NCERT Solutions for Class 9 Maths: Circles Exercise 10.4

Key Theorems and Properties for Exercise 10.4

To conquer Exercise 10.4, you need to be comfortable with a few core geometric principles. These aren't just rules to memorize; they are the logical tools you'll use to construct your proofs and find your solutions.

1. Equal chords are equidistant from the center: This is Theorem 10.6. If you have a circle with center O and two chords AB and CD such that AB = CD, then the perpendicular distances from the center to these chords will be equal. If OM ⊥ AB and ON ⊥ CD, then OM = ON. This is fundamental for problems comparing two equal chords.

1. Chords equidistant from the center are equal: This is Theorem 10.7, the converse of the previous one. If you are given that two chords are at the same perpendicular distance from the center (i.e., OM = ON), you can confidently conclude that the chords themselves are equal in length (AB = CD).

1. Property of Intersecting Circles: This is the star of the show for many problems in this exercise. When two circles intersect at two points (say A and B), they create a common chord (AB). The line segment joining the centers of these two circles is the perpendicular bisector of this common chord. This means it cuts the chord at a 90° angle and divides it into two equal halves.

Solved Example: Finding the Length of a Common Chord

1. Problem Statement — Two circles with radii 13 cm and 15 cm intersect at two points. The distance between their centers is 14 cm. Find the length of the common chord.
2. Step 1: Draw and Label the Diagram — Draw two circles with centers O and P intersecting at A and B. Let the radii be OA = 13 cm and PA = 15 cm. The distance between centers is OP = 14 cm. The common chord is AB. Let OP intersect AB at M.
3. Step 2: Apply the Perpendicular Bisector Property — We know that the line joining the centers (OP) is the perpendicular bisector of the common chord (AB). Therefore, AM = MB and ∠AMO = 90°. The length of the common chord AB will be 2 * AM.
4. Step 3: Set up Triangles using Pythagoras Theorem — In right-angled triangle ΔAMO, we have OA² = AM² + OM². So, 13² = AM² + OM², which gives 169 = AM² + OM² (Equation 1). Let OM = x, then MP = 14 - x.
5. Step 4: Apply Pythagoras Theorem to the Second Triangle — In right-angled triangle ΔAMP, we have PA² = AM² + MP². So, 15² = AM² + (14-x)². This gives 225 = AM² + 196 - 28x + x² (Equation 2).
6. Step 5: Solve the Equations — From Equation 1, AM² = 169 - OM² = 169 - x². Substitute this into Equation 2: 225 = (169 - x²) + 196 - 28x + x². The x² terms cancel out. We get 225 = 365 - 28x. Solving for x, we get 28x = 140, so x = 5 cm. Thus, OM = 5 cm.
7. Step 6: Calculate the Length of the Chord — Now substitute x = 5 back into Equation 1: 169 = AM² + 5². This gives AM² = 169 - 25 = 144. Therefore, AM = √144 = 12 cm. Since the length of the common chord AB = 2 AM, we have AB = 2 12 = 24 cm.

Exam Tips & Common Mistakes

Always Draw a Diagram: Before you write a single equation, draw a neat and labeled diagram. This visual aid is your best friend in geometry. Mark the center, radii, chords, and any perpendiculars clearly. It helps you see the relationships and form the correct triangles.

Common Mistakes to Avoid:

• Assuming Perpendicularity: Don't assume a line from the center to a chord is perpendicular unless it's given or you've proven it. The 'distance' to a chord is always the perpendicular distance.
• Pythagoras Errors: Be careful with your calculations when using the Pythagoras theorem (a² + b² = c²). Remember that 'c' is always the hypotenuse (the side opposite the right angle), which is usually the radius in these problems.
• Incomplete Proofs: When proving congruency, don't just write 'SAS' or 'SSS'. Justify each part. For example: "OA = OB (radii of the same circle)", "OM = OM (common side)". This shows the examiner you understand the logic.

