DrFrostMaths



Geometry Worksheet 1 – Angles and Circle Theorems[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?A90°B112.5° C117.5°D120°E135°If two chords AB and CD intersect at a point X, prove that AX×XB=CX×XD (i.e. prove the Intersecting Chord Theorem).[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?A6B8C9D10E12[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?A42°B69°C66°D48°E84°[Source: SMC] In the figure shown, what is the sum of the interior angles at A, B, C, D, E?A90°B135°C150°D180°Emore information required.[Source: SMC] A point P is chosen inside a square ABCD. What is the probability that ∠APD is acute? (Hint: your expression will involve π)QThe largest circle which it is possible to draw inside triangle PQR touches the triangle at S, T and U, as shown in the diagram. The size of ∠STU=55°. What is the size of ∠PQR?55°PTRSU[Source: SMC] In the diagram, O is the centre of the circle, ∠AOB=α and ∠COD=β. What is the size of ∠AXB in terms of α and β?A12α-12βB90°-12α-12βCα-βD180°-α-βEMore information needed.[Source: UKMT Mentoring] If a n-sided polygon has exactly 3 obtuse angles (i.e. 90°<θ<180°), then determine the possible values of n (Hint: determine the possible range for the sum of the interior angles, and use these inequalities to solve).Geometry Worksheet 1 - ANSWERSABCDEFGH is a regular octagon. P is the point inside the octagon such that triangle ABP is equilateral. What is the size of angle APC?B. ΔAPB is equilateral hence = 60°But BC = AB = PB so ΔPBC is isosceles with . and .If two chords AB and CD intersect at a point X, prove that AX×XB=CX×XD (i.e. prove the Intersecting Chord Theorem).Clearly ∠AXD=∠CXB (opposite angles are equal). Also, ∠DAB=∠BCD (angles are in same segment are equal). We thus have two similar triangles, and hence AXXD=CXXB, leading to the desired result.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.C. Let the points of contacts of the tangents be P Q and R as shown and let and . Then since OX is the axis of symmetry of the tangent kite OPXQ, it bisects so .Similarly . Thus, in the quadrilateral OPBRwe have2x + 2y + 2 × 90 + 48 = 360i.e. 2 (x + y) = 132i.e. x + y = 66Let a, b, c, d, e, x and y represent the sizes in degrees of certain angles in the figure, asshown 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 isan exterior angle of triangle YAC so y = a + c. In triangle EXY, e + x + y = 180, so a + b + c + d + e = 180.A point P is chosen inside a square ABCD. What is the probability that ∠APD is acute? Observe the below diagram:ABCDIf the chosen point was on the semi-circular arc, then ∠APD=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 angle) 1-π8. The largest circle which it is possible to draw inside triangle PQR touches the triangle at S, T and U, as shown in the diagram. The size of ∠STU=55°. What is the size of ∠PQR?By the Alternate Segment Theorem, ∠QUS=55°, as is ∠QSU. Thus ∠PQR=180°-2×55°=70°.∠ACB=12α (angle subtended by an arc at the centre of a circle is twice the angle subtended at the circumference) and, similarly, ∠CAD=12β. Therefore ∠AXB=12α-12β (the exterior angle of a triangle is equal to the sum of the two interior opposite angles).[Source: UKMT Mentoring] If a n-sided polygon has exactly 3 obtuse angles (i.e. 90°<θ<180°), then determine the possible values of n (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 O, has the range 270<O<540. For the n-3 angles that are not obtuse (i.e. acute or right-angled), then the sum A has the range: 0<A≤90(n-3). The total of the interior angles is 180(n-2), so 270<180n-2<540+90(n-3)Solving 270<180(n-2), we get n>3.5 and solving 180n-2<540+90(n-3), we get n<7. Thus n=4, 5 or 6. ................
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