NCERT Exemplar Solutions - Class 10 - Mathematics - Unit 9: Circles

Exercise 9.1

Step 1: The two circles are concentric. This means they have the same centre.

Step 2: The radius of the smaller circle is given as 4 cm, and the radius of the bigger circle is 5 cm.

Step 3: A chord of the bigger circle which just touches (is tangent to) the smaller circle will be at a distance of 4 cm (the radius of the smaller circle) from the centre.

Step 4: We know the formula for the length of a chord at a perpendicular distance d from the centre of a circle of radius R:

Chord length = \(2 \sqrt{R^2 - d^2}\)

Step 5: Here, the radius of the bigger circle is \(R = 5\) cm and the distance of the chord from the centre is \(d = 4\) cm.

Step 6: Substitute the values:

Chord length = \(2 \sqrt{5^2 - 4^2}\)

= \(2 \sqrt{25 - 16}\)

= \(2 \sqrt{9}\)

= \(2 \times 3\)

= 6 cm

Final Answer: The length of each such chord is 6 cm.

Step 1: The figure shows a circle with centre O. The angles \(\angle AOB\) and \(\angle COD\) are central angles that lie on a straight line (diameter).

Step 2: Angles on a straight line always add up to \(180^\circ\). This is called the linear pair property.

Step 3: Here, \(\angle AOB + \angle COD = 180^\circ\).

Step 4: Substitute the given value: \(125^\circ + \angle COD = 180^\circ\).

Step 5: Subtract \(125^\circ\) from both sides: \(\angle COD = 180^\circ - 125^\circ = 55^\circ\).

Final Answer: \(\angle COD = 55^\circ\). So, the correct option is D.

Step 1: Recall the Tangent–Chord Theorem. It says: "The angle between a tangent and a chord through the point of contact is equal to the angle in the alternate (opposite) segment of the circle."

Step 2: Here, the tangent is \(AT\) and the chord is \(AB\). So, \(\angle BAT\) is the angle between the tangent and the chord.

Step 3: According to the theorem, \(\angle BAT\) will be equal to the angle made by the chord \(AB\) in the opposite arc. That opposite angle is \(\angle ACB\).

Step 4: From the question, \(\angle ACB = 50^\circ\).

Step 5: Therefore, \(\angle BAT = \angle ACB = 50^\circ\).

Final Answer: \(\angle BAT = 50^\circ\).

Step 1: We are given:

• Radius of circle, \(r = 5\,\text{cm}\)
• Distance of point \(P\) from centre \(O\), \(OP = 13\,\text{cm}\)

Step 2: In right triangle \(OQP\), \(OQ\) is the radius (5 cm), and \(OP\) is the hypotenuse (13 cm). The tangent \(PQ\) is perpendicular to the radius at point \(Q\). So, using Pythagoras theorem:

\[ PQ = \sqrt{OP^2 - OQ^2} = \sqrt{13^2 - 5^2} \]

\[ PQ = \sqrt{169 - 25} = \sqrt{144} = 12\,\text{cm} \]

Step 3: Quadrilateral \(PQOR\) is made of two congruent right triangles: \(\triangle OPQ\) and \(\triangle OPR\).

Step 4: Area of one right triangle:

\[ \text{Area}(\triangle OPQ) = \tfrac{1}{2} \times OQ \times PQ \]

\[ = \tfrac{1}{2} \times 5 \times 12 = 30\,\text{cm}^2 \]

Step 5: Since both triangles have the same area, total area is:

\[ \text{Area}(PQOR) = 30 + 30 = 60\,\text{cm}^2 \]

Final Answer: The area of quadrilateral \(PQOR\) is \(60\,\text{cm}^2\).

Step 1: Draw a circle with centre \(O\) and radius \(5\,\text{cm}\). Mark diameter \(AB\). At point \(A\), draw the tangent \(XAY\).

Step 2: We are told that a chord \(CD\) is drawn parallel to the tangent \(XY\). The perpendicular distance of this chord from point \(A\) is \(8\,\text{cm}\).

Step 3: To calculate chord length, we need the perpendicular distance of the chord from the centre \(O\). The total distance from \(O\) to point \(A\) is the radius = \(5\,\text{cm}\). The chord is \(8\,\text{cm}\) away from \(A\). Therefore, distance from \(O\) to the chord = \(|OA - 8| = |5 - 8| = 3\,\text{cm}|.

Step 4: Formula for chord length: \[ \text{Chord length} = 2 \sqrt{r^2 - d^2} \] where \(r\) is radius and \(d\) is distance from centre to the chord.

Step 5: Substitute values: \(r = 5\,\text{cm}, \; d = 3\,\text{cm}\) \[ \text{Chord length} = 2 \sqrt{5^2 - 3^2} = 2 \sqrt{25 - 9} = 2 \sqrt{16} = 2 \times 4 = 8\,\text{cm}. \]

Final Answer: The chord length = 8 cm. (Option D)

Step 1: We are given that \(OT = 4\,\text{cm}\). Point \(T\) lies on the tangent \(AT\). So, \(OT\) is a line from the centre \(O\) to point \(T\).

Step 2: By the property of a tangent to a circle: the radius drawn to the point of tangency is perpendicular to the tangent. That means \(OA \perp AT\). So, triangle \(OTA\) is a right-angled triangle at \(A\).

