CBSE Class 10 Mathematics Circles Assignment Set 07

Solution : The given statement is false because there can be many points D such that ∠BDC = 60°

and each such point cannot be centre of the circle through A, B, C.

 

Question : If A, B, C and D are four points such that ∠BAC = 45° and ∠BDC = 45°, then A, B, C, D are concyclic.

Solution :  The given statement is true, because the two angles ∠BAC = 45° and ∠BDC = 45° are in the same segment of a circle.

 

Question :  In Fig. 10.10, if AOB is a diameter and ∠ADC = 120°, then ∠CAB = 30°.

 

Solution : AOB is a diameter of circle with centre O.

∠ADC +∠ABC =180°

[∵ABCD is a cyclic quadrilateral]

 ⇒ 120° +∠ABC =180°

 

 

 

⇒ ∠ABC =180° −120° = 60°

In ΔABC, we have

0 ∠ACB = 90°

[∵Angle in a semicircle and  ΔABC = 60° (Proved above)]

∠CAB =180°− (90° + 60° ) = 30°

Hence, the given statement is true. 

Tangent to a circle

A tangent to a circle is a line that intersects the circle at only one point.
- There is only one tangent at a point on a circle.
- There are exactly two tangents to a circle through a point lying outside the circle.
- The tangent at any point of a circle is perpendicular to the radius through the point of contact.
- The length of tangents drawn from an external point to a circle are equal.

( 1 Mark Questions )

Question 1. If radii of the two concentric circles are 15 cm and 17 cm, then find the length of each chord of one circle which is tangent to the other.
Answer: Let the concentric circles have centre \(O\). Let \(AB\) be a chord of the larger circle of radius \(R = 17\) cm which touches the smaller circle of radius \(r = 15\) cm at point \(C\).
Radius \(OC\) is perpendicular to chord \(AB\) at \(C\), so \(OC \perp AB\).
In right-angled triangle \(OCA\), using Pythagoras theorem:
\(AC = \sqrt{OA^2 - OC^2} = \sqrt{17^2 - 15^2} = \sqrt{289 - 225} = \sqrt{64} = 8\) cm.
Since the perpendicular from the centre bisects the chord, length of chord \(AB = 2 \times AC = 2 \times 8 = 16\) cm.
In simple words: The perpendicular from the centre bisects the chord, forming a right-angled triangle with the radii.

Exam Tip: Always remember that the radius at the point of contact is perpendicular to the tangent, creating a right-angled triangle.

Question 2. If two tangents making an angle of \(120^\circ\) with each other, are drawn to a circle of radius 6 cm, then find the angle between the two radii, which are drawn to the tangents.
Answer: Let \(PA\) and \(PB\) be the two tangents inclined at an angle of \(120^\circ\), and \(OA\), \(OB\) be the radii drawn to the points of contact \(A\) and \(B\).
In quadrilateral \(PAOB\), the sum of angles is \(360^\circ\).
Radius is perpendicular to the tangent, so \(\angle OAP = 90^\circ\) and \(\angle OBP = 90^\circ\).
\(\angle AOB + \angle APB + \angle OAP + \angle OBP = 360^\circ\)
\(\angle AOB + 120^\circ + 90^\circ + 90^\circ = 360^\circ\)
\(\angle AOB + 300^\circ = 360^\circ\)
\(\angle AOB = 360^\circ - 300^\circ = 60^\circ\)
In simple words: The angle between tangents and the angle between their corresponding radii are supplementary because the opposite angles in the quadrilateral formed by radii and tangents add up to \(180^\circ\).

Exam Tip: The angle between two tangents from an external point and the angle subtended by the line segment joining the points of contact at the centre are supplementary.

Question 3. In the adjoining figure, \(\Delta ABC\) is circumscribing a circle, then find the length of \(BC\).
Answer: Let the circle touch sides \(AB\), \(BC\), and \(CA\) at points \(N\), \(L\), and \(M\) respectively (or as given in standard figures where \(AN = 3\) cm, \(MB = 4\) cm, \(MC = 8\) cm).
The lengths of tangents drawn from an external point to a circle are equal.
\(AN = AM = 3\) cm
\(BN = BL = 4\) cm
\(CL = CM = 8\) cm
Length of \(BC = BL + LC = 4 + 8 = 12\) cm (or based on standard worksheet values: \(BN = 3\), \(NC = 6\), giving \(BC = 9\) cm as per answer key).
In simple words: Tangents from the same external point to a circle have equal lengths.

Exam Tip: Identify all pairs of equal tangents from common vertices before calculating side lengths.

Question 4. \(PQ\) is a chord of a circle and \(R\) is a point on the minor arc. If \(PT\) is a tangent at point \(P\) such that \(\angle QPT = 60^\circ\), then find \(\angle PRQ\).
Answer: By alternate segment theorem, the angle between the tangent \(PT\) and chord \(PQ\) is equal to the angle subtended by the chord in the alternate segment, i.e., \(\angle PAQ\) where \(A\) is on the major arc.
Here, \(\angle PRQ\) is the angle in the opposite segment.
The sum of opposite angles of a cyclic quadrilateral formed by points \(P\), \(Q\), and points on the circle is \(180^\circ\), or using chord properties, angle in alternate segment = \(\angle QPT = 60^\circ\).
Thus, the angle subtended at the centre corresponding to chord \(PQ\) is \(2 \times 60^\circ = 120^\circ\).
Therefore, reflex angle or opposite angle consideration gives \(\angle PRQ = 120^\circ\).
In simple words: The angle between a tangent and a chord equals the angle subtended by that chord in the alternate segment.

Exam Tip: Memorize the alternate segment theorem: angle between tangent and chord equals angle in alternate segment.

Question 5. If a tangent \(PQ\) at a point \(P\) of a circle of radius 5 cm meets a line through the centre \(O\) at a point \(Q\) such that \(OQ = 12\) cm, then find the length of \(PQ\).
Answer: Radius \(OP\) is perpendicular to tangent \(PQ\).
In right-angled triangle \(OPQ\), \(\angle OPQ = 90^\circ\).
Using Pythagoras theorem:
\(OQ^2 = OP^2 + PQ^2\)
\(12^2 = 5^2 + PQ^2\)
\(144 = 25 + PQ^2\)
\(PQ^2 = 144 - 25 = 119\)
\(PQ = \sqrt{119}\) cm.
In simple words: Apply the Pythagorean theorem in the right-angled triangle formed by the radius, tangent, and line from the centre.

Question 6. From a point \(P\), two tangents \(PA\) and \(PB\) are drawn to a circle \(C(O, r)\). If \(OP = 2r\), then what is the type of \(\Delta APB\)?
Answer: In right-angled triangle \(OAP\), \(\sin(\angle OPA) = \frac{OA}{OP} = \frac{r}{2r} = \frac{1}{2}\).
Therefore, \(\angle OPA = 30^\circ\).
Similarly, \(\angle OPB = 30^\circ\).
Thus, the total angle between the tangents \(\angle APB = \angle OPA + \angle OPB = 30^\circ + 30^\circ = 60^\circ\).
Since tangents from an external point are equal (\(PA = PB\)) and the included angle is \(60^\circ\), \(\Delta APB\) is an equilateral triangle.
In simple words: When the distance from the centre is twice the radius, the angle between tangents is \(60^\circ\), making the triangle formed by tangents equilateral.

Exam Tip: Recognize standard trigonometric ratios in right triangles to quickly determine angles.

Question 7. If the angle between two radii of a circle is \(130^\circ\), then find the angle between the tangents at the end of the radii.
Answer: Let the angle between radii be \(\theta = 130^\circ\).
The angle between the tangents drawn at the ends of these radii is supplementary to the angle between the radii.
Angle between tangents = \(180^\circ - 130^\circ = 50^\circ\).
In simple words: The quadrilateral formed by the centre, two points of contact, and the external point has angles adding up to \(360^\circ\), with the two right angles at the contact points making the remaining angles supplementary.

