Welcome to the ultimate revision guide for MP Board Class 9 Mathematics Chapter 9 (Circles)! This session is specially designed to strengthen your conceptual understanding for school exams while preparing you for competitive examinations like RRB, SSC, and SSC GD with Previous Year Questions (PYQs).
Important Formulas, Rules & Key Points
- Chord & Center Property: The perpendicular from the center of a circle to a chord bisects the chord.
- Angle at the Center: 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.
- Angles in the Same Segment: Angles in the same segment of a circle are equal.
- Cyclic Quadrilateral: The sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
Step-by-Step Exercise Solutions
Question 1:
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 $O$ be the center of the circle and $AB$ be the chord such that $AB = OA = OB = r$.
- Since $OA = OB = AB$, $ riangle OAB$ is an equilateral triangle.
- Therefore, $ heta = ext{∠}AOB = 60^ ext{o}$.
- The angle subtended by chord $AB$ at a point $P$ on the major arc is half the angle subtended at the center:
$ ext{∠}APB = rac{1}{2} imes ext{∠}AOB = rac{1}{2} imes 60^ ext{o} = 30^ ext{o}$. - Let $Q$ be a point on the minor arc. Since $APBQ$ is a cyclic quadrilateral, the sum of opposite angles is $180^ ext{o}$:
$ ext{∠}AQB = 180^ ext{o} - ext{∠}APB = 180^ ext{o} - 30^ ext{o} = 150^ ext{o}$.
Answer: Angle on major arc = $30^ ext{o}$, Angle on minor arc = $150^ ext{o}$.
Question 2:
If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Solution:
- Given two equal chords $AB$ and $CD$ intersecting at point $E$ inside the circle with center $O$.
- Draw perpendiculars $OM ot AB$ and $ON ot CD$. Join $OE$.
- In $ riangle OME$ and $ riangle ONE$:
• $ ext{∠}OME = ext{∠}ONE = 90^ ext{o}$
• $OE = OE$ (Common side)
• $OM = ON$ (Equal chords are equidistant from the center) - By RHS congruence, $ riangle OME ext{ ≅ } riangle ONE$, which implies $ME = NE$.
- Since $AB = CD$, their halves are also equal: $AM = CN$.
- Adding $ME$ and $NE$: $AM + ME = CN + NE ightarrow AE = CE$.
- Subtracting these from $AB$ and $CD$: $AB - AE = CD - CE ightarrow EB = ED$.
Hence, the corresponding segments are equal.
Aptitude Corner (RRB / SSC / GD PYQs)
Q1 (SSC GD PYQ):
Find the length of a chord which is at a distance of $6 ext{ cm}$ from the center of a circle of radius $10 ext{ cm}$.
Solution: In right $ riangle OMB$, $OB = 10 ext{ cm}$, $OM = 6 ext{ cm}$.
By Pythagoras theorem: $MB = ext{sqrt}(10^2 - 6^2) = ext{sqrt}(100 - 36) = ext{sqrt}(64) = 8 ext{ cm}$.
Total length of chord $AB = 2 imes MB = 2 imes 8 = 16 ext{ cm}$.
Q2 (RRB NTPC PYQ):
p>In a cyclic quadrilateral $ABCD$, if $ ext{∠}A = (2x + 4)^ ext{o}$ and $ ext{∠}C = (4x - 10)^ ext{o}$, find the value of $x$.Solution: Opposite angles of a cyclic quadrilateral sum up to $180^ ext{o}$.
$(2x + 4) + (4x - 10) = 180
ightarrow 6x - 6 = 180
ightarrow 6x = 186
ightarrow x = 31$.
Mastering Chapter 9 Circles builds a rock-solid foundation for both your MP Board Class 9 annual examinations and future competitive exams like RRB, SSC CGL, and SSC GD. Keep practicing theorem proofs and numerical problems daily to excel in mathematics!
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