Maharashtra Board Class 9 Maths Part 2 Geometry Chapter 6 Set 6.1 Circle Solutions

Question 1. Distance of chord AB from the centre of a circle is 8 cm. Length of the chord AB is 12 cm. Find the diameter of a circle. Given: In a circle with centre O, OA is radius and AB is its chord, seg OP \(\perp\) chord AB, A-P-B AB = 12 cm, OP =8 cm To Find: Diameter of the circle Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त को दर्शाता है जिसका केंद्र O है। वृत्त के अंदर एक जीवा AB है जिसकी लंबाई 12 सेमी है। केंद्र O से जीवा AB पर एक लंब OP खींचा गया है, जिसकी लंबाई 8 सेमी है, जो जीवा AB को बिंदु P पर समद्विभाजित करता है। यह त्रिज्या OA को भी दर्शाता है।
(i) AP = \(\frac{1}{2}\) AB [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
\(\therefore\) AP = \(\frac{1}{2}\) x 12 = 6 cm ...(i)
(ii) In \(\triangle\)OPA, \(\angle\)OPA = 90°
\(\therefore\) OA\(^2\) = OP\(^2\) + AP\(^2\) [Pythagoras theorem]
= 8\(^2\) + 6\(^2\) [From (i)]
= 64 + 36
\(\therefore\) OA\(^2\) = 100
\(\therefore\) OA = \(\sqrt{100}\) [Taking square root on both sides]
= 10 cm
(iii) Radius (r) = 10 cm
\(\therefore\) Diameter = 2r = 2 x 10 = 20 cm
\(\therefore\) The diameter of the circle is 20 cm.
Answer: The diameter of the circle is 20 cm.
In simple words: We used the property that a perpendicular from the center bisects the chord, then applied the Pythagorean theorem to find the radius (hypotenuse of the right-angled triangle formed), and finally doubled the radius to get the diameter.

🎯 Exam Tip: Remember to clearly state the geometric theorems used, like "perpendicular from the centre to the chord bisects the chord" and "Pythagoras theorem," as they are crucial for full marks.

Question 2. Diameter of a circle is 26 cm and length of a chord of the circle is 24 cm. Find the distance of the chord from the centre. Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त को दर्शाता है जिसका केंद्र O है। वृत्त के अंदर एक जीवा PQ है जिसकी लंबाई 24 सेमी है। केंद्र O से जीवा PQ पर एक लंब OR खींचा गया है, जो जीवा को बिंदु R पर समद्विभाजित करता है। यह त्रिज्या OP को भी दर्शाता है। Given: In a circle with centre O, PO is radius and PQ is its chord, seg OR \(\perp\) chord PQ, P-R-Q PQ = 24 cm, diameter (d) = 26 cm To Find: Distance of the chord from the centre (OR) Solution:
(i) Radius (OP) = \(\frac{d}{2}\) = \(\frac{26}{2}\) = 13 cm ......(i)
\(\therefore\) PR = \(\frac{1}{2}\) PQ [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
= \(\frac{1}{2}\) x 24 = 12 cm.....(ii)
(ii) In \(\triangle\)ORP, \(\angle\)ORP = 90°
\(\therefore\) OP\(^2\)= OR\(^2\) + PR\(^2\) [Pythagoras theorem]
\(\therefore\) 13\(^2\) = OR\(^2\) + 12\(^2\) [From (i) and (ii)]
\(\therefore\) 169 = OR\(^2\) + 144
\(\therefore\) OR\(^2\) = 169 - 144
\(\therefore\) OR\(^2\) = 25
\(\therefore\) OR = \(\sqrt{25}\) = 5 cm [Taking square root on both sides]
\(\therefore\) The distance of the chord from the centre of the circle is 5 cm.
Answer: The distance of the chord from the centre of the circle is 5 cm.
In simple words: First, we calculated the radius from the given diameter. Then, using the property that the perpendicular from the center bisects the chord, we found half the chord length. Finally, we applied the Pythagorean theorem to find the distance of the chord from the center.

🎯 Exam Tip: Clearly show the steps for calculating the radius and half the chord length before applying the Pythagorean theorem. This helps in maintaining clarity and preventing errors.