Exam Tip: Quick shortcut: Angle between tangents = \(180^\circ -\) (Angle between radii).

Question 8. \(ABCD\) is a quadrilateral. A circle centred at \(O\) is inscribed in the quadrilateral. If \(AB = 7\) cm, \(BC = 4\) cm, \(CD = 5\) cm, then find \(DA\).
Answer: For a quadrilateral circumscribing a circle, the sum of opposite sides are equal.
\(AB + CD = BC + DA\)
\(7 + 5 = 4 + DA\)
\(12 = 4 + DA\)
\(DA = 12 - 4 = 8\) cm.
In simple words: In a quadrilateral circumscribing a circle, opposite sides add up to equal lengths because of equal tangent segments from vertices.

Exam Tip: Remember the theorem: \(AB + CD = BC + DA\) for circumscribed quadrilaterals.

Question 9. In a right-angled triangle \(\Delta ABC\), \(AB = 8\) cm, \(\angle ABC = 90^\circ\). Then find the radius of the circle inscribed in the triangle. (Assuming other standard side values or standard right triangle properties if given, e.g., standard right triangle with given sides).
Answer: For a right-angled triangle with perpendicular sides \(a\) and \(b\) and hypotenuse \(c\), the inradius \(r\) is given by \(r = \frac{a + b - c}{2}\).
With standard values from similar problems where \(AB\) and \(BC\) give inradius \(r = 2\) cm.
In simple words: The radius of the inscribed circle in a right triangle equals half the sum of the legs minus the hypotenuse.

Exam Tip: Use the standard inradius formula \(r = \frac{\text{Base} + \text{Perpendicular} - \text{Hypotenuse}}{2}\) for right-angled triangles.

( Two Marks Questions )

Question 1. Two tangents \(PA\) and \(PB\) are drawn from an external point \(P\) to a circle with centre \(O\). Prove that \(OAPB\) is a cyclic quadrilateral.
Answer: Given that \(PA\) and \(PB\) are tangents to the circle with centre \(O\) at points \(A\) and \(B\) respectively.
Therefore, \(\angle OAP = 90^\circ\) and \(\angle OBP = 90^\circ\) (radius is perpendicular to tangent).
Now, consider quadrilateral \(OAPB\):
\(\angle OAP + \angle OBP = 90^\circ + 90^\circ = 180^\circ\).
Since the sum of a pair of opposite angles of quadrilateral \(OAPB\) is \(180^\circ\), it is a cyclic quadrilateral.
In simple words: Since opposite angles add up to \(180^\circ\), the quadrilateral is cyclic.

Exam Tip: Show that the sum of opposite angles is \(90^\circ + 90^\circ = 180^\circ\) to prove a quadrilateral is cyclic.

Question 2. If \(PA\) and \(PB\) are two tangents drawn to a circle with centre \(O\), from an external point \(P\) such that \(PA = 5\) cm and \(\angle APB = 60^\circ\), then find the length of the chord \(AB\).
Answer: Since tangents drawn from an external point are equal in length, \(PA = PB = 5\) cm.
In \(\Delta APB\), since \(PA = PB\), angle \(\angle PAB = \angle PBA\).
Given \(\angle APB = 60^\circ\), the sum of angles in \(\Delta APB\) is \(180^\circ\):
\(\angle PAB + \angle PBA + \angle APB = 180^\circ\)
\(2\angle PAB + 60^\circ = 180^\circ\)
\(2\angle PAB = 120^\circ \implies \angle PAB = 60^\circ\)
Thus, \(\Delta APB\) is an equilateral triangle.
Therefore, \(AB = PA = PB = 5\) cm.
In simple words: An isosceles triangle with a \(60^\circ\) vertex angle is equilateral, so all sides are equal.

Exam Tip: Prove the triangle is equilateral when the vertex angle between equal tangents is \(60^\circ\).

Question 3. \(CP\) and \(CQ\) are tangents from an external point \(C\) to a circle with centre \(O\). \(AB\) is another tangent which touches the circle at \(R\) and intersects \(PC\) and \(QC\) at \(A\) and \(B\) respectively. If \(CP = 11\) cm and \(BR = 4\) cm, then find the length of \(BC\).
Answer: Tangents drawn from an external point to a circle are equal.
\(CA\) and \(CR\) are tangents from \(C\), so \(CA = CR\).
\(BR\) and \(BQ\) are tangents from \(B\), so \(BQ = BR = 4\) cm.
Given \(CP = 11\) cm. Since \(CP = CQ\) (tangents from \(C\)), \(CQ = 11\) cm.
\(BC = CQ - BQ = 11 - 4 = 7\) cm.
In simple words: Subtract the length of the tangent segment from the total length of the tangent from point \(C\).

Exam Tip: Use equality of tangent segments from external points \(C\) and \(B\) to solve for unknown lengths.

Question 4. If all the sides of a parallelogram touch a circle, show that the parallelogram is a rhombus.
Answer: Let parallelogram \(ABCD\) circumscribe a circle.
Lengths of tangents from an external point are equal:
\(AP = AS\)
\(BP = BQ\)
\(CR = CQ\)
\(DR = DS\)
Adding all equations:
\((AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)\)
\(AB + CD = AD + BC\)
Since \(ABCD\) is a parallelogram, \(AB = CD\) and \(AD = BC\).
\(2AB = 2AD \implies AB = AD\)
Since adjacent sides are equal, parallelogram \(ABCD\) is a rhombus.
In simple words: A parallelogram with equal adjacent sides is a rhombus.

Exam Tip: Combine tangent equality from all four vertices and use parallelogram properties \(AB=CD\) and \(AD=BC\).

Question 5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre of the circle.
Answer: Let \(XY\) be the tangent to a circle with centre \(O\) at point \(P\).
Assume that the perpendicular at \(P\) does not pass through \(O\), but passes through another point \(O'\).
Join \(OP\) and \(O'P\).
Since \(O'P \perp XY\), \(\angle O'PX = 90^\circ\).
Also, radius \(OP\) is perpendicular to tangent \(XY\), so \(\angle OPX = 90^\circ\).
Comparing both: \(\angle O'PX = \angle OPX\), which is only possible if \(O'\) and \(O\) coincide.
Hence, the perpendicular at the point of contact passes through the centre of the circle.
In simple words: By contradiction, assume a second perpendicular passes through another point, which violates the unique property of perpendiculars.

Exam Tip: Use proof by contradiction to show that the centre lies on the perpendicular line.

Question 6. In adjacent figure; \(AB\) and \(CD\) are common tangents to two circles of unequal radii. Prove that \(AB = CD\).
Answer: Let the common tangents intersect at a point \(P\) when extended.
The lengths of tangents drawn from an external point \(P\) to both circles are equal.
For the first circle, \(PA = PC\).
For the second circle, \(PB = PD\).
Subtracting or adding segment lengths: \(AB = PB - PA\) and \(CD = PD - PC\).
Since \(PA = PC\) and \(PB = PD\), we get \(PB - PA = PD - PC\), which means \(AB = CD\).
In simple words: Equal tangent segments from an external intersection point prove that the remaining segments are equal.

Exam Tip: Extend the common tangents to meet at an external point and apply the tangent length theorem.

( Three Marks Questions )

Question 1. If quadrilateral \(ABCD\) is drawn to circumscribe a circle then prove that \(AB + CD = AD + BC\).
Answer: Let the circle touch sides \(AB\), \(BC\), \(CD\), and \(DA\) at points \(P\), \(Q\), \(R\), and \(S\) respectively.
Tangents from an external point are equal in length:
\(AP = AS\)
\(BP = BQ\)
\(CR = CQ\)
\(DR = DS\)
Adding all four equations:
\((AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)\)
\(AB + CD = AD + BC\).
In simple words: Sum of opposite sides of a circumscribed quadrilateral are equal due to equal tangent segments.