Practice Questions with Solutions

• Q: In a circle with a radius of 10 cm, two equal parallel chords AB and CD are drawn, each of length 12 cm. Find the distance between the chords. A: Step 1: Draw a circle with center O and radius 10 cm. Draw two parallel chords AB and CD, each 12 cm long. Draw a perpendicular from O to both chords, intersecting AB at M and CD at N. Step 2: The perpendicular from the center bisects the chord. So, AM = MB = 12/2 = 6 cm. Step 3: Consider the right-angled triangle ΔOMA. We have OA (hypotenuse) = 10 cm and AM = 6 cm. Using Pythagoras theorem, OA² = OM² + AM². So, 10² = OM² + 6². Step 4: Calculate OM. OM² = 100 - 36 = 64. So, OM = √64 = 8 cm. Since the chords are equal, their distance from the center is the same. Thus, ON = 8 cm. Step 5: The problem implies the chords could be on the same or opposite sides. If on opposite sides, the distance MN = OM + ON = 8 + 8 = 16 cm. If on the same side, the distance MN = |OM - ON| = 8 - 8 = 0 cm, which means they are the same chord. Final answer: The distance between the chords is 16 cm (assuming they are distinct parallel chords on opposite sides).
• Q: Two chords AB and CD of a circle are equidistant from the center O. If the radius is 17 cm and the distance of the chords from the center is 8 cm, find the length of the chords. A: Step 1: Since the chords are equidistant from the center, they must be equal in length (Theorem 10.7). So, AB = CD. We only need to find the length of one chord. Step 2: Let's find the length of AB. Draw a perpendicular OM from O to AB. We are given OM = 8 cm. Join OA, which is the radius, so OA = 17 cm. Step 3: In the right-angled triangle ΔOMA, apply the Pythagoras theorem: OA² = OM² + AM². Step 4: Substitute the given values: 17² = 8² + AM². This gives 289 = 64 + AM². So, AM² = 289 - 64 = 225. Step 5: Calculate AM. AM = √225 = 15 cm. The perpendicular from the center bisects the chord, so the full length of the chord AB = 2 AM. Final answer: The length of the chords is 2 15 = 30 cm each.
• Q: Three girls Reshma, Salma, and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip? A: Step 1: Let the positions of Reshma, Salma, and Mandip be R, S, and M respectively. Let the center of the circle be O. We are given RS = SM = 6m and the radius OR = OS = OM = 5m. Step 2: In ΔOSM, OS = OM = 5m, so it is an isosceles triangle. Draw a perpendicular from O to SM, let's call the point B. B is the midpoint of SM, so SB = 3m. Step 3: In right-angled ΔOBS, OS² = OB² + SB². So, 5² = OB² + 3², which gives OB² = 25 - 9 = 16. Thus, OB = 4m. Step 4: The points R, O, and B are collinear because ΔORS and ΔOMS are congruent isosceles triangles sharing a common base OS. The line joining the vertex to the midpoint of the base in an isosceles triangle is the altitude. Let the line RM intersect OS at A. Step 5: Area of ΔOSM = (1/2) base height = (1/2) SM OB = (1/2) 6 4 = 12 m². Also, Area(ΔOSM) = (1/2) OS AM = (1/2) 5 AM. So, (1/2) 5 AM = 12, which gives AM = 24/5 = 4.8m. Step 6: Since RM is a chord and OA is perpendicular to it, A is the midpoint of RM. Therefore, RM = 2 AM = 2 4.8 = 9.6m. Final answer: The distance between Reshma and Mandip is 9.6 m.
• Q: A circular park of radius 20m is situated in a colony. Two equal chords are drawn from a point on its boundary. If the length of each chord is 24m, find the distance between the ends of the two chords. A: Step 1: Let the point on the boundary be A, and the two equal chords be AB and AC, with AB = AC = 24m. Let the center of the circle be O. The radius is 20m. We need to find the length of the chord BC. Step 2: Draw perpendiculars OM and ON from O to AB and AC respectively. Since the chords are equal, they are equidistant from the center, so OM = ON. Step 3: In right-angled ΔOMA, OA (radius) = 20m. M is the midpoint of AB, so AM = 24/2 = 12m. Using Pythagoras theorem, OA² = OM² + AM², so 20² = OM² + 12². Step 4: Calculate OM. OM² = 400 - 144 = 256. So, OM = 16m. This means ON = 16m as well. Step 5: Consider the quadrilateral AMON. Since ∠AMO = ∠ANO = 90° and OM = ON, and AM = AN = 12m, ΔAMO ≅ ΔANO. By CPCTC, ∠MAO = ∠NAO. Thus, AO is the angle bisector of ∠BAC. In ΔABC, since AB=AC, AO is also the perpendicular bisector of BC. Step 6: Let AO intersect BC at P. The area of ΔABC can be calculated in two ways. Area(ΔOAB) = (1/2) AB OM = (1/2) 24 16 = 192 m². Similarly, Area(ΔOAC) = 192 m². Area(ΔABC) = Area(ΔOAB) + Area(ΔOAC) - 2Area(ΔOBC) -- This is complex. Let's use coordinates or areas differently. Area(ΔABC) = Area(ΔOAB) + Area(ΔOAC) = 2 (1/2) AM OM = 2 192 = 384? No. Area(ΔABC) = Area(ΔOAB) + Area(ΔOAC) if O is outside. Let's use Area(ΔABC) = (1/2) BC AP. Area(ΔABC) = Area(ΔOAB) + Area(ΔOAC) - Area(ΔOBC). This is getting complicated. Let's use a simpler method. Step 5 (Revised): In kite AMON, the diagonal AO is the perpendicular bisector of the diagonal MN. Let K be the intersection. Let's find cos(∠OAM) = AM/OA = 12/20 = 3/5. Then ∠BAC = 2 ∠OAM. Using the Law of Cosines in ΔABC: BC² = AB² + AC² - 2(AB)(AC)cos(∠BAC). We need cos(∠BAC) = cos(2∠OAM) = 2cos²(∠OAM) - 1 = 2(3/5)² - 1 = 2(9/25) - 1 = 18/25 - 1 = -7/25. Step 6 (Revised): BC² = 24² + 24² - 2(24)(24)(-7/25) = 2 576 (1 + 7/25) = 1152 (32/25). BC = sqrt(1152 32 / 25) = sqrt(5762 32 / 25) = (24 8) / 5 = 192 / 5 = 38.4m. Final answer: The distance between the ends of the chords is 38.4m.

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