Question 3. Radius of a circle is 34 cm and the distance of the chord from the centre is 30 cm, find the length of the chord. Given: in a circle with centre A, PA is radius and PQ is chord, seg AM \(\perp\) chord PQ, P-M-Q AP = 34 cm, AM = 30 cm To Find: Length of the chord (PQ) Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त को दर्शाता है जिसका केंद्र A है। वृत्त के अंदर एक जीवा PQ है। केंद्र A से जीवा PQ पर एक लंब AM खींचा गया है, जिसकी लंबाई 30 सेमी है। त्रिज्या AP की लंबाई 34 सेमी है। बिंदु M जीवा PQ को समद्विभाजित करता है।
I. In \(\triangle\)AMP, \(\angle\)AMP = 90°
\(\therefore\) AP\(^2\) = AM\(^2\) + PM\(^2\) [Pythagoras theorem]
34\(^2\) = 30\(^2\) + PM\(^2\)
\(\therefore\) PM\(^2\) = 34\(^2\) - 30\(^2\)
\(\therefore\) PM\(^2\) = (34 - 30)(34 + 30) [a\(^2\) - b\(^2\) = (a - b)(a + b)]
= 4 x 64
\(\therefore\) PM = \(\sqrt{4 \times 64}\) .........(i) [Taking square root on both sides]
= 2 x 8 = 16cm
ii. Now, PM = \(\frac{1}{2}\) (PQ) [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
16 = \(\frac{1}{2}\)(PQ) [From (i)]
\(\therefore\) PQ = 16 x 2
= 32cm
\(\therefore\) The length of the chord of the circle is 32cm.
Answer: The length of the chord of the circle is 32 cm.
In simple words: We used the Pythagorean theorem with the radius and the distance from the center to find half the length of the chord. Since the perpendicular from the center bisects the chord, we doubled this length to find the total length of the chord.

🎯 Exam Tip: When dealing with squares and square roots, using the difference of squares formula \((a^2 - b^2) = (a-b)(a+b)\) can simplify calculations, especially in problems like this one.

Question 4. Radius of a circle with centre O is 41 units. Length of a chord PQ is 80 units, find the distance of the chord from the centre of the circle. Given: In a circle with centre O, OP is radius and PQ is its chord, seg OM \(\perp\) chord PQ, P-M-Q OP = 41 units, PQ = 80 units, To Find: Distance of the chord from the centre of the circle(OM) Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त को दर्शाता है जिसका केंद्र O है। वृत्त के अंदर एक जीवा PQ है जिसकी लंबाई 80 इकाई है। केंद्र O से जीवा PQ पर एक लंब OM खींचा गया है। त्रिज्या OP की लंबाई 41 इकाई है। बिंदु M जीवा PQ को समद्विभाजित करता है।
i. PM = \(\frac{1}{2}\) (PQ) [Perpendicular drawn from the centre of a circle to the chord bisects the chord.]
= \(\frac{1}{2}\) (80) = 40 Units ....(i)
ii. In \(\triangle\)OMP, \(\angle\)OMP = 90°
\(\therefore\) OP\(^2\) = OM\(^2\) + PM\(^2\) [Pythagoras theorem]
\(\therefore\) 41\(^2\) = OM\(^2\) + 40\(^2\) [From (i)]
\(\therefore\) OM\(^2\) = 41\(^2\) - 40\(^2\)
= (41-40) (41 +40) [a\(^2\) - b\(^2\) = (a - b) (a + b)]
= (1)(81)
\(\therefore\) OM\(^2\) = 81
\(\therefore\) OM = \(\sqrt{81}\) = 9 units [Taking square root on both sides] [From (i)]
\(\therefore\) The distance of the chord from the centre of the circle is 9 units.
Answer: The distance of the chord from the centre of the circle is 9 units.
In simple words: We first found half the length of the chord using the property that the perpendicular from the center bisects it. Then, applying the Pythagorean theorem with the radius and half-chord length, we calculated the distance of the chord from the center.

🎯 Exam Tip: Always draw a neat and labeled diagram for geometry problems. It helps visualize the given information and apply the correct theorems. The difference of squares identity is very useful here.

