ML Aggarwal Class 9 Maths Solutions Chapter 14 Theorems on Area

Question 1. Calculate the length of a chord which is at a distance 12 cm from the centre of a circle of radius 13 cm.
Answer: In the right-angled triangle OAC, apply Pythagoras' theorem: \( OA^2 = OC^2 + AC^2 \)
\( 13^2 = 12^2 + AC^2 \)
\( AC^2 = 169 - 144 = 25 \)
\( AC = 5 \) cm. Since the perpendicular from the centre of a circle to a chord divides it into two equal parts, \( CB = AC = 5 \) cm. Therefore, \( AB = AC + CB = 5 + 5 = 10 \) cm.
In simple words: When a line from the circle's centre meets a chord at right angles, it cuts the chord in half. Using Pythagoras' theorem, we find each half is 5 cm, so the whole chord is 10 cm.

Exam Tip: Always remember that the perpendicular from the centre bisects the chord - this is a key property that appears in every circle problem involving chords and distances.

Question 2. A chord of length 48 cm is drawn in a circle of radius 25 cm. Calculate its distance from the centre of the circle.
Answer: Since the perpendicular from the centre bisects the chord, the half-length of the chord is \( \frac{48}{2} = 24 \) cm. Using the right-angled triangle OAC and Pythagoras' theorem: \( OA^2 = OC^2 + AC^2 \)
\( 25^2 = OC^2 + 24^2 \)
\( OC^2 = 625 - 576 = 49 \)
\( OC = 7 \) cm. Therefore, the chord is at a distance of 7 cm from the centre.
In simple words: Split the chord in half to get 24 cm. Then use Pythagoras to find how far the centre is from the chord - it turns out to be 7 cm.

Exam Tip: When given a chord length and radius, always find the distance by first halving the chord and then using Pythagoras' theorem in reverse (finding the unknown perpendicular distance).

Question 3. A chord of length 8 cm is at a distance 3 cm from the centre of the circle. Calculate the radius of the circle.
Answer: The perpendicular from the centre to the chord divides it into two equal parts: \( AC = BC = \frac{8}{2} = 4 \) cm. In the right-angled triangle OAC, using Pythagoras' theorem: \( OA^2 = OC^2 + AC^2 \)
\( OA^2 = 3^2 + 4^2 \)
\( OA^2 = 9 + 16 = 25 \)
\( OA = 5 \) cm. Therefore, the radius is 5 cm.
In simple words: Half the chord is 4 cm. The perpendicular from centre is 3 cm. Apply Pythagoras to find the radius is 5 cm.

Exam Tip: This is a classic 3-4-5 Pythagorean triple - recognising common triangles like this can save calculation time in exams.

Question 4. Calculate the length of a chord which is at a distance 6 cm from the center of a circle of diameter 20 cm.
Answer: First, find the radius: diameter = 20 cm, so radius = 10 cm. The perpendicular from the centre bisects the chord. Using Pythagoras' theorem in triangle OAC: \( OA^2 = OC^2 + AC^2 \)
\( 10^2 = 6^2 + AC^2 \)
\( AC^2 = 100 - 36 = 64 \)
\( AC = 8 \) cm. Since the chord is bisected, \( CB = 8 \) cm. Therefore, \( AB = AC + CB = 8 + 8 = 16 \) cm.
In simple words: First work out the radius (half the diameter). Then use Pythagoras to find half the chord is 8 cm, making the full chord 16 cm.

Exam Tip: Always convert diameter to radius at the start - forgetting this step is a common mistake that loses marks unnecessarily.

Question 5. A chord of length 16 cm is at a distance 6 cm from the center of the circle. Find the length of chord of the same circle which is at a distance of 8 cm from the center.
Answer: For the first chord AB, the half-length is \( AC = \frac{16}{2} = 8 \) cm. Using Pythagoras in triangle OAC: \( OA^2 = OC^2 + AC^2 \)
\( OA^2 = 6^2 + 8^2 = 36 + 64 = 100 \)
\( OA = 10 \) cm. So the radius is 10 cm. For the second chord at distance 8 cm from the centre (OF = 8 cm), using triangle ODF: \( OD^2 = OF^2 + DF^2 \)
\( 10^2 = 8^2 + DF^2 \)
\( DF^2 = 100 - 64 = 36 \)
\( DF = 6 \) cm. Since the perpendicular bisects the chord, \( DE = 2 \times DF = 2 \times 6 = 12 \) cm.
In simple words: First find the radius (10 cm) using the first chord. Then use that radius with the new distance (8 cm) to work out the second chord is 12 cm.

Exam Tip: Two-step chord problems require you to first find the radius using one chord, then apply it to find the unknown chord - don't skip the intermediate radius calculation.

Question 6. In a circle of radius 5 cm, AB and CD are two parallel chords of length 8 cm and 6 cm respectively. Calculate the distance between the chords, if they are on (i) the same side of the centre (ii) the opposite sides of the centre.
Answer:
(i) Same Side of the Centre: For chord AB (length 8 cm): \( AE = BE = \frac{8}{2} = 4 \) cm. Using Pythagoras in triangle OAE: \( OE^2 = OA^2 - AE^2 = 5^2 - 4^2 = 25 - 16 = 9 \), so \( OE = 3 \) cm. For chord CD (length 6 cm): \( CF = FD = \frac{6}{2} = 3 \) cm. Using Pythagoras in triangle OCF: \( OF^2 = OC^2 - CF^2 = 5^2 - 3^2 = 25 - 9 = 16 \), so \( OF = 4 \) cm. The distance between the chords is \( EF = OF - OE = 4 - 3 = 1 \) cm.

(ii) Opposite Sides of the Centre: Using the same calculations, \( OE = 3 \) cm and \( OF = 4 \) cm. Since the chords are on opposite sides of the centre, the distance between them is \( EF = OF + OE = 4 + 3 = 7 \) cm.
In simple words: Find how far each chord is from the centre using Pythagoras. If they're on the same side, subtract these distances. If they're on opposite sides, add them.

