This MCQ module is based on: Cyclic Quadrilaterals & Chapter Exercises
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Cyclic Quadrilaterals & Chapter Exercises
🎓 Class 9
Mathematics
CBSE
Theory
Ch 5 — I’m Up and Down, and Round and Round
⏱ ~50 min
🌐 Language: [gtranslate]
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This mathematics assessment will be based on: Cyclic Quadrilaterals & Chapter Exercises
Targeting Class 9 level in Geometry, with Intermediate difficulty.
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5.12 Cyclic Quadrilaterals
A quadrilateral is called a cyclic quadrilateral? if all four of its vertices lie on a single circle. The circle is called the circumcircle of the quadrilateral.
Fig. 5.29 — Cyclic quadrilateral ABCD inscribed in circle with centre O.
Theorem 9 — Cyclic Quadrilateral
The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.That is, \(\angle A + \angle C = 180°\) and \(\angle B + \angle D = 180°\).
Proof. Consider arc BCD. It subtends \(\angle\)BOD at the centre and \(\angle\)BAD = \(\angle\)A on the remaining arc. So \(\angle\)BOD = 2\(\angle\)A. Similarly, arc BAD gives reflex \(\angle\)BOD = 2\(\angle\)C. Adding: \(2\angle A + 2\angle C =\) full rotation \(= 360°\). Hence \(\angle A + \angle C = 180°\). The other pair follows similarly (or because all angles of a quadrilateral sum to 360°). ∎
Theorem 10 — Converse
If a pair of opposite angles of a quadrilateral sum to 180°, then the quadrilateral is cyclic — its four vertices lie on a circle.Worked Example
Example 12. In cyclic quadrilateral ABCD, \(\angle\)A = 80° and \(\angle\)B = 100°. Find \(\angle\)C and \(\angle\)D.
By Theorem 9: \(\angle\)C = 180° − 80° = 100°, \(\angle\)D = 180° − 100° = 80°.
Activity: Why Opposite Angles Are Supplementary
L3 ApplyMaterials: compass, ruler, protractor.
Predict: If you draw four random points on a circle and join them in order, what will the sum of opposite angles be?
- Draw a circle of any radius. Mark four points A, B, C, D on it (in order).
- Join AB, BC, CD, DA to form a quadrilateral.
- Measure all four angles with a protractor.
- Add \(\angle A + \angle C\) and \(\angle B + \angle D\). Compare with the theorem.
In every case (within measurement error), \(\angle A + \angle C = 180°\) and \(\angle B + \angle D = 180°\). The position of the points does not matter, as long as the order around the circle is preserved.
5.13 Chapter Summary
Quick Recap
- A circle is the set of points in a plane at a fixed distance (radius) from a fixed point (centre).
- Diameter = 2 × radius; the diameter is the longest chord and a line of symmetry.
- Three non-collinear points determine a unique circle.
- Equal chords ⇔ equal angles at centre ⇔ equal distances from the centre ⇔ equal arcs.
- Perpendicular from centre to a chord bisects the chord (and conversely).
- Inscribed Angle Theorem: angle at centre = 2 × angle on remaining arc.
- Angles in the same segment of a circle are equal.
- Opposite angles of a cyclic quadrilateral are supplementary.
5.14 Exercise Set
Q1. Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.
Let circles have centres O and O', each radius r. Let AB and CD be equal chords. In \(\triangle\)AOB and \(\triangle\)CO'D: OA = O'C = r, OB = O'D = r, AB = CD. By SSS, triangles are congruent, so \(\angle\)AOB = \(\angle\)CO'D.
Q2. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6 m each, what is the distance between Reshma and Mandip?
Let R, S, M be on circle, centre O. RS = SM = 6, OR = OS = OM = 5. Drop OL ⊥ RS, OK ⊥ SM. Use OL = √(25 − 9) = 4. Set up coordinates: by symmetry, the chord RM passes parallel to OS extended. Computation gives RM = 9.6 m.
Q3. A chord of a circle of radius 15 cm is at a distance of 12 cm from the centre. Find the length of the chord.
Half-chord = √(15² − 12²) = √(225 − 144) = √81 = 9. Length = 18 cm.
Q4. In a circle, two chords AB and CD intersect at a point E inside the circle. Prove that the angle subtended by AC at the centre and the angle subtended by BD at the centre have a sum equal to twice ∠AEC.
By inscribed angle theorem, \(\angle\)ABC = ½\(\angle\)AOC and \(\angle\)BAD = ½\(\angle\)BOD, where O is centre. \(\angle\)AEC is exterior angle of \(\triangle\)AEB at E, so \(\angle\)AEC = \(\angle\)EAB + \(\angle\)EBA = ½(\(\angle\)BOD + \(\angle\)AOC). Hence \(\angle\)AOC + \(\angle\)BOD = 2\(\angle\)AEC.
Q5. ABCD is a cyclic quadrilateral whose diagonals intersect at point E. If \(\angle\)DBC = 70° and \(\angle\)BAC = 30°, find \(\angle\)BCD. Further, if AB = BC, find \(\angle\)ECD.
Angles in same segment: \(\angle\)BDC = \(\angle\)BAC = 30°. So \(\angle\)BCD = 180° − \(\angle\)BAD (opposite angles). Compute: \(\angle\)BAD = 30° + (\(\angle\)CAD where \(\angle\)CAD = \(\angle\)CBD = 70°) = 100°. So \(\angle\)BCD = 80°. With AB = BC: \(\angle\)BCA = \(\angle\)BAC = 30°, so \(\angle\)ECD = 80° − 30° = 50°.
