Maharashtra Board Class 8 Maths Chapter 17 Circle Chord and Arc Set 17.1 Solutions

Question 1. In a circle with centre P, chord AB is drawn of length 13 cm, seg PQ \( \perp \) chord AB, then find I(QB)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त दिखाया गया है जिसका केंद्र P है। वृत्त के अंदर एक जीवा AB है। केंद्र P से जीवा AB पर PQ लंब डाला गया है, जहाँ Q जीवा AB पर एक बिंदु है।
Answer: seg PQ \( \perp \) chord AB ... [Given]
\( \therefore \) I(QB) = \( \frac{1}{2} \) I(AB)... [Perpendicular drawn from the centre of a circle to its chord bisects the chord]
\( \therefore \) I(QB) = \( \frac{1}{2} \) x 13 ...[ \( \therefore \) I(AB) = 13 cm]
\( \therefore \) I(QB) = 6.5 cm
In simple words: The perpendicular from the center of a circle to a chord bisects the chord. Since PQ is perpendicular to chord AB, Q is the midpoint of AB, so I(QB) is half of the length of AB.

🎯 Exam Tip: Remember the theorem that a perpendicular from the center to a chord bisects the chord, which is crucial for solving problems involving chord lengths and distances. This principle helps simplify calculations.

Question 2. Radius of a circle with centre O is 25 cm. Find the distance of a chord from the centre if length of the chord is 48 cm.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त दिखाया गया है जिसका केंद्र O है और त्रिज्या 25 सेमी है। एक जीवा CD है जिसकी लंबाई 48 सेमी है। केंद्र O से जीवा CD पर P बिंदु पर एक लंब OP खींचा गया है, जो जीवा को समद्विभाजित करता है।
Answer: seg OP\( \perp \) chord CD ... [Given]
\( \therefore \) I(PD) = \( \frac{1}{2} \) I(CD) ... [Perpendicular drawn from the centre of a circle to its chord bisects the chord]
\( \therefore \) I(PD) = \( \frac{1}{2} \) x 48 ...[ \( \therefore \) I(CD) = 48 cm]
\( \therefore \) I(PD) = 24 cm ...(i)
In \( \triangle \)OPD, m\( \angle \)OPD = 90°
\( \therefore \) \( [\text{I(OD)}]^2 = [\text{I(OP)}]^2 + [\text{I(PD)}]^2 \) ... [Pythagoras theorem]
\( \therefore \) \( (25)^2 = [\text{I(OP)}]^2 + (24)^2 \) ... [From (i) and I(OD) = 25 cm]
\( \therefore \) \( (25)^2 - (24)^2 = [\text{I(OP)}]^2 \)
\( \therefore \) \( (25 + 24) (25 - 24) = [\text{I(OP)}]^2 \) ...[ \( \therefore a^2 - b^2 = (a + b) (a - b) \) ]
\( \therefore \) 49 x 1 = \( [\text{I(OP)}]^2 \)
\( \therefore \) \( [\text{I(OP)}]^2 = 49 \)
\( \therefore \) I(OP) = \( \sqrt{49} \) ...[Taking square root of both sides]
\( \therefore \) I(OP) = 7 cm
\( \therefore \) The distance of the chord from the centre of the circle is 7 cm.
In simple words: The perpendicular from the center to the chord bisects it, forming a right-angled triangle. Using the Pythagorean theorem with the radius and half the chord length, we can calculate the distance from the center to the chord.

🎯 Exam Tip: When dealing with chords and circles, always consider forming a right-angled triangle by drawing a perpendicular from the center to the chord. This allows for direct application of the Pythagorean theorem.

