Download presentation

Presentation is loading. Please wait.

Published byJulien Amason Modified over 9 years 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

© 2024 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google