Section A (1 Mark each)
Q1. (a) 7 cm
Let radius be \(r\). Tangent \(PQ = 24\) cm, Distance \(OQ = 25\) cm.
In right \(\triangle OPQ\), \(r^2 + 24^2 = 25^2 \Rightarrow r^2 = 625 - 576 = 49 \Rightarrow r
= 7\) cm.
Q2. (b) \(70^\circ\)
\(\angle PTQ + \angle POQ = 180^\circ\) (Angles in quadrilateral with two \(90^\circ\)
angles).
\(\angle PTQ = 180^\circ - 110^\circ = 70^\circ\).
Q3. (a) 3 cm
Distance \(d = 5\) cm, Tangent \(l = 4\) cm.
Radius \(r = \sqrt{d^2 - l^2} = \sqrt{25 - 16} = \sqrt{9} = 3\) cm.
Q4. (b) Both A and R are true but R is not the correct explanation of A.
A secant intersects at two points (True). A tangent intersects at one point (True). These are independent definitions.
Section B (2 Marks each)
Q5. Prove tangents at ends of diameter are parallel.
Let \(PQ\) be diameter. Tangents \(AB\) at \(P\) and \(CD\) at \(Q\).
Radius \(\perp\) Tangent \(\Rightarrow \angle OPB = 90^\circ\) and \(\angle OQD = 90^\circ\).
Since \(\angle OPB + \angle OQD = 180^\circ\) (consecutive interior angles), lines \(AB
\parallel CD\).
Q6. Find chord length.
Let \(R = 5\) cm, \(r = 3\) cm. The chord of larger circle touches smaller circle, so it is
bisected at point of contact.
Half length = \(\sqrt{R^2 - r^2} = \sqrt{25 - 9} = 4\) cm.
Total length = \(2 \times 4 = 8\) cm.
Q7. Prove \(AB + CD = AD + BC\).
Tangents from external point are equal.
\(AP=AS, BP=BQ, CR=CQ, DR=DS\).
\(AB + CD = (AP+BP) + (CR+DR) = AS+BQ+CQ+DS = (AS+DS) + (BQ+CQ) = AD + BC\).
Section C (3 Marks each)
Q8. Prove angles are supplementary.
Let tangents be \(PA\) and \(PB\) from \(P\) to circle with centre \(O\).
In quadrilateral \(OAPB\), \(\angle OAP = \angle OBP = 90^\circ\).
Sum of angles = \(360^\circ \Rightarrow \angle APB + \angle AOB + 90^\circ + 90^\circ =
360^\circ\).
\(\angle APB + \angle AOB = 180^\circ\). Hence supplementary.
Q9. Prove \(\angle AOB = 90^\circ\).
In \(\triangle OPA\) and \(\triangle OCA\), \(OP=OC\) (radii), \(AP=AC\) (tangents), \(OA=OA\). So
\(\triangle OPA \cong \triangle OCA\).
\(\Rightarrow \angle POA = \angle COA\). Similarly \(\angle QOB = \angle COB\).
\(PQ\) is diameter \(\Rightarrow \angle POQ = 180^\circ\).
\(2\angle COA + 2\angle COB = 180^\circ \Rightarrow 2(\angle COA + \angle COB) = 180^\circ
\Rightarrow \angle AOB = 90^\circ\).
Section D (5 Marks)
Q10. Prove tangent lengths equal and Rhombus property.
Part 1: In \(\triangle OAP\) and \(\triangle OBP\) (Right angled), \(OA=OB\)
(radii), \(OP=OP\) (common). By RHS, \(\triangle OAP \cong \triangle OBP \Rightarrow
AP=BP\).
Part 2: ABCD is parallelogram \(\Rightarrow AB=CD, AD=BC\).
From Q7, \(AB+CD = AD+BC \Rightarrow 2AB = 2AD \Rightarrow AB=AD\).
Adjacent sides equal in parallelogram \(\Rightarrow\) It is a rhombus.
Section E (Case Study - 4 Marks)
(i) Length of tangent \(PT = \sqrt{OP^2 - r^2} = \sqrt{50^2 - 20^2} = \sqrt{2500 - 400} = \sqrt{2100} = 10\sqrt{21}\) m.
(ii) If angle between tangents is \(60^\circ\), then \(\angle TPO =
30^\circ\).
\(\sin 30^\circ = \frac{r}{OP} \Rightarrow \frac{1}{2} = \frac{20}{OP} \Rightarrow OP = 40\)
m.
(iii) Perimeter = \(2 \times \text{radius} + 2 \times \text{tangent}\).
Using values from (i): \(P = 2(20) + 2(10\sqrt{21}) = 40 + 20\sqrt{21}\) m.