Exam Tip: Always draw a diagram showing whether the chords are on the same side or opposite sides - this determines whether you add or subtract the perpendicular distances.

Question 7(a). In figure (i) given below, O is the center of the circle. AB and CD are two chords of the circle. OM is perpendicular to AB and ON is perpendicular to CD. AB = 24 cm, OM = 5 cm, ON = 12 cm. Find the: (i) radius of the circle (ii) length of chord CD.
Answer:
(i) Finding the Radius: Since the perpendicular from the centre bisects the chord, \( AM = BM = \frac{24}{2} = 12 \) cm. In right-angled triangle OAM, using Pythagoras' theorem: \( OA^2 = OM^2 + AM^2 \)
\( OA^2 = 5^2 + 12^2 = 25 + 144 = 169 \)
\( OA = 13 \) cm. Therefore, the radius is 13 cm.

(ii) Finding the Length of Chord CD: Since \( OC = \) radius = 13 cm, in right-angled triangle OCN: \( OC^2 = ON^2 + CN^2 \)
\( 13^2 = 12^2 + CN^2 \)
\( CN^2 = 169 - 144 = 25 \)
\( CN = 5 \) cm. Since the perpendicular bisects the chord, \( CD = 2 \times CN = 2 \times 5 = 10 \) cm.
In simple words: Use the first chord and its distance to find the radius. Then use that radius with the second chord's distance to find its length.

Exam Tip: The 5-12-13 Pythagorean triple appears here - recognising such triples speeds up calculations and reduces arithmetic errors.

Question 7(b). In the figure (ii) given below, CD is the diameter which meets the chord AB in E such that AE = BE = 4 cm. If CE = 3 cm, find the radius of the circle.
Answer: We are given that AB = 8 cm and EC = 3 cm. Let the radius be r, so OB = OC = r. Then OE = (r - 3) cm. Since chord AB is cut in half by OE, the line OE must be perpendicular to AB. In right triangle OBE, we use the Pythagorean theorem: \( r^2 = 4^2 + (r - 3)^2 \)
\( \implies r^2 = 16 + r^2 - 6r + 9 \)
\( \implies 6r = 25 \)
\( \implies r = \frac{25}{6} = 4\frac{1}{6} \) cm.
In simple words: When a line from the circle's center cuts a chord in half, it makes a right angle with that chord. Using this fact and the Pythagorean theorem, we can find the radius.

Exam Tip: Remember that a perpendicular from the centre to a chord always bisects the chord - this property is key to solving many circle problems.

Question 8. In the adjoining figure, AB and CD are two parallel chords and O is the centre. If the radius of the circle is 15 cm, find the distance MN between the two chords of length 24 cm and 18 cm respectively.
Answer: The radius is 15 cm, with chord AB = 24 cm and chord CD = 18 cm. Join OA and OC. Draw OM perpendicular to AB; this gives us AM = MB = 12 cm since the perpendicular from the centre bisects a chord. In right triangle OAM, \( OM^2 = OA^2 - AM^2 = 15^2 - 12^2 = 225 - 144 = 81 \)
\( \implies OM = 9 \) cm. Similarly, draw ON perpendicular to CD, giving CN = ND = 9 cm. In right triangle OCN, \( ON^2 = OC^2 - CN^2 = 15^2 - 9^2 = 225 - 81 = 144 \)
\( \implies ON = 12 \) cm. Therefore, MN = OM + ON = 9 + 12 = 21 cm.
In simple words: Find how far the centre is from each chord using right triangles, then add these distances together to get the gap between the chords.

Exam Tip: Always draw perpendiculars from the centre to parallel chords - these distances add together if the chords are on opposite sides of the centre.

Question 9. AB and CD are two parallel chords of a circle of lengths 10 cm and 4 cm respectively. If the chords lie on the same side of the centre and the distance between them is 3 cm, find the diameter of the circle.
Answer: Let OE = x cm, where E and F are the feet of perpendiculars from O to chords AB and CD. For chord AB with length 10 cm, we have AE = 5 cm. For chord CD with length 4 cm, we have CF = 2 cm. Since both chords are on the same side and the distance between them is 3 cm, we have OF = (x + 3) cm. Using the Pythagorean theorem in triangle OCF: \( OC^2 = (x + 3)^2 + 2^2 = x^2 + 6x + 13 \). Since OA = OC (both radii), we have \( OA^2 = x^2 + 6x + 13 \). In right triangle OAE: \( OA^2 = x^2 + 5^2 = x^2 + 25 \). Setting these equal: \( x^2 + 25 = x^2 + 6x + 13 \)
\( \implies 6x = 12 \)
\( \implies x = 2 \) cm. Then \( OC^2 = 4 + 12 + 13 = 29 \), so the radius is \( \sqrt{29} \) cm. The diameter is \( 2\sqrt{29} \) cm.
In simple words: Set up equations for the distances from the centre to each chord using the Pythagorean theorem. Since both chords touch the same circle, their radii must match, which gives you an equation to solve.

Exam Tip: When chords are on the same side of the centre, add the distance between them to one perpendicular distance to find the other - this is what "same side" means geometrically.

Question 10. ABC is an isosceles triangle inscribed in a circle. If AB = AC = 12\(\sqrt{5}\) cm and BC = 24 cm, find the radius of the circle.
Answer: Let the radius be r. Since the triangle is isosceles with AB = AC, the line from A to the midpoint D of BC is perpendicular to BC, and D lies on the line from the centre O to the circle. We have BD = DC = 12 cm. In right triangle ABD: \( (12\sqrt{5})^2 = AD^2 + 12^2 \)
\( \implies 720 = AD^2 + 144 \)
\( \implies AD = 24 \) cm. Now OD = AD - r = (24 - r) cm. In right triangle OBD, using the Pythagorean theorem: \( r^2 = (24 - r)^2 + 12^2 \)
\( \implies r^2 = 576 - 48r + r^2 + 144 \)
\( \implies 48r = 720 \)
\( \implies r = 15 \) cm.
In simple words: Draw a line from the top of the triangle to the midpoint of the base. This line passes through the centre of the circle. Use this to set up equations and find the radius.

