Ch 5 EOC Class 9 Maths Ganita Manjari Solutions | Sid Classes
Welcome to Sid Classes, your trusted destination for comprehensive academic resources! In this detailed study guide, we present complete, step-by-step solutions for the NCERT Class 9 Maths Ganita Manjari Chapter 5 End of Chapter (EOC) Exercise. Designed to reflect how top-scoring students write their proofs and solutions in school examinations, this review walkthrough covers circle theorems, cyclic quadrilaterals, chord properties, and distance calculations entirely.
🔑 Chapter Overview & Key Theorems:
- Perpendicular from Centre: The perpendicular drawn from the centre of a circle to a chord bisects the chord.
- Baudhāyana–Pythagoras Theorem: In any right-angled triangle, \(r^2 = d^2 + x^2\), connecting radius (\(r\)), perpendicular distance (\(d\)), and half-chord (\(x\)).
- Cyclic Quadrilateral Property: The opposite angles of a cyclic quadrilateral are supplementary (their sum is \(180^\circ\)), and the exterior angle equals the interior opposite angle.
- Angle in a Semicircle: The angle subtended by a diameter (or semicircle) at any point on the circumference is always a right angle (\(90^\circ\)).
NCERT Class 9 Maths Ganita Manjari Chapter 5 End of Chapter Exercise Solutions
Question 1
In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord?
Solution:
Let radius \(r = 13\) cm and perpendicular distance \(d = 5\) cm.
Using the half-chord formula \(x = \sqrt{r^2 - d^2}\):
\(x = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\) cm.
Total chord length \(= 2x = 2 \times 12 = \mathbf{24}\) cm.
Question 2
An arc of a circle subtends an angle of 70° at the centre. What is the measure of the angle subtended by the arc at a point on the circle?
Solution:
The angle subtended by an arc at the circumference is half of the angle subtended by it at the centre.
\(\text{Angle at circumference} = \frac{70^\circ}{2} = \mathbf{35^\circ}\).
Question 3
The diameter of a circle is 26 cm. A chord of length 24 cm is drawn in the circle. Find the distance from the centre of the circle to the chord.
Solution & Calculation:
Radius \(r = \frac{26}{2} = 13\) cm and half-chord \(x = \frac{24}{2} = 12\) cm.
Using Pythagoras theorem \(d = \sqrt{r^2 - x^2}\):
\(d = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = \mathbf{5}\) cm.
Question 4
A circle has a radius of 15 cm. A chord is drawn. The distance from the centre of the circle to the chord is 9 cm. What is the length of the chord?
Solution:
Radius \(r = 15\) cm, distance \(d = 9\) cm.
Half-chord \(x = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12\) cm.
Chord length \(= 2 \times 12 = \mathbf{24}\) cm.
Question 5
Prove that the perpendicular bisector of a chord passes through the centre of the circle.
Solution:
Let \(AB\) be a chord and let the perpendicular bisector pass through the midpoint \(M\). By joining the centre \(O\) to \(A\) and \(B\) and using RHS congruence on the triangles formed (\(\triangle OMA \cong \triangle OMB\)), we prove that \(OA = OB\), confirming that the centre lies on the perpendicular bisector.
Question 6
The diameter of a circle is AB. Point C is on the circumference. What is the measure of the \(\angle ACB\)? Explain your reasoning.
Solution:
\(\angle ACB = \mathbf{90^\circ}\).
Reasoning: Since \(AB\) is the diameter, it subtends a straight angle (\(180^\circ\)) at the centre. By the angle in a semicircle theorem, the angle subtended at any point on the circumference is half of the central angle, i.e., \(\frac{180^\circ}{2} = 90^\circ\).
Question 7
ABCD is a cyclic quadrilateral inscribed in a circle. If \(\angle A\) measures \(75^\circ\), what is the measure of \(\angle C\)? If \(\angle B\) measures \(110^\circ\), what is the measure of \(\angle D\)?
Solution
Since opposite angles of a cyclic quadrilateral are supplementary:
• \(\angle C = 180^\circ - 75^\circ = \mathbf{105^\circ}\)
• \(\angle D = 180^\circ - 110^\circ = \mathbf{70^\circ}\)
Question 8
Quadrilateral PQRS is inscribed in a circle. If \(\angle P = (2x + 10)^\circ\) and \(\angle R = (3x - 20)^\circ\), find the value of \(x\) and the measures of \(\angle P\) and \(\angle R\).
Solution:
Since PQRS is cyclic, \(\angle P + \angle R = 180^\circ\)
\((2x + 10) + (3x - 20) = 180\)
\(5x - 10 = 180 \implies 5x = 190 \implies \mathbf{x = 38}\).
• \(\angle P = 2(38) + 10 = \mathbf{86^\circ}\)
• \(\angle R = 3(38) - 20 = \mathbf{94^\circ}\)
Question 9
The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius of the circle.
Solution:
Half-chord \(x = \frac{16}{2} = 8\) cm, distance \(d = 6\) cm.
Radius \(r = \sqrt{x^2 + d^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = \mathbf{10}\) cm.
Question 10
A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.
Solution:
Using Brahmagupta's formula for a cyclic quadrilateral with sides \(a=5, b=5, c=12, d=12\)
Semiperimeter \(s = \frac{5 + 5 + 12 + 12}{2} = \frac{34}{2} = 17\).
\(\text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d)} = \sqrt{(12)(12)(5)(5)} = 12 \times 5 = \mathbf{60}\) sq units.
Question 11
Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find out whether the centre of the circumcircle lies inside the quadrilateral or outside? What is the best way of finding out?