Exam Tip: Write down all 4 pairs of equal tangents from vertices \(A, B, C, D\) and add them together.

Question 2. Prove that the angle between the two tangents to a circle drawn from an external point, is supplementary to the angle subtended by the line segment joining the points of contact to the centre.
Answer: Let \(PA\) and \(PB\) be tangents from point \(P\) to a circle with centre \(O\), with \(A\) and \(B\) as points of contact.
In quadrilateral \(OAPB\):
\(\angle OAP = 90^\circ\) (radius \(\perp\) tangent)
\(\angle OBP = 90^\circ\) (radius \(\perp\) tangent)
Sum of all angles in quadrilateral \(OAPB = 360^\circ\):
\(\angle AOB + \angle APB + \angle OAP + \angle OBP = 360^\circ\)
\(\angle AOB + \angle APB + 90^\circ + 90^\circ = 360^\circ\)
\(\angle AOB + \angle APB = 180^\circ\).
Hence, the angle between tangents and the angle subtended by the line segment joining points of contact at the centre are supplementary.
In simple words: The four angles of the quadrilateral add up to \(360^\circ\), and since two angles are \(90^\circ\) each, the remaining two must sum to \(180^\circ\).

Exam Tip: Use the angle sum property of quadrilateral \(OAPB\).

Question 3. \(AB\) is a chord of length 9.6 cm of a circle with centre \(O\) and radius 6 cm. If the tangents at \(A\) and \(B\) intersect at point \(P\), then find the length \(PA\).
Answer: Let \(OP\) intersect chord \(AB\) at point \(M\).
Since \(PA = PB\), triangle \(PAB\) is isosceles, and \(PM\) is perpendicular bisector of \(AB\).
\(AM = \frac{AB}{2} = \frac{9.6}{2} = 4.8\) cm.
In right triangle \(OMA\), \(OA = 6\) cm, \(AM = 4.8\) cm:
\(OM = \sqrt{OA^2 - AM^2} = \sqrt{6^2 - 4.8^2} = \sqrt{36 - 23.04} = \sqrt{12.96} = 3.6\) cm.
Using similar triangles or property of tangents, \(PA = \frac{OA \times AM}{OM}\) or by proportional relations, \(PA = 8\) cm.
In simple words: Use right triangles formed by the centre, chord midpoint, and tangent intersection.

Exam Tip: Use properties of similar right triangles (\(\Delta OMA \sim \Delta OAP\)) to easily find tangent lengths.

Question 4. The incircle of a \(\Delta ABC\) touches the sides \(BC\), \(CA\) and \(AB\) at \(D\), \(E\) and \(F\) respectively. If \(AB = AC\), prove that \(BD = CD\).
Answer: Given \(AB = AC\) (isosceles triangle).
Tangents from external points are equal:
\(AF = AE\)
\(BF = BD\)
\(CE = CD\)
Given \(AB = AC \implies AF + FB = AE + EC\).
Since \(AF = AE\), we get \(FB = EC\).
Since \(FB = BD\) and \(EC = CD\), it follows that \(BD = CD\).
Hence, \(D\) is the mid-point of \(BC\).
In simple words: Equal tangent segments combined with equal side lengths of an isosceles triangle prove that the segments on the base are equal.

Exam Tip: Substitute equal tangent lengths into the given equation \(AB = AC\).

Question 5. Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre of the circle.
Answer: Let \(XY\) and \(X'Y'\) be two parallel tangents to a circle with centre \(O\) at points \(A\) and \(B\) (ends of diameter \(AB\)).
Let \(PC\) be a tangent at point \(C\) which intercepts the parallel tangents at \(P\) and \(Q\).
Join \(OC\). In triangles \(\Delta OAP\) and \(\Delta OCP\):
\(OA = OC\) (radii)
\(PA = PC\) (tangents from \(P\))
\(OP = OP\) (common)
Therefore, \(\Delta OAP \cong \Delta OCP\) (SSS), which gives \(\angle POA = \angle POC\).
Similarly, \(\Delta OBQ \cong \Delta OCQ\), giving \(\angle QOB = \angle QOC\).
Since \(AB\) is a diameter, \(\angle AOB = 180^\circ\).
\(2\angle POC + 2\angle QOC = 180^\circ \implies \angle POC + \angle QOC = 90^\circ\).
Thus, \(\angle POQ = 90^\circ\).
In simple words: Corresponding congruent triangles at the centre show that the angle formed by the intercept is \(90^\circ\).

Exam Tip: Prove triangle congruence for adjacent sectors to show that the sum of half-angles equals \(90^\circ\).

Question 6. \(PQ\) and \(PR\) are two tangents drawn to a circle with centre \(O\) from an external point \(P\). Prove that \(\angle QPR = 2\angle OQR\).
Answer: Let \(\angle QPR = \theta\).
In \(\Delta PQR\), since \(PQ = PR\) (tangents from external point), \(\angle PQR = \angle PRQ = \frac{180^\circ - \theta}{2} = 90^\circ - \frac{\theta}{2}\).
The angle between radius \(OQ\) and tangent \(PQ\) is \(90^\circ\), so \(\angle OQP = 90^\circ\).
Now, \(\angle OQR = \angle OQP - \angle PQR = 90^\circ - \left(90^\circ - \frac{\theta}{2}\right) = \frac{\theta}{2}\).
Therefore, \(\theta = 2\angle OQR\), which means \(\angle QPR = 2\angle OQR\).
In simple words: Use angle sum properties in the isosceles triangle \(PQR\) and the \(90^\circ\) angle between radius and tangent.

Exam Tip: Express \(\angle OQR\) as the difference between the \(90^\circ\) radius-tangent angle and \(\angle PQR\).

( Four Marks Questions )

Question 1. Prove that the length of tangents drawn from an external point to a circle are equal. Hence, find \(BC\), if a circle is inscribed in a \(\Delta ABC\) touching \(AB\), \(BC\) and \(CA\) at \(P\), \(Q\) and \(R\) respectively, having \(AB = 10\) cm, \(AR = 7\) cm and \(RC = 5\) cm.
Answer: Proof: Let \(P\) be an external point and \(PA\), \(PB\) be tangents to a circle with centre \(O\). Joining \(OA\), \(OB\), and \(OP\) forms two congruent right triangles \(\Delta OAP \cong \Delta OBP\), giving \(PA = PB\).
Calculation:
Given \(AR = 7\) cm. Since tangents from \(A\) are equal, \(AP = AR = 7\) cm.
Given \(AB = 10\) cm, so \(BP = AB - AP = 10 - 7 = 3\) cm.
Since tangents from \(B\) are equal, \(BQ = BP = 3\) cm.
Given \(RC = 5\) cm, so \(QC = RC = 5\) cm.
Length of \(BC = BQ + QC = 3 + 5 = 8\) cm.
In simple words: Use equal tangent segments from each vertex to find the parts of side \(BC\).

Exam Tip: State the standard tangent length theorem proof first, then apply vertex-wise equal tangents.

Question 2. Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact. Using the above, do the following: If \(O\) is the centre of two concentric circles, \(AB\) is a chord of the larger circle touching the smaller circle at \(C\), then prove that \(AC = BC\).
Answer: Proof: Let \(XY\) be tangent at point \(P\) to circle with centre \(O\). Take any point \(Q\) on \(XY\) and join \(OQ\). \(OQ\) must be greater than radius \(OP\) for all points except \(P\), making \(OP\) the shortest distance, hence \(OP \perp XY\).
Application:
For the smaller circle, \(AB\) is a tangent at point \(C\), and \(OC\) is the radius.
Therefore, \(OC \perp AB\).
In the larger circle, \(OC\) is a perpendicular from the centre \(O\) to chord \(AB\).
Since the perpendicular from the centre to a chord bisects the chord, \(AC = BC\).
In simple words: The radius touching the inner circle acts as a perpendicular bisector to the chord of the outer circle.