Exam Tip: For isosceles triangles, the perpendicular from the equal-side vertex to the base always passes through the circle's centre - use this symmetry to simplify your work.

Question 11. An equilateral triangle of side 6 cm is inscribed in a circle. Find the radius of the circle.
Answer: Since the triangle is equilateral with side 6 cm, all sides are equal. Let the radius be r. The perpendicular from vertex A to the opposite side BC meets BC at its midpoint D, and this line passes through the circle's centre O. We have BD = DC = 3 cm. In right triangle ABD: \( 6^2 = AD^2 + 3^2 \)
\( \implies AD^2 = 27 \)
\( \implies AD = 3\sqrt{3} \) cm. Therefore, OD = AD - r = (3\sqrt{3} - r) cm. In right triangle OBD: \( r^2 = (3\sqrt{3} - r)^2 + 3^2 \)
\( \implies r^2 = 27 - 6\sqrt{3}r + r^2 + 9 \)
\( \implies 6\sqrt{3}r = 36 \)
\( \implies r = \frac{36}{6\sqrt{3}} = 2\sqrt{3} \) cm.
In simple words: In an equilateral triangle, the perpendicular from any vertex to the opposite side goes through the centre of its circle. Use this and the Pythagorean theorem to find the radius.

Exam Tip: For an equilateral triangle inscribed in a circle, the height and radius are directly related - memorizing the relationship \( r = \frac{a}{\sqrt{3}} \) where \( a \) is the side length can save time.

Question 12. AB is a diameter of a circle. M is a point in AB such that AM = 18 cm and MB = 8 cm. Find the length of the shortest chord through M.
Answer: We have AB = AM + MB = 18 + 8 = 26 cm, so the radius is 13 cm. The shortest chord passing through a point M inside a circle is the one that is perpendicular to the line OM. Let this chord be CD. We have OM = AM - AO = 18 - 13 = 5 cm. In right triangle OMC (where C is an endpoint of the chord), \( OC^2 = OM^2 + MC^2 \)
\( \implies 13^2 = 5^2 + MC^2 \)
\( \implies MC^2 = 169 - 25 = 144 \)
\( \implies MC = 12 \) cm. Since the perpendicular from the centre to a chord bisects it, M is the midpoint of CD. Therefore, CD = 2 × MC = 2 × 12 = 24 cm.
In simple words: The shortest chord through a point is the one standing at right angles to the line joining that point to the centre. Once you know this distance and the radius, the Pythagorean theorem gives you the chord length.

Exam Tip: The shortest chord through any interior point is always perpendicular to the radius ending at that point - this is a key principle that saves you from checking multiple chords.

Question 13. A rectangle with one side of length 4 cm is inscribed in a circle of diameter 5 cm. Find the area of rectangle.
Answer: Let BC = 4 cm be one side of the rectangle. The diameter is 5 cm, so the radius is 2.5 cm. Since the perpendicular from the centre O to any chord bisects the chord, we have BM = MC = 2 cm (where M is the midpoint of BC). The diagonal of the rectangle equals the diameter of the circle, which is 5 cm. In a rectangle ABCD inscribed in a circle, the diagonal AC = 5 cm (the diameter). In right triangle ABC, \( AC^2 = AB^2 + BC^2 \)
\( \implies 5^2 = AB^2 + 4^2 \)
\( \implies 25 = AB^2 + 16 \)
\( \implies AB^2 = 9 \)
\( \implies AB = 3 \) cm. Therefore, the area of the rectangle = AB × BC = 3 × 4 = 12 cm².
In simple words: The diagonal of any rectangle inscribed in a circle must equal the diameter. Use this fact along with one given side to find the other side using the Pythagorean theorem, then multiply to get the area.

Exam Tip: Always recall that for any rectangle inscribed in a circle, the diagonal equals the circle's diameter - this single fact unlocks most rectangle-in-circle problems.

Question 14. The length of the common chord of two intersecting circles is 30 cm. If the radii of the two circles are 25 cm and 17 cm, find the distance between their centres.
Answer: When a perpendicular is drawn from a circle's center to any chord, it divides that chord into two equal parts. Therefore, the chord is split into two segments of 15 cm each.

From the diagram, we apply the Pythagorean theorem to right triangle OAC:
\( OA^2 = OC^2 + AC^2 \)
\( 25^2 = OC^2 + 15^2 \)
\( 625 = OC^2 + 225 \)
\( OC^2 = 400 \)
\( OC = 20 \text{ cm} \)

Similarly, in right triangle O'AC:
\( O'A^2 = O'C^2 + AC^2 \)
\( 17^2 = O'C^2 + 15^2 \)
\( 289 = O'C^2 + 225 \)
\( O'C^2 = 64 \)
\( O'C = 8 \text{ cm} \)

The distance separating the two centers equals the sum of these two distances:
\( OO' = OC + O'C = 20 + 8 = 28 \text{ cm} \)
In simple words: The perpendicular from each circle's center to the shared chord splits it evenly. Using Pythagoras on each triangle, we find how far each center sits from the chord, then add these distances to get 28 cm between the centers.

Exam Tip: Always use the property that the perpendicular from the center bisects a chord, then apply Pythagoras theorem correctly to each right triangle formed.

Question 15. The line joining mid-points of two chords of a circle passes through its center. Prove that the chords are parallel.
Answer: Consider a circle with center O. Let AB and CD be two chords, with M and N as the midpoints of AB and CD respectively. The line MN passes through the center O.

