Download presentation

Presentation is loading. Please wait.

Published byJulien Amason Modified about 1 year ago

1
© www.teachitmaths.co.uk 2012 16973 Circle theorems workout 1

2
© www.teachitmaths.co.uk 2012 16973 Label the circle 2 radius diameter chord segment tangent sector circumference ? ? ? ? ? ? ? ? ? ? ? ? ? ? Click the rectangles to reveal each answer.

3
© www.teachitmaths.co.uk 2012 16973 Explain what we mean by ‘subtended’. 3 The blue angle is subtended by the red line. It is formed by lines starting from either end of the line. You can use the diagram to help.

4
© www.teachitmaths.co.uk 2012 16973 Complete the circle theorem. Angles in the same segment... 4... are equal.... sum to 180°.... sum to 360°. Click on the correct answer.

5
© www.teachitmaths.co.uk 2012 16973 Angles in the same segment are equal. 5 a = 26° b = 12° e = 49° c = 39° d = 39° ? ? 12° 26° a° b° 82° 38° 77° d° e° c° Find the lettered angles. Click on to reveal the answer. ? ? ? ? ? ?

6
© www.teachitmaths.co.uk 2012 16973 Complete the circle theorem. The angle subtended at the centre is... 6 half equal to twice ... the angle subtended at the circumference. Click on the correct answer.

7
© www.teachitmaths.co.uk 2012 16973 The angle subtended at the centre is twice the angle subtended at the circumference. 7 ? ? ? ? ? ? ? ? Find the lettered angles. Click on to reveal the answer. 90° a° 12° b° 53° c° a = 45°b = 78°c = 37°

8
© www.teachitmaths.co.uk 2012 16973 Complete the circle theorem. Opposite corners of a cyclic quadrilateral... 8 ... are equal.... sum to 180°.... sum to 360°. a b a + b = 180° Click on the correct answer.

9
© www.teachitmaths.co.uk 2012 16973 Opposite corners of a cyclic quadrilateral sum to 180°. 9 ? ? ? ? ? ? ? ? Find the lettered angles. Click on to reveal the answer. 127° 60° a° b° e° 42° d° 101° 114° c° a = 120° b = 53° c = 101° d = 114° e = 126°

10
© www.teachitmaths.co.uk 2012 16973 Complete the circle theorem. An angle subtended at the circumference by a diameter... 10 ... is acute.... is obtuse.... is 90°. Click on the correct answer.

11
© www.teachitmaths.co.uk 2012 16973 An angle subtended at the circumference by a diameter is 90°. 11 ? ? ? ? ? ? ? ? Find the lettered angles. Click on to reveal the answer. a° 28° 35° 28° b° c° 10° 35° e° d° a = 62°b = 62° c = 55° d = 55° e = 25°

12
© www.teachitmaths.co.uk 2012 16973 Complete the circle theorem. The angle between a chord and... 12 a tangent a diameter... is... twice equal to... the angle subtended by the chord in the opposite segment. Click on the correct answer.

13
© www.teachitmaths.co.uk 2012 16973 The angle between a chord and a tangent is equal to the angle subtended by the chord in the opposite segment. 13 ? ? ? ? ? ? ? ? Find the lettered angles. Click on to reveal the answer. a° 67° 41° b° c° d° 39° e° 64° a = 41° b = 67° c = 90° d = 51° e = 58°

14
© www.teachitmaths.co.uk 2012 16973 Exam style question AC is the diameter of a circle and B is a point on the circumference of the circle. Find angle x. 14 ? ? 110 ° 30° C x°x° D y y = 180 – 110 – 30 y = 40° (angles in a triangle) x = 90 – y x = 50° (angles subtended by the diameter)

15
© www.teachitmaths.co.uk 2012 16973 Exam style question The diagram shows a circle, centre O. PQ is a tangent to the circle at C. Angle PCA = 57° Angle AOB = 98° Calculate the size of angle OBC. 15 ? ? ABC = 57° (alternate segment theorem) ABO = ½(180 – 98) ABO = 41° (angles in isosceles triangle) OBC = 57 – 41 OBC = 16° B C A P Q O 98 ° 57 °

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google