Welcome, Class 9 students! In this guide, we present complete, step-by-step solutions for Class 9 Mathematics, Chapter 9: Circles, Exercise 9.3. Prepared strictly as per the updated NCERT guidelines and the MP Board curriculum, these solutions will help you master the key theorems of circles, cyclic quadrilaterals, and angle relationships. Let's make geometry simple and fun!
Important Formulas, Rules & Key Points
- Theorem 9.7: The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. (i.e., $\angle AOB = 2 \angle ACB$)
- Theorem 9.8: Angles in the same segment of a circle are equal.
- Angle in a Semicircle: The angle subtended by a diameter in a semicircle is a right angle ($90^\circ$).
- Cyclic Quadrilateral: A quadrilateral whose all four vertices lie on a circle.
- Theorem 9.11: The sum of either pair of opposite angles of a cyclic quadrilateral is $180^\circ$.
- Theorem 9.12: If the sum of a pair of opposite angles of a quadrilateral is $180^\circ$, then the quadrilateral is cyclic.
NCERT / MP Board Exercise 9.3 Solutions
Question 1: In the given figure, $A$, $B$ and $C$ are three points on a circle with centre $O$ such that $\angle BOC = 30^\circ$ and $\angle AOB = 60^\circ$. If $D$ is a point on the circle other than the arc $ABC$, find $\angle ADC$.
Solution:
Given details:
- $\angle AOB = 60^\circ$
- $\angle BOC = 30^\circ$
First, let's find the total angle subtended by the arc $ABC$ at the centre $O$:
$$\angle AOC = \angle AOB + \angle BOC = 60^\circ + 30^\circ = 90^\circ$$
According to Theorem 9.7, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
Therefore,
$$\angle AOC = 2 \angle ADC$$
$$\angle ADC = \frac{1}{2} \angle AOC = \frac{1}{2} \times 90^\circ = 45^\circ$$
Answer: $\angle ADC = 45^\circ$
Question 2: A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
Solution:
Let $AB$ be the chord which is equal to the radius of the circle with centre $O$.
Thus, $OA = OB = AB$ (since chord = radius).
Therefore, $\triangle OAB$ is an equilateral triangle.
Since all interior angles of an equilateral triangle are $60^\circ$, we have:
$$\angle AOB = 60^\circ$$
1. Angle on the Major Arc:
Let $C$ be a point on the major arc. By Theorem 9.7:
$$\angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 60^\circ = 30^\circ$$
2. Angle on the Minor Arc:
Let $D$ be a point on the minor arc. The quadrilateral $ACBD$ is a cyclic quadrilateral because all its vertices lie on the circle.
Since the sum of opposite angles of a cyclic quadrilateral is $180^\circ$:
$$\angle ACB + \angle ADB = 180^\circ$$
$$30^\circ + \angle ADB = 180^\circ$$
$$\angle ADB = 180^\circ - 30^\circ = 150^\circ$$
Answer: Angle on major arc = $30^\circ$, Angle on minor arc = $150^\circ$
Question 3: In the given figure, $\angle PQR = 100^\circ$, where $P$, $Q$ and $R$ are points on a circle with centre $O$. Find $\angle OPR$.
Solution:
Given $\angle PQR = 100^\circ$. $Q$ is on the major/minor arc segment opposite to the reflex angle.
The angle subtended by arc $PR$ at the centre is reflex $\angle POR$. By Theorem 9.7:
$$\text{Reflex } \angle POR = 2 \times \angle PQR = 2 \times 100^\circ = 200^\circ$$
Now, we find the interior/obtuse angle $\angle POR$:
$$\angle POR = 360^\circ - \text{Reflex } \angle POR = 360^\circ - 200^\circ = 160^\circ$$
Now consider the triangle $\triangle OPR$:
- $OP = OR$ (radii of the same circle)
- Therefore, $\triangle OPR$ is an isosceles triangle, which implies $\angle OPR = \angle ORP$.