By the fundamental property of circles, when a perpendicular is drawn from the center to a chord, it bisects that chord. Therefore:
\( OM \perp AB \) and \( ON \perp CD \)

This means:
\( \angle OMA = \angle OMB = 90° \) and \( \angle ONC = \angle OND = 90° \)

Since \( \angle OMA = \angle OND = 90° \) (these are alternate angles with respect to the transversal MN crossing the two chords), and \( \angle OMB = \angle ONC = 90° \) (also alternate angles), we can conclude that AB is parallel to CD. When two lines are both perpendicular to the same transversal, they must be parallel to each other.
In simple words: The line from the center to each chord's midpoint is perpendicular to that chord. Since both chords are perpendicular to line MN, they must run parallel to each other.

Exam Tip: The key insight is recognizing that a perpendicular to both chords means those chords are parallel to each other. Mark the 90° angles clearly in your diagram.

Question 16. If a diameter of a circle is perpendicular to one of two parallel chords of the circle, prove that it is perpendicular to the other and bisects it.
Answer: Let AB and CD be two parallel chords of a circle with center O, and suppose the diameter MN is perpendicular to AB at point M, meaning \( \angle OMA = \angle OMB = 90° \).

Since AB is parallel to CD, and MN acts as a transversal cutting across both chords, alternate angles are equal. Therefore:
\( \angle OMA = \angle OND = 90° \) (alternate angles)
\( \angle OMB = \angle ONC = 90° \) (alternate angles)

This tells us that \( ON \perp CD \) or \( MN \perp CD \).

A key theorem states that a perpendicular drawn from the center of a circle to any chord bisects that chord. Since we have shown that MN is perpendicular to CD, it must also bisect it.

Therefore, \( NC = ND \), which means the diameter is perpendicular to the other chord and bisects it at N.
In simple words: When two chords are parallel and a line is perpendicular to one, that same line must be perpendicular to the other as well. A perpendicular from the center always splits a chord into two equal pieces.

Exam Tip: Use the alternate angles property carefully when dealing with parallel lines. Show that the perpendicular condition carries from one chord to the other.

Question 17. In an equilateral triangle, prove that the centroid and the circumcentre of the triangle coincide.
Answer: Consider equilateral triangle ABC with all sides equal and all angles measuring 60°. Let AD, BE, and CF be the three medians (lines from vertices to midpoints of opposite sides), and let G be the centroid (where the medians meet).

First, examine triangles BFC and BEC:
\( BC = BC \) (common side)
\( \angle FBC = \angle ECB = 60° \)
\( BF = EC \) (since F is the midpoint of AB, E is the midpoint of AC, and AB = AC)

By the SAS (Side-Angle-Side) criterion, triangle BFC is congruent to triangle BEC. Using Corresponding Parts of Congruent Triangles (CPCT), we get \( BE = CF \).

Next, compare triangles ABE and ABD:
\( AB = AB \) (common side)
\( \angle BAE = \angle ABD = 60° \)
\( BD = AE \) (since D is the midpoint of BC, E is the midpoint of AC, and BC = AC)

By SAS, triangle ABE is congruent to triangle ABD, so \( BE = AD \) by CPCT.

From these two results: \( AD = BE = CF \)

The centroid divides each median in a 2:1 ratio from the vertex. Therefore:
\( \frac{2}{3} AD = \frac{2}{3} BE = \frac{2}{3} CF \)

This means \( GA = GB = GC \). Since G is equidistant from all three vertices, it is the circumcentre (the center of the circumscribed circle).

Thus, the centroid and circumcentre coincide in an equilateral triangle.
In simple words: All three medians of an equilateral triangle have the same length. The point where they meet (the centroid) sits the same distance from each vertex, making it also the circumcentre.

Exam Tip: Show the congruence of triangles step-by-step using SAS, and explicitly state that equal median lengths lead to equal distances from G to the vertices.

Question 18(a). In the figure (i) given below, OD is perpendicular to the chord AB of a circle whose center is O. If BC is a diameter, show that CA = 2OD.
Answer: Since the perpendicular from the center to a chord bisects that chord, we have \( AD = DB \), so D is the midpoint of AB.

Given that BC is a diameter and O is the center, we know \( OB = OC = \) radius, which means O is the midpoint of BC.

In triangle ABC, D is the midpoint of side AB and O is the midpoint of side BC. By the midpoint theorem (also called the midsegment theorem), the line segment joining the midpoints of two sides of a triangle is parallel to the third side and equals half its length.

Therefore:
\( OD \parallel AC \) and \( OD = \frac{1}{2} AC \)

Rearranging: \( AC = 2OD \)

Thus, \( CA = 2OD \).
In simple words: D divides AB in half, and O divides BC in half. The segment OD connecting these midpoints is half the length of AC and runs parallel to it.

Exam Tip: Identify the midpoints clearly and apply the midpoint theorem directly. This is a classic application showing why the theorem is useful in circle geometry.

Question 18(b). In the figure (ii) given below, O is the center of a circle. If AB and AC are chords of the circle such that AB = AC and OP ⊥ AB, OQ ⊥ AC, prove that PB = QC.
Answer: Let \( AB = AC = x \).

Since a perpendicular from the center to a chord bisects the chord:
\( AM = MB = \frac{x}{2} \) and \( AN = NC = \frac{x}{2} \)

Therefore, \( MB = NC \) ... (1)

Equal chords in a circle are equidistant from the center. Since AB = AC, we have:
\( ON = OM = y \) (let's call this distance y)

Let the radius of the circle be r. From the diagram:
\( OQ = OP = r \)
\( QN = OQ - ON = r - y \)
\( PM = OP - OM = r - y \)

Therefore, \( QN = PM \) ... (2)

Now consider triangles QNC and PMB:
\( NC = MB \) (from equation 1)
\( QN = PM \) (from equation 2)
\( \angle QNC = \angle PMB = 90° \) (both equal to 90°)

By the SAS (Side-Angle-Side) criterion, \( \triangle QNC \cong \triangle PMB \).

Using CPCT (Corresponding Parts of Congruent Triangles), we obtain \( PB = QC \).
In simple words: Equal chords sit the same distance from the center. The perpendiculars to these chords create two triangles that are identical in size and shape, so their remaining sides PB and QC must be equal too.

