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CIRCLES CLASS 9 QUESTIONS FOR CBSE STUDENTS
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21. In the given figure, O is the centre of a circle and A, B, C, D and E are points on the circle such that AB = BC = CD = DE = EA. Find the value of < AOB.
Sol. Given: O is the center of the circle and AB = BC = CD= DE = EA.
Construction: Join OC, OD, OE
To find:
AOB
Proof: A, B, C, D and E are the points which lie on the circle.
Also AB = BC = CD = DE = EA
All are the chords of a circle
As we know that equal chords subtend equal angles at the center of the circle.
AOB = BOC = COD
= DOE = OE
Also, AOB + BOC + COD + DOE + EOA = 360°
(Sum of angles at the center of a circle)
Using (i) AOB + AOB + AOB = 360°
5 AOB = 360°
AOB = 72°
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22. Justify your statement
“The angles subtended by a chord at any two points of a circle are equal”
Sol. The angles subtended by a chord at any two points of a circle are equal if the two points lie in the same segment (larger or smaller), otherwise they are not equal.
23. Justify your statement
“Two chords of a circle of length 10 cm and 8 cm are at a distance of 8 cm and 3.5 cm respectively from the center”
Sol. The statement is not correct as the longer chord will be at a lesser distance from the centre.
24. In figure,
∠AOB = 90° and ∠ABC = 30° then ∠CAO is equal to
(a) 30°
(b) 45°
(c) 90°
(d) 60°
25. In the given figure, O is the center of the circle,
∠AOC = 50° and ∠COB = 30°. Find the measure of
ALSO VISIT : 1.quadrilaterals class 9
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