Welcome to AssamSchool. This article contains 30 Extra Questions from Circles from Class 9 Mathematics Chapter 9. These Circles Class 9 Extra Questions are helpful to test your understanding and knowledge of the chapter. These additional questions will not only help you to excel in the chapter but also will give you a strong foundation in Mathematics.
We have also provided step-by-step answers to the questions for reference. So, let’s start…
30 Circles Class 9 Extra Questions
1. In a circle, if chord \(AB\) subtends an angle of \(50°\) at the center, what angle does it subtend at a point on the remaining part of the circle?
2. True or False: Congruent arcs of a circle have equal chords.
3. The radius of a circle is \(10 cm\). What is the length of a chord located \(6 cm\) from the center?
4. Prove that the perpendicular from the center of a circle to a chord bisects the chord.
5. In a circle, two parallel chords of lengths \(8 cm\) and \(6 cm\) are on opposite sides of the center. If the distance between them is \(7 cm\), find the radius.
6. If two equal chords of a circle intersect, prove that their segments are equal.
7. Prove that the angle subtended by a chord at the center is twice the angle subtended at any point on the remaining circle.
8. In a cyclic quadrilateral \(ABCD\), \(∠A = 70°\) and \(∠B = 110°\). Find \(∠C\) and \(∠D\).
9. In a semicircle, what is the measure of the angle subtended by the diameter?
10. \(ABCD\) is a cyclic quadrilateral with \(∠A = 80°\) and \(∠D = 100°\). Show that it is a rectangle if and only if its diagonals are equal.
11. Two circles intersect at points \(A\) and \(B\). A line through \(A\) meets the circles at \(C\) and \(D\). Prove that \(∠CBD\) is constant, regardless of the line’s position.
12. If a chord is equal to the radius of the circle, find the angle it subtends at the center.
13. Prove that if the sum of a pair of opposite angles in a quadrilateral is \(180°\), the quadrilateral is cyclic.
14. In a circle, chords \(AB\) and \(CD\) intersect at \(E\). If \(AE = 4 cm\), \(EB = 6 cm\), and \(CE = 3 cm\), find \(ED\).
15. True or False: All parallelograms are cyclic quadrilaterals.
16. In a cyclic trapezium, prove that the non-parallel sides are equal.
17. \(AB\) and \(CD\) are two chords of a circle intersecting at \(E\). Prove that \(AE × EB = CE × ED\).
18. The angle subtended by an arc at the center is \(120°\). What is the angle subtended at the circumference?
19. Prove that the line joining the midpoints of two equal chords in a circle is perpendicular to the line joining their midpoints to the center.
20. \(ABCD\) is a cyclic quadrilateral. Diagonals \(AC\) and \(BD\) intersect at \(E\). If \(∠ABD = 30°\) and \(∠BAC = 40°\), find \(∠BDC\).
21. In a circle, if two chords are equidistant from the center, what can be said about their lengths?
22. Prove that the angle bisectors of a cyclic quadrilateral form another cyclic quadrilateral.
23. Two concentric circles have radii \(13 cm\) and \(5 cm\). Find the length of a chord of the larger circle that is tangent to the smaller circle.
24. If a cyclic quadrilateral has one angle of \(85°\), find its opposite angle.
25. In a circle, chord \(AB\) is \(12 cm\), and its distance from the center is \(8 cm\). Find the radius.
26. Prove that the four points formed by the internal angle bisectors of any quadrilateral are concyclic.
27. What is the measure of the angle in a semicircle?
28. In a circle, chords \(PQ\) and \(RS\) are equal. If \(∠POQ = 100°\), find \(∠ROS\).
29. \(ABCD\) is a cyclic quadrilateral with \(AB\) as diameter. If \(∠ADC = 130°\), find \(∠BAC\).
30. If the diagonals of a cyclic quadrilateral are equal, prove it is a rectangle.
Answers
- Solution:
By Theorem 9.7, angle at the center is twice the angle at the circumference:
\((50° / 2 = 25°)\) - Solution:
True. Congruent arcs correspond to equal chords (Theorem 9.8 converse). - Solution:
Using Pythagoras: \( \sqrt{10^2 – 6^2} = \sqrt{64} = 8 \, \text{cm} \). - Solution:
Let \(O\) be the center, \(AB\) the chord, and \(OM ⊥ AB\).
In triangles \(OMA\) and \(OMB\), \(OA = OB\) (radii), \(OM\) common, \(∠OMA = ∠OMB = 90°\).
Hence, \(ΔOMA ≅ ΔOMB ⇒ AM = BM\). - Solution:
Let radius \(= ( r )\).