Exam Tip: Use the property that equal chords are equidistant from the center, then set up the congruent triangle pairs correctly using all three measurements (sides and angles).

Question 19(a). In the figure (i) given below, a line l intersects two concentric circles at the points A, B, C and D. Prove that AB = CD.
Answer: Draw OM perpendicular from the center O to the line l.

Since a perpendicular from the center to a chord bisects the chord, M serves as the midpoint of BC for the smaller circle. Let \( BM = MC = x \).

Similarly, for the larger circle, M is the midpoint of AD. Let \( AM = MD = y \).

From the diagram, the segment AB can be expressed as:
\( AB = AM - BM = y - x \)

The segment CD can be expressed as:
\( CD = MD - MC = y - x \)

Therefore, \( AB = CD \).
In simple words: A single perpendicular from the center bisects both chords formed by the line's intersection with the two circles. The "outer" portions (what's left after removing the inner chord parts) are equal.

Exam Tip: Draw the perpendicular OM first and mark the two midpoints clearly. Express AB and CD in terms of the same variables to show they are equal.

Question 19(b). In the figure (ii) given below, chords AB and CD of a circle with centre O intersect at E. If OE bisects ∠AED, prove that AB = CD.
Answer: Draw perpendiculars from O to chords AB and CD, meeting them at points M and N respectively.

Consider triangles OME and ONE:
\( \angle OME = \angle ONE = 90° \) (by construction, since OM and ON are perpendiculars)
\( \angle OEM = \angle OEN \) (since OE bisects ∠AED)
\( OE = OE \) (common side)

By the AAS (Angle-Angle-Side) criterion, \( \triangle OME \cong \triangle ONE \).

Using CPCT, we get \( OM = ON \).

This means chords AB and CD are equidistant from the center O. A fundamental theorem states that in the same circle, chords that are equidistant from the center are equal in length.

Therefore, \( AB = CD \).
In simple words: The angle bisector OE creates two identical right triangles. These show that the perpendicular distances from O to the two chords are equal, which means the chords themselves must be equal.

Exam Tip: Establish the congruence using all available angle and side information, then apply the property that equidistant chords from the center are equal.

Question 20(a). In the figure (i) given below, AD is a diameter of a circle with center O. If AB || CD, prove that AB = CD.
Answer: Draw OM ⊥ AB and ON ⊥ CD. In triangles OAM and ODN, we have OA = OD (both are radii of the circle), angle AOM = angle DON (vertically opposite angles are equal), and angle OMA = angle OND (both equal 90°). By the A.S.A. axiom, triangle OAM is congruent to triangle ODN. Using the Corresponding Parts of Congruent Triangles (C.P.C.T.), we get ND = AM. Since a perpendicular drawn from the center of a circle to a chord bisects that chord, we can write AM = AB/2 and ND = CD/2. Therefore, AB/2 = CD/2, which gives us AB = CD.
In simple words: When you draw perpendiculars from the center to two parallel chords, the two right triangles formed are congruent. This means the two chords must be equal in length.

Exam Tip: Highlight the use of the congruence criterion (A.S.A.) and the property that perpendiculars from the center bisect chords - these are the key reasoning steps examiners look for in circle geometry proofs.

Question 20(b). In the figure (ii) given below, AB and CD are equal chords of a circle with center O. If AB and CD meet at E (outside the circle) prove that (i) AE = CE (ii) BE = DE.
Answer: Draw ON ⊥ CD and OM ⊥ AB. Join OE. (i) Since equal chords are the same distance from the center of a circle, we have ON = OM. In triangles ONE and OME, we know ON = OM, angle ONE = angle OME (both equal to 90°), and OE = OE (common side). By the R.H.S. (Right angle - Hypotenuse - Side) congruence rule, triangle ONE is congruent to triangle OME. Using C.P.C.T., we get NE = ME = y (let). Now, let AB = CD = x. Since the perpendicular from the center bisects a chord, we have CN = ND = CD/2 = x/2 and AM = MB = AB/2 = x/2. From the figure, AE = AM + ME = x/2 + y and CE = CN + NE = x/2 + y. Therefore, AE = CE. (ii) From the figure, BE = ME - MB = y - x/2 and DE = NE - ND = y - x/2. Therefore, BE = DE.
In simple words: When two equal chords extended outside a circle meet at a point, that point is equidistant from the endpoints of each chord on both chords.

Exam Tip: The key is recognizing that equal chords are equidistant from the center, which leads to congruent triangles and equal segments. Show both sub-parts (i) and (ii) clearly with the algebraic setup for full marks.

Exercise 14.2

Question 1. If arcs APB and CQD of a circle are congruent, then find the ratio of AB : CD.
Answer: We are given that arc APB = arc CQD. There is a basic property in circle geometry: when two arcs are equal, their corresponding chords are also equal. Therefore, AB = CD = x (let). The ratio is AB/CD = x/x = 1/1. Hence, AB : CD = 1 : 1.
In simple words: Equal arcs always produce equal chords. So if two arcs are the same size, the line segments connecting their endpoints will also be the same length.

Exam Tip: Remember the relationship: equal arcs → equal chords. This is a fundamental theorem and forms the basis for many circle geometry proofs. State it clearly in your answer.

Question 2. A and B are points on a circle with center O. C is a point on the circle such that OC bisects ∠AOB, prove that OC bisects the arc AB.
Answer: Let C be a point on the circle such that OC bisects angle AOB. This means angle AOC = angle BOC. There is a key property: equal angles at the center of a circle subtend equal arcs. Since the angles are equal, the arcs they subtend must also be equal. Therefore, arc AC = arc BC. When the two arcs AC and BC are equal, point C must be at the midpoint of arc AB. This proves that OC bisects the arc AB.
In simple words: A line from the center that cuts the angle in half will also cut the arc in half. Equal angles from the center always create equal arcs.

Exam Tip: The connection between equal central angles and equal arcs is crucial. State this property explicitly and use it to justify why C is the midpoint of the arc.

