3 obtuse angles (i.e. ππ° < π < 180°), then determine the possible values of π (Hint: determine the possible range for the sum of the interior angles, and use these inequalities to solve). If 3 angles are obtuse, the sum of these, say π, has the range 270 < π < 540. For the π − 3 angles that are not obtuse (i.e. acute or right-angled), then the sum π΄ has the range: 0 < π΄ ≤ 90(π − 3). The total of the interior angles is 180(π − 2), so 270 < 180(π − 2) < 540 + 90(π − 3) Solving 270 < 180(π − 2), we get π > 3.5 and solving 180(π − 2) < 540 + 90(π − 3), we get π < 7. Thus π = 4, 5 ππ 6. www.drfrostmaths.com/rzc