Question 5. In the adjoining figure, centre of two circles is O. Chord AB of bigger circle intersects the smaller circle in points P and Q. Show that AP = BQ.
ℹ️ चित्र व्याख्या (Diagram Explanation): यह दो संकेंद्रित वृत्त दर्शाता है, जिनका केंद्र O है। बड़ा वृत्त एक जीवा AB रखता है, जो छोटे वृत्त को P और Q बिंदुओं पर काटता है। केंद्र O से जीवा AB पर एक लंब OM खींचा गया है, जो AB को M पर और PQ को M पर समद्विभाजित करता है। Given: Two concentric circles having centre O. To prove: AP = BQ Construction: Draw seg OM \(\perp\) chord AB, A-M-B Solution: Proof: For smaller circle, seg OM \(\perp\) chord PQ [Construction, A-P-M, M-Q-B]
\(\therefore\) PM = MQ .....(i) [Perpendicular drawn from the centre of the circle to the chord bisects the chord.] For bigger circle, seg OM \(\perp\) chord AB [Construction]
\(\therefore\) AM = MB [Perpendicular drawn from the centre of the circle to the chord bisects the chord.]
\(\therefore\) AP + PM = MQ + QB [A-P-M, M-Q-B]
\(\therefore\) AP + MQ = MQ + QB [From (i)]
\(\therefore\) AP = BQ
Answer: AP = BQ is proven.
In simple words: We proved AP = BQ by drawing a perpendicular from the common center to the chord. This perpendicular bisects both the smaller and larger chord segments. By subtracting the equal segments from the larger chord's bisected parts, the remaining segments AP and BQ are shown to be equal.

🎯 Exam Tip: For proof-based questions, clearly state Given, To Prove, Construction (if any), and Proof steps. Each step in the proof must be justified by a geometric property or a previous statement.

Question 6. Prove that, if a diameter of a circle bisects two chords of the circle then those two chords are parallel to each other. Solution:
ℹ️ चित्र व्याख्या (Diagram Explanation): यह एक वृत्त को दर्शाता है जिसका केंद्र O है। एक व्यास PQ है। दो जीवाएँ AB और CD हैं। व्यास PQ, जीवा AB को बिंदु M पर और जीवा CD को बिंदु N पर समद्विभाजित करता है। Given: O is the centre of the circle. seg PQ is the diameter. Diameter PQ bisects the chords AB and CD in points M and N respectively. To prove: chord AB || chord CD. Proof: Diameter PQ bisects the chord AB in point M [Given]
\(\therefore\) seg AM \(\cong\) seg BM
\(\therefore\) seg OM \(\perp\) chord AB [Segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord, P-M-O, O-N-Q]
\(\therefore\) \(\angle\)OMA = 90° .....(i) Also, diameter PQ bisects the chord CD in point N [Given]
\(\therefore\) seg CN \(\cong\) seg DN
seg ON \(\perp\) chord CD [Segment joining the centre of a circle and the midpoint of its chord is perpendicular to the chord, P-M-O, O-N-Q]
\(\therefore\) \(\angle\)ONC = 90° .....(ii) Now, \(\angle\)OMA + \(\angle\)ONC = 90° + 90° [From (i) and (ii)]
= 180° But, \(\angle\)OMA and \(\angle\)ONC form a pair of interior angles on lines AB and CD when seg MN is their transversal.
\(\therefore\) chord AB || chord CD [Interior angles test]
Answer: It is proven that if a diameter bisects two chords, then those two chords are parallel to each other.
In simple words: When a diameter bisects a chord, the line segment from the center to the midpoint of the chord is perpendicular to the chord. If a diameter bisects two chords, then the lines connecting the center to their midpoints are perpendicular to both chords. If the sum of the interior angles on one side of a transversal is 180 degrees, the lines are parallel. Here, the angles formed by the perpendiculars are 90 degrees each, summing to 180 degrees, proving the chords are parallel.

🎯 Exam Tip: This proof relies on the key property that a line segment joining the center of a circle to the midpoint of a chord is perpendicular to the chord. Also, remember the interior angles test for parallel lines.