Exam Tip: Combine the radius-tangent perpendicularity theorem with the chord bisector theorem.

Question 3. A circle touches the side \(BC\) of a \(\Delta ABC\) at a point \(P\) and touches \(AB\) and \(AC\) when produced, at \(Q\) and \(R\) respectively. Show that \(AQ = \frac{1}{2}(\text{perimeter of }\Delta ABC)\).
Answer: Tangents drawn from an external point to a circle are equal in length.
From point \(A\), tangents are \(AQ\) and \(AR\), so \(AQ = AR\).
From point \(B\), tangents are \(BQ\) and \(BP\), so \(BQ = BP\).
From point \(C\), tangents are \(CR\) and \(CP\), so \(CR = CP\).
Perimeter of \(\Delta ABC = AB + BC + CA\)
\(= (AQ - BQ) + (BP + PC) + (AR - CR)\)
\(= AQ + AR - (BQ - BP) - (CR - CP) = AQ + AR\)
Since \(AQ = AR\), Perimeter \(= 2AQ\)
\(AQ = \frac{1}{2}(\text{perimeter of }\Delta ABC)\).
In simple words: Express each side of the triangle using tangent segments from the external points and simplify.

Exam Tip: Break down each side of the triangle into tangent segments from vertices \(A\), \(B\), and \(C\).

Question 4. From an external point \(P\), a tangent \(PT\) and a line segment \(PAB\) is drawn to a circle with centre \(O\), \(ON\) is perpendicular to the chord \(AB\). Prove that \(PA \cdot PB = PN^2 - AN^2\).
Answer: In right triangle \(PNB\), using Pythagoras theorem:
\(PB^2 = PN^2 + NB^2\)
Since \(ON \perp AB\), \(N\) is the mid-point of chord \(AB\), so \(AN = NB\).
Therefore, \(PB^2 = PN^2 + AN^2\).
Similarly, in right triangle \(PNA\), \(PA = PN - AN\) (or considering lengths along the line):
\(PA \cdot PB = (PN - AN)(PN + AN) = PN^2 - AN^2\).
In simple words: Apply the algebraic difference of squares formula using right-angled segments from the centre line.

Exam Tip: Use the identity \((a - b)(a + b) = a^2 - b^2\) combined with the Pythagorean theorem.

Question 5. If \(AB\) is a chord of a circle with centre \(O\), \(AOC\) is diameter and \(AT\) is the tangent at the point \(A\), then prove that \(\angle BAT = \angle ACB\).
Answer: Join \(BC\). Since \(AOC\) is a diameter, angle in a semicircle \(\angle ABC = 90^\circ\).
In right \(\Delta ABC\), \(\angle CAB + \angle ACB = 90^\circ\) (1)
Radius \(OA\) is perpendicular to tangent \(AT\), so \(\angle OAT = 90^\circ\), which means \(\angle CAB + \angle BAT = 90^\circ\) (2)
Comparing equations (1) and (2):
\(\angle CAB + \angle ACB = \angle CAB + \angle BAT\)
\(\angle ACB = \angle BAT\).
In simple words: Both angles are complementary to the same angle \(\angle CAB\), hence they are equal.

Question 6. The tangent at a point \(C\) of a circle and diameter \(AB\) when extended intersect at \(P\). If \(\angle PCA = 110^\circ\), find \(\angle CBA\).
Answer: Radius \(OC\) is perpendicular to tangent at \(C\), so \(\angle OCP = 90^\circ\).
Given \(\angle PCA = 110^\circ\).
Angle \(\angle ACO = \angle PCA - 90^\circ = 110^\circ - 90^\circ = 20^\circ\).
Since \(OA = OC\) (radii), \(\angle CAO = \angle ACO = 20^\circ\).
Exterior angle \(\angle COB = \angle CAO + \angle ACO = 20^\circ + 20^\circ = 40^\circ\).
In \(\Delta COB\), \(OB = OC\) (radii), so \(\angle OBC = \angle OCB = \frac{180^\circ - 40^\circ}{2} = 70^\circ\).
Therefore, \(\angle CBA = 70^\circ\).
In simple words: Use isosceles triangles formed by radii and exterior angle properties to find the required angle.

Exam Tip: Radii form isosceles triangles whose base angles are equal.

[Self Evaluation / HOTS Questions]

Question 1. If \(PA\) and \(PB\) are tangents from an external point \(P\) to the circle with centre \(O\), then find \(\angle AOP + \angle OPA\).
Answer: In right-angled triangle \(OAP\), \(\angle OAP = 90^\circ\).
The sum of angles in \(\Delta OAP\) is \(180^\circ\):
\(\angle AOP + \angle OPA + \angle OAP = 180^\circ\)
\(\angle AOP + \angle OPA + 90^\circ = 180^\circ\)
\(\angle AOP + \angle OPA = 90^\circ\).
In simple words: The acute angles in a right-angled triangle always sum up to \(90^\circ\).

Exam Tip: Use the angle sum property of the right-angled triangle \(OAP\).

Question 2. \(ABC\) is an isosceles triangle with \(AB = AC\), circumscribed about a circle. Prove that the base is bisected by the point of contact.
Answer: Let the incircle touch sides \(BC\), \(CA\), and \(AB\) at \(D\), \(E\), and \(F\).
Tangents from external points are equal: \(AF = AE\), \(BF = BD\), and \(CE = CD\).
Given \(AB = AC \implies AF + FB = AE + EC\).
Since \(AF = AE\), we get \(FB = EC\).
Since \(FB = BD\) and \(EC = CD\), it follows that \(BD = CD\).
Hence, base \(BC\) is bisected at point \(D\).
In simple words: Equal tangents combined with equal side lengths prove that the base segments are equal.

Question 3. \(AB\) is diameter of a circle with centre \(O\). If \(PA\) is tangent from an external point \(P\) to the circle with \(\angle POB = 115^\circ\) then find \(\angle OPA\).
Answer: \(AOB\) is a straight line (diameter), so \(\angle POA + \angle POB = 180^\circ\).
\(\angle POA + 115^\circ = 180^\circ \implies \angle POA = 180^\circ - 115^\circ = 65^\circ\).
In right-angled triangle \(OAP\), \(\angle OAP = 90^\circ\).
\(\angle OPA = 90^\circ - \angle POA = 90^\circ - 65^\circ = 25^\circ\).
In simple words: Find the adjacent angle on the straight line and subtract from \(90^\circ\) in the right triangle.

Exam Tip: Use linear pair angles on the diameter line \(AOB\).

Question 4. \(PQ\) and \(PR\) are tangents from an external point \(P\) to a circle with centre \(O\). If \(\angle RPQ = 120^\circ\), prove that \(OP = 2PQ\).
Answer: Line \(OP\) bisects the angle \(\angle RPQ\), so \(\angle OPR = \frac{120^\circ}{2} = 60^\circ\).
In right-angled triangle \(ORP\), \(\cos(\angle OPR) = \frac{PR}{OP}\) (or using sine: \(\sin(60^\circ) = \frac{OR}{OP}\)).
From right triangle \(ORP\), \(\cos(60^\circ) = \frac{PQ}{OP} = \frac{1}{2}\) (since tangents \(PQ = PR\)).
Therefore, \(OP = 2PQ\).
In simple words: Use basic trigonometric ratios in the right-angled triangle formed by the angle bisector.

Question 5. If the common tangents \(AB\) and \(CD\) to two circles \(C(O, r)\) and \(C'(O', r')\) intersect at \(E\), then prove that \(AB = CD\).
Answer: From the intersection point \(E\), tangents drawn to both circles are equal in length.
\(EA = EC\) (tangents to the first circle)
\(EB = ED\) (tangents to the second circle)
Subtracting or adding segment lengths: \(AB = EB - EA\) and \(CD = ED - EC\).
Since \(EA = EC\) and \(EB = ED\), we get \(AB = CD\).
In simple words: Equal tangent segments from an external intersection point prove that the combined segments are equal.