Solution
By checking the angles of the quadrilateral. If the quadrilateral has an obtuse angle (\(> 90^\circ\)), the circumcentre lies outside; if all angles are acute (\(< 90^\circ\)), it lies inside. Alternatively, constructing perpendicular bisectors of any two sides will intersect at the circumcentre.
Question 12
When two chords intersect, each of them is divided into two line segments. Show that if the intersecting chords are of equal length, then the line segments of one chord are equal to the corresponding line segments of the other chord.
Solution:
Using intersecting chord properties and perpendicular distances from the centre to equal chords, symmetry dictates that the respective segments formed by the intersection point are equal.
Question 13
Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre. (Hint: Is it a circumcircle of a suitable triangle?)
Solution:
Radius \(r = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}\) cm. Construct a right-angled triangle with half-chord 3 cm and height 3 cm to find the radius and draw the circle.
Question 14
Show that rectangle is the only parallelogram that can be inscribed in a circle.
Solution:
In a cyclic parallelogram, opposite angles are equal and supplementary (\(x + x = 180^\circ \implies x = 90^\circ\)). Since all interior angles are \(90^\circ\), the parallelogram must be a rectangle.
Question 15
Show that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.
Solution:
The diagonals of a rectangle are equal and bisect each other. In a circle, diagonals that pass through the centre subtend \(90^\circ\) at the circumference, making them diameters whose intersection point coincides with the circle's centre.
Question 16
Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords
Solution:
The locus of the midpoints of all chords of a fixed length in a circle is a concentric circle inside the original circle
Question 17
In a circle with centre O, chords AB and AC are congruent. Explain why this statement is true: "The centre of the circle lies on the angle bisector of \(\angle BAC\)"
Solution:
Since \(AB = AC\) (congruent chords), they are equidistant from centre \(O\). The perpendicular bisectors of both chords intersect at \(O\), which forms the locus of points equidistant from the arms of \(\angle BAC\), proving \(O\) lies on the angle bisector.
Question 18
Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre of a circle. The distance between the chords is 7 cm. Find the radius of the circle.
Solution:
Let half-chords be \(x_1 = 5\) cm and \(x_2 = 12\) cm. Let distance of smaller chord from centre be \(d\), so distance of larger chord is \(d - 7\).
\(r^2 = 5^2 + d^2 = 12^2 + (d - 7)^2\)
\(25 + d^2 = 144 + d^2 - 14d + 49\)
\(14d = 193 - 25 = 168 \implies d = 12\) cm.
Radius \(r = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = \mathbf{13}\) cm.
Question 19
A regular hexagon is inscribed in a circle of radius \(r\). Find the length of the sides of the hexagon and the distance of each side from the centre.
Solution:
• Side length of regular hexagon \(= \mathbf{r}\) (since each central angle is \(60^\circ\)).
• Distance from centre (\(d\)) \(= \sqrt{r^2 - \left(\frac{r}{2}\right)^2} = \mathbf{\frac{\sqrt{3}}{2}r}\).
Question 20
A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about \(\angle MOP\) and \(\angle MNP\)? Explain your reasoning.
Solution:
• \(\angle MNP = 90^\circ\) (angle in a semicircle).
• \(\angle MOP\) is the central angle corresponding to arc MP, which is twice the inscribed angle \(\angle MNP\), so \(\angle MOP = 2 \times \angle MNP\).
Question 21
Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., \(\angle CDE = \angle ABC\), where E is a point on the extension of side CD).
Solution:
\(\angle CDE + \angle ADC = 180^\circ\) (linear pair). Also, \(\angle ABC + \angle ADC = 180^\circ\) (opposite angles of cyclic quadrilateral). Therefore, \(\angle CDE = \angle ABC\).
Question 22
"There is no chord of a circle that is longer than its diameter." How do you justify this statement?
Solution:
The diameter passes through the centre and represents the maximum possible straight-line distance between any two points on a circle. Any other chord is a chord segment shorter than or equal to the diameter.
Question 23
Let A be any point within a given circle with centre O. Show that the shortest chord of the circle that passes through point A is the one that is perpendicular to OA.
Solution:
Chords closer to the centre are longer, and chords farther are shorter. Among all chords passing through an internal point \(A\), the shortest chord is perpendicular to the radius vector \(OA\).
Question 24
How would you use Fig. 5.30 to justify the statement that the angle in a semicircle is 90°?
Solution:
As shown in Fig. 5.30, the radius splits the triangle into two isosceles triangles where base angles are equal (\(a\) and \(b\)). The sum of angles of \(\triangle OAP\) and \(\triangle OBP\) gives \(\angle A = a + b = 90^\circ\).
Question 25
In a circle, two chords CC' and DD' are drawn perpendicular to diameter AB. Prove that the segment MM' joining the midpoints of the chords CD and C' D' is perpendicular to AB.
Solution:
By circle symmetry and perpendicular properties from the diameter to parallel chords, the line joining midpoints MM' is parallel to the chords and hence perpendicular to diameter AB.
Question 26
How would you use Fig. 5.31 to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is 180°?
Solution:
Using the four triangles formed by the diagonals meeting at centre \(O\) (Fig. 5.31), the base angles corresponding to arcs yield \(p + q + u + v = 180^\circ\), proving opposite angles are supplementary.
Frequently Asked Questions (FAQs)
Q1: How do you find the distance of a chord from the centre?
A: Use the Pythagorean relation \(d = \sqrt{r^2 - x^2}\), where \(r\) is the radius and \(x\) is half the chord length.
Q2: What is the rule for opposite angles in a cyclic quadrilateral?
A: Opposite angles are supplementary, meaning their sum equals \(180^\circ\).
Q3: What does the angle in a semicircle theorem state?
A: Any angle subtended by a diameter at the circumference of a circle is always a right angle (\(90^\circ\)).
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