Q6. How would you use the result that the angle in a semicircle is 90° to construct the perpendicular from a point P outside a circle to a chord AB?
Use the fact that an angle inscribed in a semicircle is a right angle. Draw a circle with PA (or PB) as diameter; its second intersection with the line AB gives the foot of perpendicular.
Q7. In a circle, two chords CC' and DD' are drawn perpendicular to a diameter AB. M is the midpoint of CC' and N is the midpoint of DD'. Prove that MN joining the midpoints lies on AB.
Perpendicular from the centre to a chord bisects it (Theorem 4). Since CC' ⊥ AB and AB passes through O, OM ⊥ CC' so M lies on AB. Similarly N. Hence M and N both lie on AB.
Q8. Prove that the sum of opposite angles of a cyclic quadrilateral is 180°.
Already proven as Theorem 9 above. Use the inscribed-angle theorem on the two arcs cut by the diagonal: \(\angle A + \angle C = \tfrac12(\text{arc BCD}) + \tfrac12(\text{arc DAB}) = \tfrac12(360°) = 180°\).
Competency-Based Questions
Scenario: A traditional rangoli is drawn by inscribing a four-pointed star inside a circle of radius 20 cm. The four tips A, B, C, D of the star lie on the circle. The artist measures \(\angle A = 95°\) and \(\angle B = 85°\), and is told the figure ABCD is cyclic.
Q1. Find \(\angle C\).
L3 Apply(b) 85°. Opposite angles of a cyclic quadrilateral are supplementary: \(\angle C = 180° - 95° = 85°\).
Q2. The artist sketches a second quadrilateral PQRS with \(\angle P = 110°\) and \(\angle R = 60°\). Analyse: can PQRS be inscribed in a circle?
L4 AnalyseNo. By the converse theorem, opposite angles must sum to 180°. Here \(110° + 60° = 170° \neq 180°\), so PQRS is NOT cyclic.
Q3. Evaluate the design rule: "If a quadrilateral is a rectangle, it can always be inscribed in a circle."
L5 EvaluateTrue. Each angle is 90°, so opposite angles sum to 90° + 90° = 180°. By Theorem 10, every rectangle is cyclic. Its circumcircle has the diagonal as diameter.
Q4. Design: choose four angles that produce a cyclic quadrilateral but NOT a rectangle, with all angles different.
L6 CreateMany answers. Example: \(\angle A = 80°, \angle B = 70°, \angle C = 100°, \angle D = 110°\). Check: \(\angle A + \angle C = 180°\) ✓, \(\angle B + \angle D = 180°\) ✓, and the four angles sum to 360°. They are all different, so the quadrilateral is not a rectangle.
Assertion–Reason Questions
A: A rectangle can always be inscribed in a circle.
R: All four angles of a rectangle are right angles, so each pair of opposite angles sums to 180°.
R: All four angles of a rectangle are right angles, so each pair of opposite angles sums to 180°.
(a) R is the converse of Theorem 9 applied to a rectangle.
A: A parallelogram is always cyclic.
R: Opposite angles of a parallelogram are equal.
R: Opposite angles of a parallelogram are equal.
(d) A is FALSE — only rectangles among parallelograms are cyclic. R is true: \(\angle A = \angle C\), so for the parallelogram to be cyclic we need \(\angle A + \angle C = 180°\) ⇒ each is 90° ⇒ rectangle.
A: The angle in a semicircle is 90°.
R: A diameter subtends 180° at the centre, and the inscribed angle is half of the central angle.
R: A diameter subtends 180° at the centre, and the inscribed angle is half of the central angle.
(a) Half of 180° is 90°. R explains A perfectly.
Frequently Asked Questions
What is a cyclic quadrilateral in Class 9 Maths?
A cyclic quadrilateral is a quadrilateral whose four vertices all lie on the circumference of a single circle. Examples include rectangles, squares and isosceles trapeziums; a parallelogram is cyclic only if it is a rectangle.
Why do opposite angles of a cyclic quadrilateral sum to 180 degrees?
Opposite vertices of the quadrilateral cut the circle into two arcs that together make up the full circle. The inscribed-angle theorem says each opposite angle is half the central angle of the opposite arc. Adding gives half of 360 degrees = 180 degrees.
What is the inscribed angle theorem used in Class 9 circles?
The inscribed angle theorem says: the angle subtended by an arc at the centre of a circle is twice the angle subtended by the same arc at any point on the remaining (major) arc. It is the key tool for proving cyclic-quadrilateral and angle-in-a-semicircle results.
Why is the angle in a semicircle a right angle?
If AB is a diameter and P is any point on the circle (not A or B), then arc AB is a semicircle so the central angle is 180 degrees. By the inscribed angle theorem, angle APB = 180/2 = 90 degrees, so the angle in a semicircle is always a right angle.
How can you check that four points are concyclic?
Four points are concyclic if the quadrilateral they form has opposite angles summing to 180 degrees, or equivalently if any two of them subtend equal angles at the other two on the same side of the chord. Either condition is necessary and sufficient.
In a cyclic quadrilateral ABCD, if angle A = 110 deg, what is angle C?
Opposite angles of a cyclic quadrilateral are supplementary, so angle A + angle C = 180 degrees. Therefore angle C = 180 - 110 = 70 degrees.
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