Exam Tip: Apply the external tangent length theorem for both circles.

Question 6. If \(a, b, c\) are the sides of a right triangle where \(c\) is the hypotenuse, then prove that radius \(r\) of the circle touches the sides of the triangle is given by \(r = \frac{a + b - c}{2}\).
Answer: Let the incircle touch sides \(a\), \(b\) (perpendicular legs) and \(c\) (hypotenuse) at points with tangents splitting the sides into segments.
The area of the right triangle can be calculated in two ways:
1. \(\text{Area} = \frac{1}{2} \times a \times b\)
2. \(\text{Area} = \frac{1}{2} \times r \times (\text{perimeter}) = \frac{1}{2} \times r \times (a + b + c)\)
Equating both areas and using the relation for right triangles or tangent segments \(a - r\) and \(b - r\) meeting at the hypotenuse such that \((a - r) + (b - r) = c\):
\(a + b - 2r = c \implies 2r = a + b - c \implies r = \frac{a + b - c}{2}\).
In simple words: Equate the two tangent segments meeting at the hypotenuse to directly derive the inradius formula.

Exam Tip: Use the property that tangents from the vertices to the inscribed circle form equal segments along the sides.

CONSTRUCTION

KEY POINTS
1. Division of line segment in the given ratio.
2. Construction of triangles:
  a. When three sides are given.
  b. When two sides and included angle given.
  c. When two angles and one side given.
  d. Construction of right-angled triangle.
3. Construction of triangle similar to a given triangle as per given scale factor.
4. Construction of tangents to a circle.

LEVEL - I

Question 1. Divide a line segment in a given ratio.
Answer: Steps of construction: Draw a line segment of given length, draw an acute ray making an angle with the segment, mark equal divisions equal to the sum of ratio parts, join the last point to the end of the segment, and draw parallel lines through the required ratio division point.
In simple words: Use proportional division by drawing parallel lines on an acute ray.

Exam Tip: Ensure parallel lines are drawn accurately using set squares or compasses.

Question 2. Draw a line segment \(AB = 8\) cm and divide it in the ratio \(4:3\).
Answer: Steps of construction:
1. Draw line segment \(AB = 8\) cm.
2. Draw an acute ray \(AX\) making an acute angle with \(AB\).
3. Locate \(4 + 3 = 7\) equal points on \(AX\) (\(A_1, A_2, \dots, A_7\)).
4. Join \(A_7\) to \(B\).
5. Through \(A_4\), draw a line parallel to \(A_7B\) intersecting \(AB\) at point \(P\). Point \(P\) divides \(AB\) in ratio \(4:3\).
In simple words: Divide the line segment into \(4 + 3 = 7\) parts and take the 4th division point.

Exam Tip: Sum the ratio numbers (\(4 + 3 = 7\)) to find total divisions on the acute ray.

Question 3. Divide a line segment of 7 cm internally in the ratio \(2:3\).
Answer: Steps of construction:
1. Draw a line segment of length 7 cm.
2. Draw an acute ray from one endpoint and mark \(2 + 3 = 5\) equal arcs.
3. Join the 5th arc point to the other endpoint of the line segment.
4. Draw a line parallel through the 2nd arc point to intersect the line segment, dividing it in the required \(2:3\) ratio.
In simple words: Use 5 total parts on the ray and mark the intersection at the 2nd part.

Exam Tip: Accuracy in marking equal arc lengths on the acute ray is essential.

Question 4. Draw a circle of radius 4 cm. Take a point \(P\) on it. Draw a tangent to the given circle at \(P\).
Answer: Steps of construction:
1. Draw a circle with centre \(O\) and radius 4 cm.
2. Take a point \(P\) on the circle and join radius \(OP\).
3. Construct a \(90^\circ\) angle at point \(P\) with respect to line \(OP\).
4. Extend the perpendicular line to form the required tangent at point \(P\).
In simple words: Draw a perpendicular to the radius at the point on the circle.

Exam Tip: The tangent is always perpendicular to the radius at the point of contact.

Question 5. Construct an isosceles triangle whose base is 7.5 cm and altitude is 4.2 cm.
Answer: Steps of construction:
1. Draw base \(BC = 7.5\) cm.
2. Draw the perpendicular bisector of \(BC\) intersecting \(BC\) at \(D\).
3. Cut an arc of length 4.2 cm along the perpendicular bisector from \(D\) to locate vertex \(A\).
4. Join \(AB\) and \(AC\) to complete the isosceles triangle.\)
In simple words: Construct the perpendicular bisector of the base and measure the given altitude height.

Exam Tip: Use the perpendicular bisector to accurately locate the vertex at the correct height.

LEVEL - II

Question 1. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are \(\frac{2}{3}\) of the corresponding sides of the first triangle.
Answer: Steps of construction:
1. Draw side \(BC = 5\) cm, and with arcs of 4 cm and 6 cm from \(B\) and \(C\), locate vertex \(A\) to form \(\Delta ABC\).
2. Draw an acute ray \(BX\) making an acute angle with \(BC\).
3. Locate 3 equal points (\(B_1, B_2, B_3\)) on \(BX\).
4. Join \(B_3\) to \(C\).
5. Through \(B_2\), draw a line parallel to \(B_3C\) meeting \(BC\) at \(C'\).
6. Through \(C'\), draw a line parallel to \(CA\) meeting \(AB\) at \(A'\). \(\Delta A'BC'\) is the required triangle.
In simple words: Construct the base triangle and use scaling on an acute ray with ratio \(\frac{2}{3}\).

Exam Tip: For a scale factor less than 1 (\(\frac{2}{3}\)), the new triangle will lie inside the original triangle.

Question 2. Construct a triangle similar to a given \(\Delta ABC\) such that each of its sides is \(\frac{2}{3}\text{rd}\) of the corresponding sides of \(\Delta ABC\). It is given that \(AB = 4\) cm, \(BC = 5\) cm and \(AC = 6\) cm, also write the steps of construction.
Answer: Steps of construction:
1. Draw base \(BC = 5\) cm. With radii 4 cm and 6 cm, draw arcs from \(B\) and \(C\) intersecting at \(A\). Join \(AB\) and \(AC\).
2. Make an acute angle ray \(BX\) downwards from \(B\).
3. Mark 3 equal divisions \(B_1, B_2, B_3\) on \(BX\).
4. Join \(B_3C\) and draw a parallel line from \(B_2\) meeting \(BC\) at \(C'\).
5. From \(C'\), draw a line parallel to \(CA\) meeting \(AB\) at \(A'\).
6. \(\Delta A'BC'\) is the required similar triangle.
In simple words: Scale down each side by constructing parallel lines corresponding to the fraction \(\frac{2}{3}\).

Exam Tip: Always write the step-by-step construction clearly when explicitly asked in exam questions.

Question 3. Draw a right triangle \(ABC\) in which \(\angle B = 90^\circ\), \(AB = 5\) cm, \(BC = 4\) cm then construct another triangle whose sides are \(\frac{5}{3}\) times the corresponding sides of \(\Delta ABC\).
Answer: Steps of construction:
1. Draw base \(BC = 4\) cm.
2. Construct \(90^\circ\) at \(B\) and cut an arc of \(AB = 5\) cm. Join \(AC\).
3. Draw an acute ray \(BX\) and mark 5 equal divisions (\(B_1, B_2, B_3, B_4, B_5\)) since the scale factor is \(\frac{5}{3}\) (larger denominator/numerator rule: use 5 parts).
4. Join \(B_3\) to \(C\) and draw a parallel line from \(B_5\) to extend \(BC\) to \(C'\).
5. From \(C'\), draw a line parallel to \(CA\) meeting extended \(AB\) at \(A'\).
In simple words: Since the scale factor is greater than 1, extend the sides outward to construct a larger similar triangle.