Distance from center to \(8 cm\) chord: \( \sqrt{r^2 – 4^2} \)
Distance to \(6 cm\) chord: \( \sqrt{r^2 – 3^2} \)
Total distance: \( \sqrt{r^2 – 16} + \sqrt{r^2 – 9} = 7 \)
Solving gives \( r = 5 \, \text{cm} \) - Solution:
Let chords \(AB\) and \(CD\) intersect at \(E\)
Since \(AB = CD\), by Theorem 9.5, they are equidistant.
By Theorem 9.6, \(AE = ED\) and \(BE = EC\) - Solution:
Refer to Theorem 9.7. Let arc \(PQ\) subtend \(∠POQ\) at center and \(∠PAQ\) at circumference.
Draw \(AO\) extended to \(B\)
Use exterior angle property to show \(∠POQ = 2∠PAQ\) - Solution:
In cyclic quadrilateral, \(∠A + ∠C = 180° ⇒ ∠C = 110°\)
Similarly, \(∠B + ∠D = 180° ⇒ ∠D = 70°\) - Solution:
\(90°\) (Angle in a semicircle is a right angle). - Solution:
If diagonals are equal in a cyclic quadrilateral, all angles are \(90°\), making it a rectangle. - Solution:
\(∠CBD = ∠CBA + ∠ABD\)
Since angles subtended by arc \(AD\) are equal, \(∠CBD\) remains constant. - Solution:
Chord = radius ⇒ Central angle = \(60°\) (equilateral triangle). - Solution:
By Theorem 9.11, if \(∠A + ∠C = 180°\) or \(∠B + ∠D = 180°\), the quadrilateral is cyclic. - Solution:
By intersecting chords theorem: \(AE \times EB = CE \times ED \Rightarrow 4 \times 6 = 3 \times ED \Rightarrow ED = 8 \, \text{cm} \) - Solution:
False. Only rectangles (special parallelograms) are cyclic. - Solution:
In cyclic trapezium, base angles are equal. If non-parallel sides are unequal, angles would not sum to \(180°\), contradicting cyclic property. - Solution:
By the intersecting chords theorem, the products of segments are equal. - Solution:
\( 120° / 2 = 60° \) - Solution:
Let midpoints be \(M\) and \(N\). \(OM\) and \(ON\) are perpendicular to chords. Equal chords \(⇒ OM = ON\). Line \(MN\) is perpendicular bisector. - Solution:
\(∠BDC = ∠BAC = 40°\) (Angles in the same segment) - Solution:
They are equal (Theorem 9.6) - Solution:
Let angle bisectors form quadrilateral EFGH.
The sum of opposite angles = \(180°\) ⇒ cyclic (Theorem 9.11) - Solution:
Chord length = \( 2\sqrt{R^2 – r^2} = 2\sqrt{13^2 – 5^2} = 24 \, \text{cm} \) - Solution:
Opposite angle = \( 180° – 85° = 95° \) - Solution:
Radius = \( \sqrt{6^2 + 8^2} = 10 \, \text{cm} \) - Solution:
As in Example 5 of the PDF, sum of opposite angles in the formed quadrilateral = \(180°\), making it cyclic. - Solution:
\(90°\) - Solution:
Equal chords subtend equal angles \(⇒ ∠ROS = 100°\) - Solution:
\(∠BAC = ∠BDC = 180° – 130° = 50°\) (since \(AB\) is diameter, \(∠ACB = 90°\)). - Solution:
Diagonals equal ⇒ all angles \(90°\) (by Theorem 9.10 and properties of rectangles).
Related Posts
- Number Systems Class 9 Extra Questions | Mathematics Chapter 1
- Polynomials Class 9 Extra Questions | Mathematics Chapter 2
- Coordinate Geometry Class 9 Extra Questions | Mathematics Chapter 3
The Bottom Line
I hope these extra questions have helped you a lot. Remember, math is all about practice, so don’t hesitate to revisit these questions and explore more on your own. If you found this post, Circles Class 9 Extra Questions helpful, feel free to share it with your friends or classmates who might benefit from it.
If you find any mistake in the above questions and answers or have any other issues regarding the content of this website, feel free to contact us through the contact us section.
You can also check the related MCQs for this chapter here: MCQ on Circles
Thank you for visiting our blog, have a great day.
Abdur Rohman is an Electrical Engineer from Charaideo, Assam, who wears multiple hats as a part-time teacher, blogger, entrepreneur, and digital marketer. Passionate about education, he founded The Assam School blog to provide free, comprehensive textbook solutions, MCQs (Multiple Choice Questions), and other academic content for students from Class V to XII.