NCERT Solutions for Class 10 Maths Chapter 10 Exercise 10.2

1. Radius \(\perp\) tangent: \(r^2 = 25^2 - 24^2 = 49\).

1. In quadrilateral OPTQ, \(\angle OPT = \angle OQT = 90^\circ\).
2. \[\begin{aligned}\angle PTQ &= 360^\circ - 90^\circ - 90^\circ - 110^\circ \\ &= 70^\circ\end{aligned}\]

1. \[\begin{aligned}\angle AOB &= 180^\circ - 80^\circ \\ &= 100^\circ\end{aligned}\], and OP bisects it (the triangles OAP and OBP are congruent).

1. Let AB be a diameter, with tangents \(l\) at A and \(m\) at B.
2. Each tangent is perpendicular to the radius: \(l \perp AB\) and \(m \perp AB\).
3. Two lines perpendicular to the same line are parallel (co-interior angles \(90^\circ + 90^\circ = 180^\circ\)).

1. Let the tangent touch at P and suppose the perpendicular to it at P does not pass through the centre O.
2. But OP is also perpendicular to the tangent at P (tangent \(\perp\) radius). Through P there is only one line perpendicular to the tangent, so the perpendicular must be OP, which passes through O.

1. \(r^2 = 5^2 - 4^2 = 9\).

1. The radius of the small circle to the point of contact is perpendicular to the chord and bisects it.
2. Half-chord \(= \sqrt{5^2 - 3^2} = 4\).

1. Let the circle touch AB, BC, CD, DA at P, Q, R, S.
2. Tangents from a point are equal: \(AP = AS\), \(BP = BQ\), \(CR = CQ\), \(DR = DS\).
3. Add: \[\begin{aligned}AB + CD &= (AP + BP) + (CR + DR) \\ &= (AS + BQ) + (CQ + DS) \\ &= AD + BC\end{aligned}\]

1. Join OC and the points of contact P (on XY) and Q (on X′Y′).
2. \(\triangle OPA \cong \triangle OCA\) (RHS), so OA bisects \(\angle POC\); similarly OB bisects \(\angle QOC\).
3. P, O, Q are collinear (both tangents are perpendicular to the diameter PQ), so \(\angle POC + \angle QOC = 180^\circ\).
4. Hence \[\begin{aligned}\angle AOB &= \tfrac12(\angle POC + \angle QOC) \\ &= 90^\circ\end{aligned}\]

1. Let PA, PB touch the circle (centre O) at A, B.
2. In quadrilateral OAPB, \(\angle OAP = \angle OBP = 90^\circ\).
3. So \[\begin{aligned}\angle APB + \angle AOB &= 360^\circ - 180^\circ \\ &= 180^\circ\end{aligned}\]

1. By Q8 (tangents from a point are equal), \(AB + CD = AD + BC\).
2. In a parallelogram \(AB = CD\) and \(AD = BC\), so \(2AB = 2BC\), i.e. \(AB = BC\).
3. A parallelogram with two adjacent sides equal is a rhombus.

1. Tangent lengths: \(BF = BD = 8\), \(CE = CD = 6\), \(AE = AF = x\). Sides: \(a = 14\), \(b = x + 6\), \(c = x + 8\); semi-perimeter \(s = x + 14\).
2. Area \(= rs = 4(x + 14)\), and by Heron \(= \sqrt{(x + 14)\cdot x \cdot 6 \cdot 8}\).
3. \[\begin{aligned}&16(x + 14)^2 = 48x(x + 14) \\ \Rightarrow\ &x + 14 = 3x \\ \Rightarrow\ &x = 7\end{aligned}\]

Where marks slip: Reject \(x = -14\): a length is positive.

1. Let the circle touch AB, BC, CD, DA at P, Q, R, S. Congruent right triangles give \(\angle AOP = \angle AOS\), \(\angle BOP = \angle BOQ\), \(\angle COQ = \angle COR\), \(\angle DOR = \angle DOS\).
2. The eight angles add to \(360^\circ\), so \[2(\angle AOP + \angle BOP + \angle COR + \angle DOR) = 360^\circ\]
3. Hence \(\angle AOB + \angle COD = 180^\circ\) (and likewise \(\angle BOC + \angle AOD = 180^\circ\)).