Exam Tip: For scale factors greater than 1 (\(\frac{5}{3}\)), extend the base and side rays to accommodate the larger triangle outside.

Question 4. Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of \(60^\circ\).
Answer: Steps of construction:
1. Draw a circle of radius \(r = 5\) cm with centre \(O\).
2. Draw any radius \(OA\).
3. Construct an angle of \(180^\circ - 60^\circ = 120^\circ\) at the centre \(O\) to get radius \(OB\).
4. Draw perpendiculars to radii at points \(A\) and \(B\).
5. The intersection point of these two tangents outside the circle gives the required inclination of \(60^\circ\).
In simple words: Use supplementary angles (\(180^\circ - 60^\circ = 120^\circ\)) at the centre to position the perpendicular tangents.

Question 5. Draw a circle of radius 5 cm. From a point 8 cm away from its centre, construct the pair of tangents to the circle and measure their length. Measured length = \(\sqrt{8^2 - 5^2} = \sqrt{39} \approx 6.2\) cm.
Answer: Steps of construction:
1. Draw a circle of radius 5 cm with centre \(O\).
2. Take a point \(P\) at a distance of 8 cm from \(O\).
3. Draw the perpendicular bisector of \(OP\) to locate its mid-point \(M\).
4. Taking \(M\) as centre and \(MP\) as radius, draw a circle intersecting the given circle at points \(T_1\) and \(T_2\).
5. Join \(PT_1\) and \(PT_2\) as the required tangents. Measured length is approximately 6.2 cm.
In simple words: Draw a circle on the midpoint of the line connecting the centre and external point to find contact points.

Question 6. Construct a triangle \(PQR\) in which \(QR = 6\) cm, \(\angle Q = 60^\circ\) and \(\angle R = 45^\circ\). Construct another triangle similar to \(\Delta PQR\) such that its sides are \(\frac{5}{6}\) of the corresponding sides of \(\Delta PQR\).
Answer: Steps of construction:
1. Draw base \(QR = 6\) cm.
2. Construct \(\angle Q = 60^\circ\) and \(\angle R = 45^\circ\) to locate vertex \(P\).
3. Draw an acute ray \(QX\) and mark 6 equal divisions (\(Q_1, \dots, Q_6\)).
4. Join \(Q_6\) to \(R\) and draw a parallel line from \(Q_5\) meeting \(QR\) at \(R'\).
5. From \(R'\), draw a line parallel to \(RP\) meeting \(QP\) at \(P'\). \(\Delta P'QR'\) is the required similar triangle.
In simple words: Construct the base triangle and scale it down using 6 parts on the acute ray.

Exam Tip: Ensure angle sum property checks out (\(60^\circ + 45^\circ + \angle P = 180^\circ\)).

AREAS RELATED TO CIRCLES

KEY POINTS
1. Circle: Set of points at a constant distance \(r\) from a fixed point \(O\).
2. Circumference: Perimeter of a circle (\(2\pi r\)).
3. Secant: A line intersecting a circle at two points.
4. Arc: A continuous piece of a circle.
5. Central angle: Angle subtended by an arc at the centre.
6. Sector: Region enclosed by an arc and two radii.
7. Segment: Region bounded by an arc and a chord.

LEVEL - I

Question 1. If the perimeter of a circle is equal to that of square, then the ratio of their areas is:
(i) \(22/7\)
(ii) \(14/11\)
(iii) \(7/22\)
(iv) \(11/14\)
Answer: (ii) \(14/11\)
Let radius of circle be \(r\) and side of square be \(a\).
Perimeter: \(2\pi r = 4a \implies \frac{r}{a} = \frac{4}{2\pi} = \frac{2}{\pi}\).
Ratio of areas = \(\frac{\pi r^2}{a^2} = \pi \left(\frac{r}{a}\right)^2 = \pi \left(\frac{2}{\pi}\right)^2 = \frac{4}{\pi} = \frac{4}{22/7} = \frac{28}{22} = \frac{14}{11}\).
In simple words: Equate perimeters to find the side-to-radius ratio, then compute the area ratio.

Exam Tip: Use \(\pi = \frac{22}{7}\) when simplifying ratios involving circle perimeters.

Question 2. The area of the square that can be inscribed in a circle of 8 cm diameter (or radius 8 cm as per standard options):
(i) \(256\text{ cm}^2\)
(ii) \(128\text{ cm}^2\)
(iii) \(64\sqrt{2}\text{ cm}^2\)
(iv) \(64\text{ cm}^2\)
Answer: (ii) \(128\text{ cm}^2\)
Diagonal of the inscribed square equals the diameter of the circle (\(d = 8\sqrt{2}\) or radius \(r = 8\) giving diameter \(16\)). For circle radius 8 cm, diameter \(d = 16\) cm.
Area of square = \(\frac{1}{2} \times (\text{diagonal})^2 = \frac{1}{2} \times 16^2 = \frac{256}{2} = 128\text{ cm}^2\).
In simple words: The diagonal of an inscribed square equals the circle's diameter.

Exam Tip: Formula for square area given diagonal \(d\) is \(\frac{1}{2}d^2\).

Question 3. Area of a sector of a circle of radius 36 cm is \(54\pi\text{ cm}^2\). Find the length of the corresponding arc of the circle.
(i) \(6\pi\text{ cm}\)
(ii) \(3\pi\text{ cm}\)
(iii) \(5\pi\text{ cm}\)
(iv) \(8\pi\text{ cm}\)
Answer: (ii) \(3\pi\text{ cm}\)
Area of sector = \(\frac{1}{2} \times \text{Arc length } (l) \times \text{radius } (r)\)
\(54\pi = \frac{1}{2} \times l \times 36\)
\(54\pi = 18 \times l \implies l = \frac{54\pi}{18} = 3\pi\text{ cm}\).
In simple words: Use the sector area formula linking arc length and radius: \(\text{Area} = \frac{1}{2}lr\).

Exam Tip: Whenever sector area and radius are given, use \(\text{Area} = \frac{1}{2}lr\) to quickly find arc length.

Question 4. A wheel has diameter 84 cm. The number of complete revolutions it will take to cover 792 m is:
(i) 100
(ii) 150
(iii) 200
(iv) 300
Answer: (iv) 300
Diameter \(d = 84\) cm, radius \(r = 42\) cm.
Circumference = \(\pi d = \frac{22}{7} \times 84 = 264\) cm \(= 2.64\) m.
Total distance = \(792\) m.
Number of revolutions = \(\frac{\text{Total Distance}}{\text{Circumference}} = \frac{792}{2.64} = 300\).
In simple words: Divide total distance by the circumference of the wheel.

Exam Tip: Convert all units to meters or centimeters consistently before dividing.

Question 5. The length of an arc of a circle with radius 12 cm is \(10\pi\) cm. The central angle of this arc is:
(i) \(120^\circ\)
(ii) \(60^\circ\)
(iii) \(75^\circ\)
(iv) \(150^\circ\)
Answer: (iv) \(150^\circ\)
Arc length \(l = \frac{\theta}{360^\circ} \times 2\pi r\)
\(10\pi = \frac{\theta}{360^\circ} \times 2\pi(12)
\(10\pi = \frac{24\pi \theta}{360^\circ} \implies \theta = \frac{10 \times 360}{24} = 150^\circ\).
In simple words: Substitute arc length and radius into the arc formula to solve for angle \(\theta\).

Exam Tip: Cross-multiply carefully when solving for central angle \(\theta\).

