Latur School Exam Papers 2026
Raja Narayanlal Lahoti English School
Class 9 | Term Test-I
38 Circle Geometry (Cyclic Quadrilaterals)
Four Friends Rima, Mohan, Sohan and Sita are sitting on the circumference of a circular park full of water. Their locations are marked by points A, P, Q and R such that the APQR is a quadrilateral with greenery. Rohit joins them and sits at the centre of the circular park, so he is equidistant from all the other friends. His position is marked as O. They are sitting in such a way that \(\angle PQR=110^\circ\).[cite: 50]
Based on this information, answer the following:
(i) What is the measure of reflex \(\angle POR\)? (2 Marks)
(ii) What is the measure of \(\angle PAR\)? (2 Marks)
(iii) (b) What is measure of interior \(\angle POR\)? (1 Mark)
(i) What is the measure of reflex \(\angle POR\)? (2 Marks)
(ii) What is the measure of \(\angle PAR\)? (2 Marks)
OR
(iii) (a) Find \(\angle OPR\)? (1 Mark)(iii) (b) What is measure of interior \(\angle POR\)? (1 Mark)
Concept:
1
Cyclic Quadrilateral: When all four vertices of a quadrilateral touch a circle (like APQR), the opposite angles sum to \(180^\circ\) (\(\angle PQR + \angle PAR = 180^\circ\)).
2
Central Angle Theorem: The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle. (Reflex \(\angle POR = 2 \times \angle PQR\)).
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Hint:
1
For (i), look at the major arc PAR. It subtends reflex \(\angle POR\) at the center and \(\angle PQR\) (\(110^\circ\)) at the circumference. Multiply by 2!
2
For the OR part, triangle OPR is an isosceles triangle because OP and OR are both radii of the same circle. Find interior \(\angle POR\) first (\(360^\circ - \text{reflex}\)), then use the \(180^\circ\) triangle sum rule!
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Mistakes to Avoid:
!
Mixing up Reflex and Interior Angles: Reflex \(\angle POR\) is the large angle on the "outside" (greater than \(180^\circ\)). The interior \(\angle POR\) is the smaller angle inside triangle OPR. Don't swap them!
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PART (i) SOLUTION:
PART (ii) SOLUTION:
The angle subtended by major arc PAR at the center is Reflex \( \angle POR \).
The angle subtended by it at the circumference is \( \angle PQR \).
By the Central Angle Theorem:
Reflex \( \angle POR = 2 \times \angle PQR \) [1 M]
Reflex \( \angle POR = 2 \times 110^\circ = 220^\circ \) [1 M]
Reflex \(\angle POR\) = 220°
PART (ii) SOLUTION:
APQR is a cyclic quadrilateral because all its vertices lie on the circle.
The sum of opposite angles in a cyclic quadrilateral is \( 180^\circ \).
\( \angle PQR + \angle PAR = 180^\circ \) [1 M]
\( 110^\circ + \angle PAR = 180^\circ \)
\( \angle PAR = 180^\circ - 110^\circ = 70^\circ \) [1 M]
\(\angle PAR\) = 70°
OR
OR PART (iii) SOLUTION:
(b) Finding interior \( \angle POR \):
Angles around the center O add up to \( 360^\circ \).
Interior \( \angle POR = 360^\circ - \text{Reflex } \angle POR \)
Interior \( \angle POR = 360^\circ - 220^\circ = 140^\circ \) [1 M]
(a) Finding \( \angle OPR \):
In \( \Delta OPR \), \( OP = OR \) (radii of the same circle).
Therefore, \( \Delta OPR \) is an isosceles triangle, so \( \angle OPR = \angle ORP \).
Sum of angles in a triangle is \( 180^\circ \):
\( \angle OPR + \angle ORP + \text{Interior } \angle POR = 180^\circ \)
\( 2\angle OPR + 140^\circ = 180^\circ \)
\( 2\angle OPR = 40^\circ \implies \angle OPR = 20^\circ \) [1 M]
\(\angle OPR\) = 20° | \(\angle POR\) = 140°
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