Vardaan Learning Institute

In $\triangle OQT$, $\angle T = 90^\circ$.
$OT^2 = OQ^2 - QT^2 = 25^2 - 24^2$
$= 625 - 576 = 49$.
$OT = 7$ cm.

PTQO is a cyclic quad. $\angle P + \angle Q = 90+90=180$.
$\angle PTQ + \angle POQ = 180^\circ \Rightarrow \angle PTQ = 180 - 110 = 70^\circ$.

$r^2 = 5^2 - 4^2 = 25 - 16 = 9 \Rightarrow r = 3$ cm.

Parallel tangents are at the ends of a diameter. Distance = Diameter = $2r = 2 \times 4 = 8$ cm.

$r = \sqrt{26^2 - 24^2} = \sqrt{676 - 576} = \sqrt{100} = 10$ cm.

Chord is bisected by perpendicular from centre.
Half length $L = \sqrt{5^2 - 3^2} = \sqrt{16} = 4$.
Total length $= 2 \times 4 = 8$ cm.

$\angle AOB = 180 - 50 = 130^\circ$.
In $\triangle OAB$, $OA=OB$ (radii) so $\angle OAB = \angle OBA$.
$\angle OAB = (180 - 130)/2 = 25^\circ$.

Wait, distance 8cm from A? A is on the circle. Diameter AB is 10cm. Max distance in circle is 10cm.
If CD is parallel to tangent at A, then CD is perpendicular to diameter AB.
A is one end, B is other. CD is at distance 8cm from A.
Distance from center O? O is 5cm from A. So CD is $8-5=3$ cm from O.
Half chord = $\sqrt{r^2 - 3^2} = \sqrt{25-9} = 4$.
Chord length = $2 \times 4 = 8$ cm.

Sum of opposite sides must be equal.
$AB + CD = BC + DA$.

Angle between tangents is supplementary to angle at center.
$180 - 60 = 120 \ne 60$.
So A is false. R is true.

Let chord be AB touching small circle at P. $OP \perp AB$.
In $\triangle OPA$, $AP^2 = OA^2 - OP^2 = 5^2 - 3^2 = 16$.
$AP = 4$ cm. Total length AB = $2 \times 4 = 8$ cm.

Tangents from ext point are equal.
$AP=AS, BP=BQ, CR=CQ, DR=DS$.
$AB+CD = (AP+BP) + (CR+DR) = (AS+BQ) + (CQ+DS)$
$= (AS+DS) + (BQ+CQ) = AD + BC$.

Let ABCD be ||gm. $AB=CD$ and $AD=BC$.
From previous result, $AB+CD = AD+BC$.
$2AB = 2AD \Rightarrow AB = AD$.
Adjacent sides equal $\Rightarrow$ Rhombus.

$\angle BQR = 70^\circ \Rightarrow \angle BAQ = 70^\circ$ (Alternate Segment Theorem).
Since $AB \parallel PR$, $\angle ABQ = \angle BQR = 70^\circ$ (Alternate interior angles).
In $\triangle ABQ$, $\angle AQB = 180 - (70+70) = 40^\circ$.

Given: Circle(O), Point P outside. Tangents PQ, PR.
Join OQ, OR, OP. $\angle OQP = \angle ORP = 90^\circ$ (Radius $\perp$ Tangent).
In $\triangle OQP$ and $\triangle ORP$:
$OQ=OR$ (Radii), $OP=OP$ (Common).
By RHS, $\triangle OQP \cong \triangle ORP$.
So $PQ = PR$ (CPCT).

Join OC. $\triangle OPA \cong \triangle OCA \Rightarrow \angle POA = \angle COA$.
$\triangle OQB \cong \triangle OCB \Rightarrow \angle QOB = \angle COB$.
POQ is diameter (straight line). $2\angle COA + 2\angle COB = 180^\circ$.
$2(\angle COA + \angle COB) = 180^\circ \Rightarrow \angle AOB = 90^\circ$.

Let angles at center subtended by parts of sides divided by contact points be $a,a, b,b, c,c, d,d$.
Total angle = $2(a+b+c+d) = 360 \Rightarrow a+b+c+d = 180$.
Angle by AB is $a+b$, by CD is $c+d$. Sum $= 180$.
Angle by BC is $b+c$, by DA is $d+a$. Sum $= 180$.

Let $\angle PTQ = \theta$. TP=TQ (tangents). $\triangle TPQ$ is isosceles.
$\angle TPQ = \angle TQP = (180-\theta)/2 = 90 - \theta/2$.
Since Radius $\perp$ Tangent, $\angle OPT = 90$.
$\angle OPQ = \angle OPT - \angle TPQ = 90 - (90 - \theta/2) = \theta/2$.
So $\theta = 2\angle OPQ \Rightarrow \angle PTQ = 2\angle OPQ$.

$AQ = AR$ (tangents from A).
$BP = BQ$ (tangents from B), $CP = CR$ (tangents from C).
Perimeter $= AB + BC + AC = AB + (BP+PC) + AC$
$= (AB+BQ) + (AC+CR) = AQ + AR = 2AQ$.
$AQ = \frac{1}{2} \text{Perimeter}$.

Let OT intersect PQ at R. $PR=RQ=4cm, OR=3cm$.
Let TP = y, TR = x.
In $\triangle TRP$, $y^2 = x^2 + 16$.
In $\triangle TOP$, $TP^2 + OP^2 = OT^2 \Rightarrow y^2 + 25 = (x+3)^2$.
$x^2+16+25 = x^2+6x+9 \Rightarrow 41 = 6x+9 \Rightarrow 6x=32 \Rightarrow x=16/3$.
$y^2 = (16/3)^2 + 16 = 256/9 + 144/9 = 400/9$.
$y = 20/3$ cm.

Let $AF=AE=x$. Side lengths: $c=x+8, b=x+6, a=14$.
$s = 14+x$.
Area via Heron's: $A = \sqrt{(14+x) \cdot x \cdot 8 \cdot 6} = \sqrt{48x(14+x)}$.
Area via triangles: $A = rs = 4(14+x)$.
$16(14+x)^2 = 48x(14+x) \Rightarrow 14+x = 3x \Rightarrow 2x = 14 \Rightarrow x=7$.
$AB = 7+8=15$ cm, $AC = 7+6=13$ cm.

Part 1: Standard theorem proof. Take point Q on tangent. $OQ > radius \Rightarrow OQ > OP$. OP is shortest distance, so $\perp$.
Part 2: Congruency of $\triangle OPA$ and $\triangle OPB$ (RHS).
$\Rightarrow \angle AOP = \angle BOP$.

(i) In $\triangle OPT$, $\angle T = 90^\circ$. $OP=120, OT=50$.
$PT = \sqrt{120^2 - 50^2} = \sqrt{14400 - 2500}$
$= \sqrt{11900} = 10 \sqrt{119} \approx 109.1$ m.
(ii) Tangents from external point are equal. $PS = PT = 10\sqrt{119}$ m.
(iii) Tangent is perpendicular to radius at point of contact.