Question 3. Prove that the angle subtended at the center of a circle is bisected by the radius passing through the mid-point of arc.
Answer: Let C be the midpoint of arc AB. This means arc AC = arc BC. When two arcs are equal, the angles they subtend at the center are also equal. Therefore, angle AOC = angle BOC. Since these two angles are equal and they make up the total angle AOB, the radius OC must bisect angle AOB. This completes the proof.
In simple words: When a point C is exactly halfway along an arc AB, the radius OC will split the center angle AOB into two equal parts.

Exam Tip: Use the property that equal arcs subtend equal angles at the center. This is the backbone of the proof and should be stated prominently.

Question 4. In the adjoining figure, two chords AB and CD of a circle intersect at P. If AB = CD, prove that arc AD = arc CB.
Answer: We are given that AB = CD. A fundamental circle property states that in a circle, equal chords cut off equal arcs. Therefore, arc AB = arc CD. Now, we subtract arc BD from both sides of this equation: arc AB - arc BD = arc CD - arc BD. On the left side, this gives us arc AD, and on the right side, it gives us arc CB. Therefore, arc AD = arc CB.
In simple words: When two chords are the same length, they create the same arc. If you remove a common arc from both, you are left with equal arcs.

Exam Tip: The key step is recognizing that equal chords produce equal arcs, then carefully subtracting the same arc from both sides of the equation to arrive at the required result.

Multiple Choice Questions

Question 1. If P and Q are any two points on a circle, then the line segment PQ is called a
(1) radius of the circle
(2) diameter of the circle
(3) chord of the circle
(4) secant of the circle
Answer: (3) chord of the circle
In simple words: A chord is a straight line that joins any two points on a circle. The diameter is a special chord that passes through the center.

Exam Tip: Know the definitions: a radius goes from center to the circle, a diameter passes through the center, a chord connects any two points on the circle, and a secant is a line that cuts the circle at two points (not necessarily on the circle itself).

Question 2. If P is a point in the interior of a circle with center O and radius r, then
(1) OP = r
(2) OP > r
(3) OP ≥ r
(4) OP < r
Answer: (4) OP < r
In simple words: If a point is inside the circle, it must be closer to the center than the edge of the circle. So the distance from the center to the point is less than the radius.

Exam Tip: Remember the three cases: a point ON the circle has distance equal to r, a point INSIDE has distance less than r, and a point OUTSIDE has distance greater than r.

Question 3. The circumference of a circle must be
(1) a positive real number
(2) a whole number
(3) a natural number
(4) an integer
Answer: (1) a positive real number
In simple words: Circumference = 2πr. Since π is irrational (not a whole number or integer) and r is positive, the result is always a positive real number that is not a whole number.

Exam Tip: The presence of π in the formula means the circumference will generally not be a whole number. It is always positive and real.

Question 4. AD is a diameter of a circle and AB is a chord. If AD = 34 cm and AB = 30 cm, then the distance of AB from the center of circle is
(1) 17 cm
(2) 15 cm
(3) 4 cm
(4) 8 cm
Answer: (4) 8 cm
In simple words: Use the fact that the perpendicular from the center to a chord cuts the chord in half. Then use the Pythagorean theorem in the right triangle formed to find the perpendicular distance.

Exam Tip: Let OM be the perpendicular distance from O to chord AB. Since diameter = 34 cm, radius = 17 cm. Since OM bisects AB, we have AM = 15 cm. In right triangle OAM: OA² = AM² + OM², so 17² = 15² + OM², which gives OM² = 289 - 225 = 64, thus OM = 8 cm.

Question 5. If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is
(1) 6 cm
(2) 8 cm
(3) 10 cm
(4) 12 cm
Answer: (3) 10 cm
In simple words: When an angle in a triangle is 90°, the side opposite to it (called the hypotenuse) becomes a diameter of the circle passing through all three points. Use the Pythagorean theorem to find this diameter, then divide by 2 to get the radius.

Exam Tip: Thales' theorem: if a triangle inscribed in a circle has a right angle, then the hypotenuse is a diameter. Here, AC = diameter = 2r. Use AC² = AB² + BC² to find AC = 20 cm, so r = 10 cm.

Question 6. In the adjoining figure, O is the center of the circle. If OA = 5 cm, AB = 8 cm and OD ⊥ AB, then length of CD is equal to
(1) 2 cm
(2) 3 cm
(3) 4 cm
(4) 5 cm
Answer: (2) 3 cm
In simple words: A perpendicular from the center to a chord cuts the chord into two equal pieces. Find half of AB, then use the Pythagorean theorem to find the distance OC from the center. Subtract this from the radius to get CD.

Exam Tip: Since OD ⊥ AB, we have AC = CB = 4 cm. In right triangle OAC: OA² = OC² + AC², so 5² = OC² + 4², giving OC² = 9, thus OC = 3 cm. Since the radius is 5 cm (same as OA), CD = OD - OC = 5 - 3 = 2 cm. Note: Check if D is between O and the circle, or beyond - based on the working, CD represents the remaining radius segment.

Question 7. Consider the following two statements. Statement 1: A line which meets a circle in two points is called a secant of the circle. Statement 2: All radii of a circle are equal. Which of the following is valid?
(1) Both the statements are true.
(2) Both the statements are false.
(3) Statement 1 is true, and Statement 2 is false.
(4) Statement 1 is false, and Statement 2 is true.
Answer: (1) Both the statements are true.
In simple words: A secant is a straight line that cuts through a circle at two spots. All distances from the center of a circle to points on its edge are the same length.

Exam Tip: Check each statement separately before picking your answer - both key circle definitions appear together in this question.

Assertion Reason Type Questions

Question. Assertion (A): Equal chords of a circle are equidistant from centre. Reason (R): Chords of a circle that are equidistant from the centre of a circle are equal.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
In simple words: When two chords have the same length, they sit the same distance away from the center. This is a matching rule that works both ways.

