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High School Mathematics - 2
8.4 Theorems - Circles

Theorem: Angle is a semi circle is a right angle.

Given : Angle AOB is in semi circle.
To prove: Angle ACB = 90o
Proof: Angle ACB = 1/2 angle AOB (angle subtended at the centre is twice angle at circumference)
Angle AOB = 180o (straight line) Hence ACB = 90o

Theorem: Angles in the same segment of a circle are equal.


Directions: Solve the following

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Date:___________________

High School Mathematics - 2
8.4 Theorems - Circles

Q 1: Find the angle x.

90 degrees
17.5 degrees
70 degrees

Q 2: find x.

90 degrees
55 degrees
110 degrees

Q 3: Find x.

140 degrees
70 degrees
35 degrees

Q 4: Two circles intersect each other at points A and B. If AP and BQ be the respective diameters, show that PBQ is a straight line.
Answer:

Q 5: If diagonals AC and BD of a quadrilateral intersect at M. Prove that a line drawn through M to bisect any side of the quadrilateral is perpendicular to the opposite side.
Answer:

Q 6: In the figure, CAD and CBE are straight lines. If CA is the diameter of the circle ABC, find angle ADE

180 degrees
90 degrees
45 degrees

Q 7: Angles in a semicircle is always ____.
90 degrees
180 degrees
270 degrees

Q 8: In the figure, PAQ and RAS are straight lines. Show that angPXR = angQYS.

Answer:

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Question 10: This question is available to subscribers only!


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