(i) (b): OA²=AB²+OB²

\(\Rightarrow \quad O A=\sqrt{10^{2}+5^{2}}=5 \sqrt{5} \mathrm{~cm}\)
(ii) (a) : \(O A=\sqrt{O P^{2}-A P^{2}} \text { (Given) }\)
\(=\sqrt{17^{2}-15^{2}}=\sqrt{64}=8 \mathrm{~cm}\)
(iii) (b): Length of tangent \(A P=\sqrt{O P^{2}-O A^{2}} \text { (Given) }\)
\(=\sqrt{25^{2}-7^{2}}=\sqrt{576}=24 \mathrm{~cm}\)
(iv) (b):

\(\text { Since, } O P=\sqrt{(P T)^{2}+(O T)^{2}}=\sqrt{21^{2}+20^{2}}=29 \mathrm{~cm}\)
(v) (a): Since, OP² + PQ² = OQ²
\(\Rightarrow\) 8² + x² = (x + 2)²\(\Rightarrow\) 64 = 4x + 4\(\Rightarrow\) x = 15 cm
So, length of tangent, PQ = 15 cm.

(i) (b)
(ii) (c): In \(\Delta\)OAP and \(\Delta\)OBP,
\(\angle\)OAP = \(\angle\)OBP = 90°
[Since, radius at the point of contact is perpendicular to tangent]
OP = OP (Common)
OA = OB (Radii of circle)
So, \(\angle\)OAP == \(\angle\)OBP (By RHS congruency criterion)
(iii) (b): In quadrilateral OAPB, \(\angle\)AOB = 150°  [Given]
\(\angle\)OAP = \(\angle\)OBP = 90°
\(\therefore\) \(\angle\)APB = 360° - 90° - 90° - 150° = 30°
(iv) (d): We have, \(\angle\)APB = 40°

Now,PA =PB [Since, length of tangents drawn  from an external point are equal]
In \(\Delta\)PAB, \(\angle\)PAB = \(\angle\)PBA = 70° [Angles opposite to equal sides are equal]
Also, \(\angle\)PAB + \(\angle\)BAO = 90°
[Since, radius at the point of contact is perpendicular to tangent]
\(\Rightarrow\) \(\angle\)BAO = 90° - 70° = 20°
(v) (d): We have, \(\angle\)ABO = 45°

\(\because\) AO = OB (Radii of circle)
\(\therefore\) \(\angle\)BAO = \(\angle\)ABO = 45°  [Angles opposite to equal sides are equal]
Now, in \(\Delta\)OAB,
\(\angle\)AOB = 180° - 45° - 45° = 90°
Since, \(\angle\)APB = 360° - 90° - 90° - 90° = 90° i.e., AP\(\perp\) BP
So, OAPB is a square.

Here, AS = 5 cm, BT = 4 cm  [\(\therefore\)Radii of circles]
(i) (c): Since, radius at point of contact is perpendicular to tangent.
\(\therefore\) By Pythagoras theorem, we have
\(P A=\sqrt{P S^{2}+A S^{2}}=\sqrt{12^{2}+5^{2}}=\sqrt{169}=13 \mathrm{~cm}\)
(ii) (b): Again by Pythagoras theorem, we have
\(B Q=\sqrt{T Q^{2}+B T^{2}}=\sqrt{3^{2}+4^{2}}=\sqrt{25}=5 \mathrm{~cm}\)
(iii) (d): PK = PA + AK = 13 + 5 = 18 cm
(iv) (c): QY = BQ - BY = 5 - 4 = 1 cm
(v) (c): PS² = PA² - AS² = PA² - AK²
= (PA + AK)(PA - AK) = PK.PX [\(\because\) AK = AX]

Here in right angled triangle ABC, AB = 6 m and BC= 8 cm.
\(\therefore\) By Pythagoras theorem \(A C=\sqrt{(A B)^{2}+(B C)^{2}}\)
\(=\sqrt{(8)^{2}+(6)^{2}}=\sqrt{100}=10 \mathrm{~m}\)
Also, AP = x m.
(i) (c):AR=AP = xm      ..(1)
[Since, length of tangents drawn from an external point are equal]
(ii) (b): BQ = BR = AB - AR = (6 - x) m         (Using (1))
(iii) (d): CQ = CP = AC - AP = (10 - x) m
Also, CQ = BC - BQ = BC - BR = 8 - (6 - x) = 2 + x
(iv) (b): Since, CQ = 10 - x = 2 + x
\(\Rightarrow\) 8 = 2x \(\Rightarrow\) x = 4
\(\therefore\)  AR = AP = 4 m, BR = BQ = 2 m
and CP = CQ = 6 m
Also, OQ.\(\perp\)BQ and OR.l BR
\(\therefore\) BROQ is a square.
(v) (a): Radius of the circle, OR = BR = 2 cm

(i) (a)
(ii) (d)
(iii) (b)
(iv) (a)
(v) (c)