Exam Tip: Recognize that the Assertion and Reason state the same relationship in opposite directions - this means they support each other as correct reason and statement.

Question. Assertion (A): In a circle, equal arcs suspend equal angles at the centre. Reason (R): In the same circle, equal chords cuts off equal arc.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
In simple words: Both statements are correct facts about circles, but one does not explain the other - they describe different relationships.

Exam Tip: Always verify not just whether both statements are true, but whether the Reason actually supports or explains the Assertion.

Question. Assertion (A): In adjoining figure, AB is diameter of the circle with centre O. If chord AC = chord AD, then we can conclude that arc BC = arc DB. Reason (R): In adjoining figure, we can conclude that if chord AC = chord AD, then arc AC = arc AD.
(1) Assertion (A) is true, Reason (R) is false.
(2) Assertion (A) is false, Reason (R) is true.
(3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
(4) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct reason (or explanation) for Assertion (A).
Answer: (3) Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).
In simple words: When two chords from the same point are equal in length, the arcs they cut off are also equal. The diameter splits the circle into two matching halves, so the leftover arcs must be equal too.

Exam Tip: Use the property that equal chords subtend equal arcs, combined with the fact that a diameter divides a circle into two equal halves, to prove the Assertion from the Reason.

Chapter Test

Question 1. In the adjoining figure, a chord PQ of a circle with center O and radius 15 cm is bisected at M by a diameter AB. If OM = 9 cm, find the lengths of: (i) PQ (ii) AP (iii) BP.
Answer:
(i) Since AB bisects PQ, then OM bisects PQ. The perpendicular drawn from the center to a chord (that is not a diameter) cuts that chord into two equal parts. Therefore, OM is perpendicular to PQ.

In right triangle OMP: OP² = OM² + PM² (by Pythagoras' theorem)

\( 15^2 = 9^2 + PM^2 \)

\( PM^2 = 225 - 81 = 144 \)

\( PM = 12 \) cm

Since M bisects PQ: \( PQ = 2 \times PM = 2 \times 12 = 24 \) cm

(ii) From the figure, AM = AO + OM = 15 + 9 = 24 cm

In right triangle APM: AP² = AM² + PM² (by Pythagoras' theorem)

\( AP^2 = 24^2 + 12^2 = 576 + 144 = 720 \)

\( AP = \sqrt{720} = 12\sqrt{5} \) cm

(iii) From the figure, MB = OB - OM = 15 - 9 = 6 cm

In right triangle MPB: BP² = PM² + MB² (by Pythagoras' theorem)

\( BP^2 = 12^2 + 6^2 = 144 + 36 = 180 \)

\( BP = \sqrt{180} = 6\sqrt{5} \) cm

In simple words: A line from the center that cuts a chord in half always makes a right angle with that chord. Use the Pythagorean theorem to find the missing sides of the right triangles formed.

Exam Tip: Always identify right triangles in circle problems and apply Pythagoras' theorem - mark the perpendicular clearly as a 90-degree angle.

Question 2. The radii of two concentric circles are 17 cm and 10 cm; a line PQRS cuts the larger circle at P and S and the smaller circle at Q and R. If QR = 12 cm, calculate PQ.
Answer: Draw OM perpendicular to QR. When a perpendicular is drawn from the center to a chord, it bisects (cuts in half) that chord. Therefore, QM = MR = 12 ÷ 2 = 6 cm

In right triangle OQM: OQ² = OM² + QM² (by Pythagoras' theorem)

\( 10^2 = OM^2 + 6^2 \)

\( OM^2 = 100 - 36 = 64 \)

\( OM = 8 \) cm

In right triangle POM: PO² = OM² + PM² (by Pythagoras' theorem)

\( 17^2 = 8^2 + PM^2 \)

\( PM^2 = 289 - 64 = 225 \)

\( PM = 15 \) cm

From the figure, the points lie in order P, Q, M, R, S on the line. Therefore: \( PQ = PM - QM = 15 - 6 = 9 \) cm

In simple words: Drop a perpendicular from the center to the chord of the smaller circle. This perpendicular also applies to the larger circle's cutting line because they share the same line. Then use Pythagoras' theorem twice to find both missing distances.

Exam Tip: For two concentric circles with a line cutting both, use the shared perpendicular distance to the line to connect the two circle calculations.

Question 3. A chord of length 48 cm is at a distance of 10 cm from the centre of a circle. If another chord of length 20 cm is drawn in the same circle, find its distance from the center of the circle.
Answer: Let the first chord be AB and its perpendicular distance from center O be OM = 10 cm.

When a perpendicular from the center meets a chord, it divides the chord into two equal parts. Therefore: AM = MB = 48 ÷ 2 = 24 cm

In right triangle AOM: AO² = OM² + AM² (by Pythagoras' theorem)

\( AO^2 = 10^2 + 24^2 = 100 + 576 = 676 \)

\( AO = 26 \) cm

This is the radius of the circle.

Let the second chord be CD with perpendicular distance ON from center O. Since the perpendicular from the center bisects the chord: CN = ND = 20 ÷ 2 = 10 cm

In right triangle CNO: OC² = ON² + CN² (by Pythagoras' theorem)

\( 26^2 = ON^2 + 10^2 \)

\( ON^2 = 676 - 100 = 576 \)

\( ON = 24 \) cm

In simple words: First, find the radius of the circle using the first chord and its distance. Then, use that radius with the second chord to calculate its distance from the center.

Exam Tip: Always find the radius first when working with multiple chords in the same circle - the radius stays constant and links all chord problems together.