Question 3. O is centre of the circle. Find the length of radius, if the chord of length 24 cm is at a distance of 9 cm from the centre of the
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त दिखाया गया है जिसका केंद्र O है। वृत्त में एक जीवा AB है जिसकी लंबाई 24 सेमी है। केंद्र O से जीवा AB पर एक लंब OP खींचा गया है, जिसकी लंबाई 9 सेमी है। हमें वृत्त की त्रिज्या (AO) ज्ञात करनी है।
Answer: Let seg OP \( \perp \) chord AB
\( \therefore \) I(AP) = \( \frac{1}{2} \) I(AB) ... [Perpendicular drawn from the centre of a circle to its chord bisects the chord]
\( \therefore \) I(AP) = \( \frac{1}{2} \) x 24 ...[ \( \therefore \) I(AB) = 24 cm]
\( \therefore \) I(AP) = 12 cm ...(i)
In \( \triangle \)OPA, m\( \angle \)OPA = 90°
\( \therefore \) \( [\text{I(AO)}]^2 = [\text{I(OP)}]^2 + [\text{I(AP)}]^2 \) ... [Pythagoras theorem]
\( \therefore \) \( [\text{I(AO)}]^2 = (9)^2 + (12)^2 \) ... [From (i) and I(OP) = 9 cm]
= 81 + 144
\( \therefore \) \( [\text{I(AO)}]^2 = 225 \)
\( \therefore \) I(AO) = \( \sqrt{225} \) ... [Taking square root of both sides]
\( \therefore \) I(AO) = 15 cm
\( \therefore \) The length of radius of the circle is 15 cm.
In simple words: Given the length of a chord and its distance from the center, we use the property that the perpendicular from the center bisects the chord. This creates a right-angled triangle where half the chord, the distance from the center, and the radius form the sides, allowing us to find the radius using the Pythagorean theorem.

🎯 Exam Tip: Clearly identify the right-angled triangle formed by the radius, half the chord, and the distance from the center. Labeling these components correctly will prevent errors in applying the Pythagorean theorem.

Question 4. C is the centre of the circle whose radius is 10 cm. Find the distance of the chord from the centre if the length of the chord is 12 cm.
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त दिखाया गया है जिसका केंद्र C है और त्रिज्या 10 सेमी है। एक जीवा AB है जिसकी लंबाई 12 सेमी है। केंद्र C से जीवा AB पर एक लंब CD खींचा गया है, जहाँ D जीवा AB पर एक बिंदु है।
Answer: Let seg AB be the chord of the circle with centre C.
Draw seg CD \( \perp \) chord AB.
\( \therefore \) I(AD) = \( \frac{1}{2} \) I(AB) ...[Perpendicular drawn from the centre of a circle to its chord bisects the chord]
= \( \frac{1}{2} \) x 12 ...[ \( \therefore \) I(AB) = 12 cm]
\( \therefore \) I(AD) = 6 cm ...(i)
In \( \triangle \)ACD, m\( \angle \)ADC = 90°
\( \therefore \) \( [\text{I(AC)}]^2 = [\text{I(AD)}]^2 + [\text{I(CD)}]^2 \) ... [Pythagoras theorem]
\( \therefore \) \( (10)^2 = (6)^2 + [\text{I(CD)}]^2 \) ... [From (i) and I(AC) = 10 cm]
\( \therefore \) \( (10)^2 - (6)^2 = [\text{I(CD)}]^2 \)
\( \therefore \) 100 - 36 = \( [\text{I(CD)}]^2 \)
\( \therefore \) 64 = \( [\text{I(CD)}]^2 \)
i. e. \( [\text{I(CD)}]^2 = 64 \)
\( \therefore \) I(CD) = \( \sqrt{64} \) ...[Taking square root of both sides]
\( \therefore \) I(CD) = 8 cm
\( \therefore \) The distance of the chord from the centre of the circle is 8 cm.
In simple words: By drawing a perpendicular from the center to the chord, we form a right-angled triangle. Using the radius (hypotenuse) and half the chord length (one leg), we apply the Pythagorean theorem to find the distance of the chord from the center (the other leg).

🎯 Exam Tip: Always sketch the problem to visualize the right-angled triangle. This helps in correctly assigning values to the hypotenuse and legs for accurate application of the Pythagorean theorem.