Using the angle sum property of a triangle in $\triangle OPR$:
$$\angle OPR + \angle ORP + \angle POR = 180^\circ$$
$$2\angle OPR + 160^\circ = 180^\circ$$
$$2\angle OPR = 180^\circ - 160^\circ = 20^\circ$$
$$\angle OPR = 10^\circ$$
Answer: $\angle OPR = 10^\circ$
Question 4: In the given figure, $\angle ABC = 69^\circ$, $\angle ACB = 31^\circ$, find $\angle BDC$.
Solution:
Consider the triangle $\triangle ABC$. By the angle sum property of a triangle:
$$\angle BAC + \angle ABC + \angle ACB = 180^\circ$$
$$\angle BAC + 69^\circ + 31^\circ = 180^\circ$$
$$\angle BAC + 100^\circ = 180^\circ$$
$$\angle BAC = 80^\circ$$
We know that angles in the same segment of a circle are equal (Theorem 9.8).
Therefore, $\angle BDC = \angle BAC = 80^\circ$.
Answer: $\angle BDC = 80^\circ$
Question 5: In the given figure, $A, B, C$ and $D$ are four points on a circle. $AC$ and $BD$ intersect at a point $E$ such that $\angle BEC = 130^\circ$ and $\angle ECD = 20^\circ$. Find $\angle BAC$.
Solution:
Angles on a straight line ($BD$) sum up to $180^\circ$:
$$\angle DEC + \angle BEC = 180^\circ \quad \text{(Linear pair)}$$
$$\angle DEC + 130^\circ = 180^\circ$$
$$\angle DEC = 180^\circ - 130^\circ = 50^\circ$$
Now, consider the triangle $\triangle DEC$. Using the angle sum property of a triangle:
$$\angle EDC + \angle DEC + \angle ECD = 180^\circ$$
$$\angle EDC + 50^\circ + 20^\circ = 180^\circ$$
$$\angle EDC + 70^\circ = 180^\circ$$
$$\angle EDC = 110^\circ$$
This can also be written as $\angle BDC = 110^\circ$.
Since angles in the same segment are equal:
$$\angle BAC = \angle BDC = 110^\circ$$
Answer: $\angle BAC = 110^\circ$
Question 6: $ABCD$ is a cyclic quadrilateral whose diagonals intersect at a point $E$. If $\angle DBC = 70^\circ$, $\angle BAC = 30^\circ$, find $\angle BCD$. Further, if $AB = BC$, find $\angle ECD$.
Solution:
Part 1: Find $\angle BCD$
Angles subtended by the same segment arc $BC$ are equal:
$$\angle BDC = \angle BAC = 30^\circ$$
Now look at triangle $\triangle BCD$. Using the angle sum property:
$$\angle BCD + \angle DBC + \angle BDC = 180^\circ$$
$$\angle BCD + 70^\circ + 30^\circ = 180^\circ$$
$$\angle BCD + 100^\circ = 180^\circ$$
$$\angle BCD = 80^\circ$$
Part 2: Find $\angle ECD$ if $AB = BC$
In $\triangle ABC$, since $AB = BC$:
$$\angle BCA = \angle BAC = 30^\circ \quad \text{(Angles opposite to equal sides)}$$
We know that:
$$\angle BCD = \angle BCA + \angle ECD$$
$$80^\circ = 30^\circ + \angle ECD$$
$$\angle ECD = 80^\circ - 30^\circ = 50^\circ$$
Answer: $\angle BCD = 80^\circ$ and $\angle ECD = 50^\circ$
Aptitude Corner - Quick Tricks for Competitive Exams (NTSE/Olympiads)
In competitive exams like NTSE and Olympiads, geometry problems on cyclic quadrilaterals are common. Keep these quick tricks handy:
- Ptolemy's Theorem: For any cyclic quadrilateral $ABCD$, the product of the diagonals is equal to the sum of the products of the opposite sides:
$$\mathbf{AC \times BD = (AB \times CD) + (BC \times AD)}$$ - Brahmagupta's Formula: The area of a cyclic quadrilateral with sides $a, b, c, d$ is given by:
$$\mathbf{\text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d)}}$$ where $s = \frac{a+b+c+d}{2}$ is the semi-perimeter.
We hope these detailed solutions help you build strong foundations in circle theorems. Practicing these problems will boost your speed and accuracy in exams! Feel free to explore more resources, and keep practicing!
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