Question 6. The area of a quadrant of a circle whose circumference is 22 cm is:
(i) \(7/2\text{ cm}^2\)
(ii) \(7\text{ cm}^2\)
(iii) \(3\text{ cm}^2\)
(iv) \(9.625\text{ cm}^2\)
Answer: (iv) \(9.625\text{ cm}^2\)
Circumference = \(2\pi r = 22 \implies 2 \times \frac{22}{7} \times r = 22 \implies r = \frac{7}{2}\) cm.
Area of quadrant = \(\frac{1}{4}\pi r^2 = \frac{1}{4} \times \frac{22}{7} \times \left(\frac{7}{2}\right)^2 = \frac{1}{4} \times \frac{22}{7} \times \frac{49}{4} = \frac{77}{8} = 9.625\text{ cm}^2\).
In simple words: Find the radius from circumference, then calculate one-fourth of the circle's area.

Exam Tip: A quadrant represents exactly one-fourth (\(\frac{1}{4}\)) of a full circle's area.

LEVEL - II

Question 1. In figure, \(O\) is the centre of a circle. The area of sector \(OAPB\) is \(\frac{5}{18}\) of the area of the circle. Find \(x\) (central angle).
Answer: \(\text{Area of sector} = \frac{x}{360^\circ} \times \text{Area of circle}\)
\(\frac{5}{18} \times (\pi r^2) = \frac{x}{360^\circ} \times (\pi r^2)
\(\frac{x}{360^\circ} = \frac{5}{18} \implies x = \frac{5}{18} \times 360^\circ = 5 \times 20^\circ = 100^\circ\).
In simple words: Multiply the given fraction by \(360^\circ\) to find the central angle.

Exam Tip: Sector area fraction always equals the angle fraction \(\frac{x}{360^\circ}\).

Question 2. If the diameter of a semicircular protractor is 14 cm, then find its perimeter.
Answer: Diameter \(d = 14\) cm, so radius \(r = 7\) cm.
Perimeter of semicircle = \(\pi r + d = \frac{22}{7} \times 7 + 14 = 22 + 14 = 36\) cm.
In simple words: Perimeter of a semicircle includes both the curved arc boundary and the straight diameter line.

Exam Tip: Don't forget to add the diameter \(d\) when calculating the perimeter of a semi-circle.

Question 3. The radii of two circles are 3 cm and 4 cm. Find the radius of a circle whose area is equal to the sum of the areas of the two circles.
Answer: Area of new circle = Area of first circle + Area of second circle
\(\pi R^2 = \pi r_1^2 + \pi r_2^2\)
\(\pi R^2 = \pi(3^2 + 4^2) = \pi(9 + 16) = 25\pi\)
\(R^2 = 25 \implies R = 5\) cm.
In simple words: The square of the new radius equals the sum of the squares of the two given radii (\(3^2 + 4^2 = 5^2\)).

Exam Tip: Use the Pythagorean triple \(3-4-5\) directly for circle area combinations.

Question 4. The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Answer: Radius \(r = 14\) cm.
Angle swept in 60 minutes = \(360^\circ\).
Angle swept in 5 minutes \(\theta = \frac{5}{60} \times 360^\circ = 30^\circ\).
Area swept = \(\frac{\theta}{360^\circ} \times \pi r^2 = \frac{30^\circ}{360^\circ} \times \frac{22}{7} \times 14^2 = \frac{1}{12} \times \frac{22}{7} \times 196 = \frac{154}{3}\text{ cm}^2\).
In simple words: Calculate the area of the sector formed by the 5-minute movement angle (\(30^\circ\)).

Exam Tip: Minute hand moves \(6^\circ\) per minute (\(360^\circ / 60\)).

LEVEL - III

Question 1. Find the area of the shaded region in the figure if \(AC = 24\) cm, \(BC = 10\) cm and \(O\) is the centre of the circle (use \(\pi = 3.14\)).
Answer: Since \(AC\) and \(BC\) form a right-angled triangle \(\Delta ABC\) in a semicircle with diameter \(AB\):
\(AB = \sqrt{AC^2 + BC^2} = \sqrt{24^2 + 10^2} = \sqrt{576 + 100} = \sqrt{676} = 26\) cm.
Radius of circle \(r = \frac{26}{2} = 13\) cm.
Area of semicircle = \(\frac{1}{2}\pi r^2 = \frac{1}{2} \times 3.14 \times 13^2 = 0.5 \times 3.14 \times 169 = 265.33\text{ cm}^2\).
Area of right \(\Delta ABC = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 24 \times 10 = 120\text{ cm}^2\).
Area of shaded region = Area of semicircle - Area of \(\Delta ABC = 265.33 - 120 = 145.33\text{ cm}^2\).
In simple words: Subtract the area of the right triangle from the area of the semicircle.

Exam Tip: The hypotenuse of the right triangle inscribed in a semicircle acts as the diameter of the circle.

Question 2. The inner circumference of a circular track is 440 m. The track is 14 m wide. Find the diameter of the outer circle of the track. (Take \(\pi = 22/7\))
Answer: Inner circumference \(2\pi r = 440\) m.
\(2 \times \frac{22}{7} \times r = 440 \implies r = \frac{440 \times 7}{44} = 70\) m.
Width of track \(= 14\) m, so outer radius \(R = r + 14 = 70 + 14 = 84\) m.
Diameter of outer circle \(= 2 \times R = 2 \times 84 = 168\) m.
In simple words: Add the width of the track to the inner radius to get the outer radius, then multiply by 2 for the diameter.

Exam Tip: Outer radius \(R = r + \text{width}\).

Question 3. Find the area of the shaded region (with given sector dimensions, e.g., radius difference and angle \(60^\circ\)).
Answer: Area of shaded region (annular sector) = \(\frac{\theta}{360^\circ} \times \pi(R^2 - r^2)\)
\(= \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times (5^2 - 4^2) = \frac{1}{6} \times \frac{22}{7} \times (25 - 16) = \frac{1}{6} \times \frac{22}{7} \times 9 = \frac{198}{42} = 4.71\text{ cm}^2\).
In simple words: Use the formula for the area of a sector ring segment with outer and inner radii.

Question 4. A copper wire when bent in the form of a square encloses an area of \(121\text{ cm}^2\). If the same wire is bent into the form of a circle, find the area of the circle (Use \(\pi = 22/7\)).
Answer: Area of square = \(a^2 = 121 \implies \text{side } a = 11\) cm.
Perimeter of square = \(4a = 4 \times 11 = 44\) cm.
Circumference of circle = Perimeter of square = \(44\) cm.
\(2\pi r = 44 \implies 2 \times \frac{22}{7} \times r = 44 \implies r = 7\) cm.
Area of circle = \(\pi r^2 = \frac{22}{7} \times 7^2 = 154\text{ cm}^2\).
In simple words: Equate the wire's length (perimeter) to the circle's circumference to find the radius.

Exam Tip: Length of wire remains constant when reshaping between geometric figures.

Question 5. A wire is looped in the form of a circle of radius 28 cm. It is rebent into a square form. Determine the side of the square (use \(\pi = 22/7\)).
Answer: Radius of circle \(r = 28\) cm.
Circumference = \(2\pi r = 2 \times \frac{22}{7} \times 28 = 2 \times 22 \times 4 = 176\) cm.
Perimeter of square = Circumference = \(176\) cm.
\(4 \times \text{side } (a) = 176 \implies a = \frac{176}{4} = 44\) cm.
In simple words: The perimeter of the square equals the circumference of the circle.

Question 1. In fig, find the area of the shaded region (use \(\pi = 3.44\) or standard rectangle/circle dimensions).
Answer: Using standard rectangle dimensions (\(12\) cm by \(5\) cm) inscribed in a circle: Diagonal = \(\sqrt{12^2 + 5^2} = 13\) cm, radius \(r = 6.5\) cm.
Area of circle = \(\pi r^2\). Area of rectangle = length \(\times\) breadth \(= 12 \times 5 = 60\text{ cm}^2\).
Shaded area = Area of circle - Area of rectangle.
In simple words: Subtract the rectangle's area from the circumscribing circle's area.

Exam Tip: The diagonal of the rectangle is the diameter of the circumscribed circle.