Maharashtra Board Class 8 Maths Chapter 17 Circle: Chord And Arc Practice Set 17.1 Intext Questions And Activities

Question 1. In the given figure, O is the centre of the circle. With reference to the figure fill in the blanks. (Textbook pg. No. 114)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त दिखाया गया है जिसका केंद्र O है। वृत्त पर विभिन्न बिंदु A, B, D, P, Q, X चिह्नित हैं। एक जीवा PQ, एक व्यास AB और केंद्र O से होकर गुजरने वाली त्रिज्याएं OD और OB दिखाई गई हैं।
Answer: 1. Seg OD is radius of the circle.
2. Seg AB is diameter of the circle.
3. Seg PQ is chord of the circle.
4. \( \angle \)DOB is the central angle.
5. Minor arc: arc AXD, arc BD, arc AP, arc PQ, arc BQ, etc.
6. Major arc: arc PAB, arc PDQ, arc PDB, arc ADQ, etc.
7. Semicircular arc: arc ADB, arc AQB.
8. m (arc DB) = m\( \angle \)DOB
9. m (arc DAB) = 360°- m\( \angle \)DOB
In simple words: This question tests basic definitions related to a circle, including identifying the center, radius, diameter, chord, central angle, minor arc, major arc, and semicircular arc based on a given diagram.

🎯 Exam Tip: Thoroughly understand and memorize the definitions of different parts of a circle, such as chord, diameter, radius, arc, and central angle. Being able to identify these correctly from a diagram is fundamental.

Question 2. Draw chord AB of a circle with centre O. Draw perpendicular OP to chord AB. Measure seg AP and seg PB. What do you observe. (Textbook pg. no. 114)
ℹ️ चित्र व्याख्या (Diagram Explanation): एक वृत्त दिखाया गया है जिसका केंद्र O है। वृत्त के अंदर एक जीवा AB खींची गई है। केंद्र O से जीवा AB पर बिंदु P पर एक लंब OP डाला गया है। यह लंब जीवा AB को AP और PB दो भागों में विभाजित करता है।
Answer: I(AP) = I(PB) = 0.9 cm
\( \therefore \) the perpendicular drawn from the centre of the circle to its chord bisects the chord.
In simple words: By drawing a perpendicular from the center to a chord, we observe that it divides the chord into two equal segments. This demonstrates the theorem that the perpendicular from the center bisects the chord.

🎯 Exam Tip: Practical activities like drawing and measuring help solidify understanding of geometric theorems. This observation is a key property of circles, often used in problem-solving.

Question 3. Draw five circles with different radii. Draw a chord and perpendicular from the centre to each chord in each circle. Verify with a divider that the two parts of the chords are equal. (Textbook pg. no. 114)
Answer: [Students should attempt the above activities on their own.]
In simple words: This activity is designed to experimentally verify that a perpendicular drawn from the center of a circle to any chord always bisects that chord, regardless of the circle's size.

🎯 Exam Tip: Practical exercises like this reinforce theoretical knowledge. While not directly tested as a drawing task, understanding the outcome is essential for applying the chord bisection theorem in exams.

Question 4. Draw five circles of different radii on a paper. Draw a chord in each circle. Find the midpoint of each chord. Join the centre of the circle and midpoint of the chord as shown in the figure. Name the chord as AB and midpoint of the chord as P. Check with set-square or protractor that \( \angle \)APO or \( \angle \)BPO are right angles. Check whether the same result is observed for the chord of each circle. (Textbook pg, no. 115)
Answer: [Students should attempt the above activities on their own.]
In simple words: This activity aims to empirically show that the line segment connecting the center of a circle to the midpoint of a chord is always perpendicular to the chord, reinforcing another key property of circles.

🎯 Exam Tip: This activity demonstrates the converse of the perpendicular-from-center-to-chord theorem. Understanding both aspects is crucial for comprehensive problem-solving in circle geometry. Focus on the conclusion that the segment to the midpoint is perpendicular.