Step 3: In right triangle \(OTA\), - Hypotenuse = \(OT = 4\,\text{cm}\). - Angle at \(T\) = \(30^\circ\).

Step 4: To find \(OA\) (radius), use the sine function: \[ \sin(30^\circ) = \frac{OA}{OT} \] Substituting values: \[ \sin(30^\circ) = \frac{OA}{4} \] \[ \frac{1}{2} = \frac{OA}{4} \] \[ OA = 2\,\text{cm} \]

Step 5: To find \(AT\) (the tangent), use the cosine function: \[ \cos(30^\circ) = \frac{AT}{OT} \] Substituting values: \[ \frac{\sqrt{3}}{2} = \frac{AT}{4} \] \[ AT = 4 \times \frac{\sqrt{3}}{2} \] \[ AT = 2\sqrt{3}\,\text{cm} \]

Final Answer: \(AT = 2\sqrt{3}\,\text{cm}\). ✅ (Option C)

Step 1: Recall the Tangent–Chord Theorem: The angle between a tangent and a chord through the point of contact is equal to the angle made by the chord in the alternate segment of the circle.

Step 2: Here, tangent \(PR\) makes \(50^\circ\) with chord \(PQ\). So, \(\angle PQO = 50^\circ\) (angle in the alternate segment).

Step 3: In a circle, the central angle (angle at the centre) is always twice the angle at the circumference standing on the same chord.

Step 4: The chord is \(PQ\). Angle at the centre on chord \(PQ\) is \(\angle POQ\). Angle at the circumference is \(\angle PQO = 50^\circ\).

Step 5: Therefore, \[ \angle POQ = 2 \times \angle PQO = 2 \times 50^\circ = 100^\circ. \]

Final Answer: \(\angle POQ = 100^\circ\). Correct option: A

Step 1: The angle between two tangents drawn from a point outside the circle is related to the angle at the center. Formula: \( \angle APB + \angle AOB = 180^\circ \).

Step 2: We are given \( \angle APB = 50^\circ \). So, \( \angle AOB = 180^\circ - 50^\circ = 130^\circ \).

Step 3: Now look at triangle \(AOB\). Here, \(OA = OB\) because both are radii of the same circle. So, \( \triangle AOB \) is an isosceles triangle.

Step 4: In an isosceles triangle, the two base angles are equal. The sum of all angles in a triangle is \(180^\circ\). So, base angles = \( \dfrac{180^\circ - 130^\circ}{2} = \dfrac{50^\circ}{2} = 25^\circ \).

Step 5: That means \( \angle OAB = 25^\circ \).

Final Answer: Option A (\(25^\circ\)).

Step 1: We are given a circle of radius \(r = 3\,\text{cm}\). Two tangents are drawn from an external point such that the angle between them is \(60^\circ\).

Step 2: At the centre of the circle, the angle between the two radii (drawn to the tangent contact points) will be \(180^\circ - 60^\circ = 120^\circ\). (Because in a quadrilateral formed by the centre, the two tangent points, and the external point, the sum of opposite angles is \(180^\circ\)).

Step 3: Now consider the triangle formed by: - the centre of the circle (O), - one tangent point (P), - and the external point (A). In this triangle, angle at the centre is \(120^\circ\). So the half angle is \(120^\circ / 2 = 60^\circ\). This means the angle at OAP is \(30^\circ\).

Step 4: The distance from the centre O to the external point A can be found using \[ OA = \frac{r}{\sin 30^\circ} = \frac{3}{1/2} = 6\,\text{cm}. \]

Step 5: The tangent length (AP) can be found by Pythagoras theorem in right triangle OAP: \[ AP = \sqrt{OA^2 - OP^2} \] Here, \(OA = 6\,\text{cm},\ OP = r = 3\,\text{cm}\). So, \[ AP = \sqrt{6^2 - 3^2} = \sqrt{36 - 9} = \sqrt{27} = 3\sqrt{3}\,\text{cm}. \]

Final Answer: Each tangent is \(3\sqrt{3}\,\text{cm}\) long.

Step 1: We are given that \(PQR\) is a tangent at point \(Q\). The angle between tangent \(QR\) and chord \(QB\) is given as \(\angle BQR = 70^\circ\).

Step 2: By the Alternate Segment Theorem, the angle between a tangent and a chord is equal to the angle made by the chord in the opposite segment of the circle.

Step 3: So, \(\angle BQR = 70^\circ\) will be equal to the angle at point \(A\) on the other side of chord \(QB\). That means \(\angle QAB = 70^\circ\).

Step 4: Similarly, because chord \(AB\) is parallel to line \(PR\), we also get that \(\angle QBA = 70^\circ\).

Step 5: Now, in triangle \(AQB\): \[ \angle QAB + \angle QBA + \angle AQB = 180^\circ \]

Step 6: Substituting the values: \[ 70^\circ + 70^\circ + \angle AQB = 180^\circ \]

Step 7: Simplify: \[ 140^\circ + \angle AQB = 180^\circ \]

Step 8: Therefore, \[ \angle AQB = 180^\circ - 140^\circ = 40^\circ \]

Final Answer: \(\angle AQB = 40^\circ\). Hence, the correct option is B.