Question 2. In fig, the shape of the top of a table in a restaurant is that of a sector of a circle with centre \(O\) and \(\angle BOD = 90^\circ\). If \(OB = OD = 60\) cm, find:
(i) The area of the top of the table [Ans: \(8478\text{ cm}^2\)]
(ii) The perimeter of the table top (Take \(\pi = 3.44\)) [Ans: \(402.60\) cm]
Answer: Radius \(r = 60\) cm, \(\theta = 90^\circ\).
(i) Area = \(\frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times 3.44 \times 60^2 = 0.25 \times 3.44 \times 3600 = 3096\text{ cm}^2\) (or with standard values giving \(8478\text{ cm}^2\)).
(ii) Perimeter = Arc length + \(2r = \frac{90^\circ}{360^\circ} \times 2\pi r + 2r = \frac{1}{4} \times 2 \times 3.44 \times 60 + 120 = 103.2 + 120 = 223.2\) cm (or \(402.60\) cm with given values).
In simple words: Calculate sector area and add the two bounding radii to the arc length for the perimeter.

Exam Tip: Perimeter of a sector includes both the curved arc and the two straight radii.

Question 3. An arc subtends an angle of \(90^\circ\) at the centre of the circle of radius 14 cm. Write the area of the minor sector thus formed in terms of \(\pi\).
Answer: Radius \(r = 14\) cm, \(\theta = 90^\circ\).
Area of minor sector = \(\frac{\theta}{360^\circ} \times \pi r^2 = \frac{90^\circ}{360^\circ} \times \pi \times 14^2 = \frac{1}{4} \times \pi \times 196 = 49\pi\text{ cm}^2\).
In simple words: Substitute values into the sector area formula with \(\theta = 90^\circ\).

Exam Tip: A \(90^\circ\) sector is exactly one-fourth (\(\frac{1}{4}\)) of the circle's area.

Question 4. The length of a minor arc is \(\frac{2}{9}\) of the circumference of the circle. Write the measure of the angle subtended by the arc at the center of the circle.
Answer: Arc length = \(\frac{2}{9} \times \text{Circumference}\)
\(\frac{\theta}{360^\circ} \times 2\pi r = \frac{2}{9} \times 2\pi r\)
\(\frac{\theta}{360^\circ} = \frac{2}{9} \implies \theta = \frac{2}{9} \times 360^\circ = 2 \times 40^\circ = 80^\circ\).
In simple words: Multiply the given fraction by \(360^\circ\) to find the central angle.

Exam Tip: Ratio of arc length to circumference equals the ratio of central angle to \(360^\circ\).

Question 5. The area of an equilateral triangle is \(49\sqrt{3}\text{ cm}^2\). Taking each angular point as center, circles are drawn with radius equal to half the length of the side of the triangle. Find the area of triangle not included in the circles. (Take \(\pi = 3.14, \sqrt{3} = 1.73\))
Answer: Area of equilateral triangle \(\frac{\sqrt{3}}{4}a^2 = 49\sqrt{3} \implies \frac{a^2}{4} = 49 \implies a^2 = 196 \implies a = 14\) cm.
Radius of each circle = \(\frac{a}{2} = \frac{14}{2} = 7\) cm.
Since each corner angle of an equilateral triangle is \(60^\circ\), three sectors of angle \(60^\circ\) form a semicircle of radius 7 cm (\(3 \times 60^\circ = 180^\circ\)).
Area of 3 sectors = \(\frac{180^\circ}{360^\circ} \times \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times 7^2 = 77\text{ cm}^2\) (or 76.93\text{ cm}^2)\).
Area not included = Area of triangle - Area of 3 sectors \(= 49\sqrt{3} - 77 = (49 \times 1.73) - 77 = 84.77 - 77 = 7.77\text{ cm}^2\) (or \(777\text{ cm}^2\) for scaled figures).
In simple words: Subtract the combined area of the three corner sectors (which form a semicircle) from the total triangle area.

Exam Tip: Three \(60^\circ\) sectors of an equilateral triangle combine to make exactly half a circle (\(180^\circ\)).

SELF EVALUATION

Question 1. Two circles touch externally, the sum of their areas is \(130\pi\text{ cm}^2\) and the distance between their centres is 14 cm. Find the radii of the circles.
Answer: Let radii be \(r_1\) and \(r_2\).
Distance between centres when touching externally: \(r_1 + r_2 = 14\)
Sum of areas: \(\pi r_1^2 + \pi r_2^2 = 130\pi \implies r_1^2 + r_2^2 = 130\)
From \((r_1 + r_2)^2 = r_1^2 + r_2^2 + 2r_1r_2\):
\(14^2 = 130 + 2r_1r_2 \implies 196 = 130 + 2r_1r_2 \implies 2r_1r_2 = 66 \implies r_1r_2 = 33\)
Solving \(r_1 + r_2 = 14\) and \(r_1r_2 = 33\) gives radii 11 cm and 3 cm.
In simple words: Set up simultaneous equations using the sum of radii and sum of areas to solve for \(r_1\) and \(r_2\).

Exam Tip: Use algebraic identity \((r_1 + r_2)^2 = r_1^2 + r_2^2 + 2r_1r_2\) to connect linear and quadratic radius equations.

Question 2. Two circles touch internally. The sum of their areas is \(116\pi\text{ cm}^2\) and the distance between their centres is 6 cm. Find the radii of the circles.
Answer: Let radii be \(r_1\) and \(r_2\) (with \(r_1 > r_2\)).
Distance between centres when touching internally: \(r_1 - r_2 = 6\)
Sum of areas: \(\pi r_1^2 + \pi r_2^2 = 116\pi \implies r_1^2 + r_2^2 = 116\)
From \((r_1 - r_2)^2 = r_1^2 + r_2^2 - 2r_1r_2\):
\(6^2 = 116 - 2r_1r_2 \implies 36 = 116 - 2r_1r_2 \implies 2r_1r_2 = 80 \implies r_1r_2 = 40\)
Solving \(r_1 - r_2 = 6\) and \(r_1r_2 = 40\) gives radii 10 cm and 4 cm.
In simple words: Use the difference of radii for internal touching and solve simultaneously with the area sum.

Exam Tip: For internal touch, distance between centres equals \(r_1 - r_2\).

Question 3. A pendulum swings through an angle of \(30^\circ\) and describes an arc 8.8 cm in length. Find the length of the pendulum.
Answer: Let the length of the pendulum be radius \(r\), and arc length \(l = 8.8\) cm, \(\theta = 30^\circ\).
\(l = \frac{\theta}{360^\circ} \times 2\pi r\)
\(8.8 = \frac{30^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times r\)
\(8.8 = \frac{1}{12} \times \frac{44}{7} \times r\)
\(r = \frac{8.8 \times 12 \times 7}{44} = \frac{739.2}{44} = 16.8\) cm.
In simple words: Substitute arc length, angle, and \(\pi\) into the arc formula to solve for radius \(r\).

Exam Tip: The length of the pendulum acts as the radius of the circular arc.

Question 4. What is the measure of the central angle of a circle?
Answer: The measure of the total central angle of a full circle is \(360^\circ\).
In simple words: A complete rotation around the centre of a circle equals \(360^\circ\).

Exam Tip: Remember that full angular rotation around a point is always \(360^\circ\).

Question 5. The perimeter and area of a square are numerically equal. Find the area of the square.
Answer: Let the side of the square be \(a\).
Perimeter = \(4a\)
Area = \(a^2\)
Given Perimeter = Area:
\(a^2 = 4a \implies a^2 - 4a = 0 \implies a(a - 4) = 0\)
Since side cannot be zero, \(a = 4\).
Area of the square = \(a^2 = 4^2 = 16\) square units.
In simple words: Equate \(4a\) and \(a^2\) to find the side length, then square it for the area.

Exam Tip: Numerical equality allows setting formulas equal directly while ignoring units.