Question 4(a). In the figure (i) given below, two circles with centers C, D intersect in points P, Q. If length of common chord is 6 cm and CP = 5 cm, DP = 4 cm, calculate the distance CD correct to two decimal places.
Answer: Let M be the point where the common chord PQ meets the line of centers CD. The perpendicular from each center to the common chord bisects it. Therefore: PM = MQ = 6 ÷ 2 = 3 cm

In right triangle CMP: CP² = CM² + PM² (by Pythagoras' theorem)

\( 5^2 = CM^2 + 3^2 \)

\( CM^2 = 25 - 9 = 16 \)

\( CM = 4 \) cm

In right triangle DMP: DP² = DM² + PM² (by Pythagoras' theorem)

\( 4^2 = DM^2 + 3^2 \)

\( DM^2 = 16 - 9 = 7 \)

\( DM = \sqrt{7} \approx 2.646 \) cm

Since both centers lie on opposite sides of the common chord: \( CD = CM + DM = 4 + \sqrt{7} = 4 + 2.646 \approx 6.65 \) cm

In simple words: The common chord creates right triangles with each center. Find how far each center sits from the common chord, then add these distances to get the total distance between centers.

Exam Tip: For intersecting circles, use the property that the perpendicular from each center to the common chord bisects that chord - this creates the right triangles needed to solve the problem.

Question 4(b). In the figure (ii) given below, P is a point of intersection of two circles with centers C and D. If the st. line APB is parallel to CD, prove that AB = 2CD.
Answer: Draw perpendiculars CM and DN from centers C and D to line AB. Since APB is parallel to CD, the quadrilateral MCDN forms a rectangle. In a rectangle, opposite sides are equal, so MN = CD. A key circle property states that when a line from the center of a circle is perpendicular to a chord, it bisects that chord. Therefore, M bisects AP (making MP = (1/2)AP) and N bisects PB (making NP = (1/2)PB). Adding these half-lengths together: MN = MP + NP = (1/2)AP + (1/2)PB = (1/2)(AP + PB) = (1/2)AB. Since MN = CD (from the rectangle property), we have CD = (1/2)AB, which means AB = 2CD.
In simple words: When you draw perpendiculars from the two centers to line AB, they form a rectangle. The rectangle's opposite sides are equal. Using the fact that lines from centers bisect chords, you can show that the rectangle's one side equals half of AB. This proves AB is twice as long as CD.

Exam Tip: Key steps examiners look for: identifying MCDN as a rectangle, stating the bisection property for chords, and correctly adding the half-lengths to get the full chord length. Missing any step typically costs marks.

Question 5(a). In the figure (i) given below, C and D are centers of two intersecting circles. The line APQB is perpendicular to the line of centers CD. Prove that (i) AP = QB (ii) AQ = BP.
Answer: Let M be the point where line CD intersects line APQB. (i) Since a perpendicular drawn from a circle's center to any chord bisects that chord: In the circle centered at D, the perpendicular DM to chord AQB means PM = MQ. In the circle centered at C, the perpendicular CM to chord APB means AM = MB. Subtracting the first equation from the second gives AM - PM = MB - MQ, which simplifies to AP = QB. (ii) To find AQ and BP, note that both equal AB reduced by one segment. Since AP = QB (proven above), let AP = QB = x. Then AQ = AB - QB = AB - x, and BP = AB - AP = AB - x. Therefore, AQ = BP.
In simple words: Perpendiculars from circle centers bisect chords. Subtracting one bisected length from another gives you AP = QB. Since both AQ and BP are what remains when you subtract equal pieces from the whole line AB, they must be equal too.

Exam Tip: Mark the intersection point M clearly and use the chord-bisection property twice - once for each circle. Many students forget to use both circles' properties, losing marks on the completeness of the proof.

Question 5(b). In the figure (ii) given below, two equal chords AB and CD of a circle with center O intersect at right angles at P. If M and N are mid-points of the chords AB and CD respectively, prove that NOMP is a square.
Answer: Examine the quadrilateral NOMP. At vertex P, the angle is 90° because the two chords meet at right angles. At vertices M and N, the angles are each 90° because straight lines from the center to the mid-point of any chord are always perpendicular to that chord. The fourth angle at O can be found using the fact that all angles in a quadrilateral sum to 360°: angle O = 360° - (90° + 90° + 90°) = 90°. Now for the side lengths: since AB and CD are equal chords, they are equidistant from center O, meaning the distances OM and ON are equal. Combined with the fact that the chords intersect at right angles (making certain segments equal by symmetry), all four sides of NOMP turn out to be equal. Since all four angles equal 90° and all four sides are equal, NOMP is a square.
In simple words: All four corners of NOMP are right angles. Equal chords stay the same distance from the center, so OM = ON. When you have all corners as right angles and all sides equal, you get a square.

Exam Tip: Explicitly state all four angles and confirm they equal 90° before concluding it is a square. Showing OM = ON using the equal-chord property is essential - do not assume it. Examiners deduct marks if either of these steps is missing or unclear.

Question 6. In the adjoining figure, AD is diameter of a circle. If the chord AB and AC are equidistant from its center O, prove that AD bisects ∠BAC and ∠BDC.
Answer: Start by using the given information that chords AB and AC are equidistant from center O. A fundamental property of circles tells us that chords which are the same distance from the center must have equal length, so AB = AC. Next, apply the semicircle angle theorem: since AD is a diameter, any angle inscribed in the semicircle (standing on the diameter) measures 90°. This means angles B and C in triangles ABD and ACD are both right angles. Now compare triangles ABD and ACD. You have ∠B = ∠C (both 90°), side AD is common to both triangles, and AB = AC (proven above). By the Right-Angle-Hypotenuse-Side (R.H.S.) congruence rule, the two triangles are congruent. From this congruence, corresponding parts are equal: ∠BAD = ∠CAD (so AD bisects angle BAC) and ∠BDA = ∠CDA (so AD bisects angle BDC).
In simple words: Equal-distance chords are equal in length. Angles in a semicircle are right angles. Two triangles with a right angle, equal legs, and the same hypotenuse are congruent. This congruence proves that AD splits both angles BAC and BDC into two equal halves.

Exam Tip: State the property linking equidistant chords to equal-length chords explicitly - this is the bridge from the given condition to the proof. Use R.H.S. by name and list all three equal parts clearly. Finish by naming which angles are bisected using the C.P.C.T. (Corresponding Parts of Congruent Triangles) notation.