(4)

Draw CM perpendicular to AB.

Now, AM = MC and MB = MC (tangents drawn from external point are equal).

⇒ AM = MC

⇒ $\angle MAC$ = $\angle MCA$ = 45°

(Since Δ AMC is right triangle)

∴  Also, MB = MC ⇒ $\angle MBC$ = $\angle MCB$ = 45° (Since Δ MBC is right angle triangle)

∴ $\angle ACB$ = $\angle MCA$ + $\angle MCB$ = 45° + 45° = 90° ⇒ $\angle ACB$ = 90°

(2)

Area of the circle > Area of the square

(4)

ABCD is a square of side 6 cm. PQ is a diameter of the given circle such that $PQ=AB=6cm$

$\therefore$ $\text{Radius }\left(r\right)=\frac{6}{2}=3\text{ cm}$

$\text{Area of the circle}=\pi r^2=\pi {\left(3\right)}^2=9\pi {\text{ cm}}^2.$

(4)

Angle between two tangents = $60°$     (given)

$\therefore$  Tangents are equally inclined to each other

$\therefore$  $\angle OPA=\angle OPB=30°$

and $\angle OAP=90°$

(Tangent is perpendicular to the radius)

In $∆PAO,$ $\tan 30°=\frac{OA}{AP}\Rightarrow \frac{1}{\sqrt{3}}=\frac{3}{AP}\Rightarrow AP=3\sqrt{3}$

Hence, the length of each tangent is $3\sqrt{3}$ cm

(2)

PA = PB (Tangents from an external point P)

$∆\mathrm{APB}$ is a Right angle Triangle

$A{B}^2=A{P}^2+P{B}^2$$\text{[AP = PB]}$

$A{B}^2=2A{P}^2$

$A{B}^2=2\times 5^2$

$\Rightarrow AB=5\sqrt{2}\text{ cm}$

(2)     $5\sqrt{3}cm$

(4)

$\angle APD=\angle CPB$ [vertically opposite angle]

$\angle ADP=\angle CBP$ [Angle subtends on the same segment]

$∆ADP~∆CBP$ (by AA similarity)

(4)

Given, $\angle CAT=40°$

$\angle BAT=90°$

$\Rightarrow \angle BAC+\angle CAT=90°$

$\Rightarrow \angle BAC=50°$

$\Rightarrow \angle ACB=90° \text{[Angle in semi-circle]}$

$\text{In }∆ABC, \angle A+\angle B+\angle C=180°$

$\Rightarrow 50°+\angle B+90°=180°$

$\Rightarrow \angle B=180°-140°=40°$

(3)

The maximum number of common tangents that can be drawn to two circles intersecting at two distinct points is 2, as shown in

(2)

As ML is tangent at M and OM is radius of the circle, therefore ∠OML = 90°

⇒ ∠OMN + ∠NML = 90°
⇒ ∠OMN + 70° = 90° ⇒ ∠OMN = 20°

In ΔOMN, we have OM = ON (? radii of circle)
⇒ ∠OMN = ∠ONM = 20° (Angles opposite to equal sides are equal)

Also, ∠OMN + ∠MON + ∠ONM = 180°
⇒ 20° + ∠MON + 20° = 180°
⇒ ∠MON = 180° – 40° = 140°

(2)

Since PQ is tangent at Q to the circle and OQ is radius,
∴ OQ ? PQ ⇒ ∠OQP = 90°

In ΔPOQ, we have

∠OQP + ∠OPQ + ∠POQ = 180°  [Angle Sum Property of triangle]

⇒ 90° + x + y = 180°
⇒ x + y = 180° – 90° = 90°

(1)

Since TA is tangent at A and OA is radius of the circle,
∴ $OA ⟂ TA \Rightarrow \angle OAT = 90°$

$$\begin{gathered} \therefore In right angled \Delta OAT, we have \\ cos30°=\frac{TA}{OT}=\frac{\sqrt{3}}{2} \\ \frac{TA}{4}=\frac{\sqrt{3}}{2} \\ \Rightarrow TA=2\sqrt{3} \mathrm{cm} \end{gathered}$$

(1)

Let O be the centre of the circle and P be the external point and PA is tangent at A as shown in Fig.

$$\begin{gathered} \therefore OA ⟂ PA \\ Given, OA = 9 cm and OP = 41 cm \\ In right \Delta OAP, we have \\ O{P}^2=O{A}^2+P{A}^2(\mathrm{Using} \mathrm{Pythagoras} \mathrm{theorem}) \\ (41{)}^2=(9{)}^2+(PA{)}^2 \\ 1681-81=P{A}^2 \\ 1600=P{A}^2\Rightarrow PA=40 \mathrm{cm} \end{gathered}$$

(1)

Given, OQ = 5 cm

Since PQ is tangent to a circle with centre O, at point Q

$$\begin{gathered} \therefore OQ ⟂ PQ \\ \Rightarrow \angle OQP = 90° \\ In right \Delta OPQ, we have \\ tan30°=\frac{OQ}{PQ} \\ \frac{1}{\sqrt{3}}=\frac{5}{PQ} \\ \Rightarrow PQ=5\sqrt{3} \mathrm{cm} \end{gathered}$$

(2)

Construction: Join OR as shown in

$$\begin{gathered} We have, \Delta OQP \cong \Delta ORP (by SSS congruency rule) \\ \therefore \angle OPQ = \angle OPR = 45° \\ (by CPCT) (\because \angle QPR = 90°) \\ In \Delta POQ, \angle OQP = 90°, \angle OPQ = 45° \\ \angle POQ=180°-(90°+45°)=45° \\ \therefore \angle OPQ = \angle POQ = 45° \\ \therefore \angle OPQ = \angle POQ = 45° \Rightarrow OQ = PQ (Opposite sides of equal angles are equal) \\ \Rightarrow PQ=4 \mathrm{cm} \end{gathered}$$

(4)

Let P be an external point and O be the centre of the circle, such that OP = 2r units where r is radius of the circle as shown in Fig
Let PA be the tangent at point A on circle.

$$\begin{gathered} \therefore OA ⟂ PA and OA = r units \\ Now, in right \Delta OAP, we have \\ A{P}^2+O{A}^2=O{P}^2 \\ A{P}^2+r^2=(2r{)}^2 \\ A{P}^2=4r^2-r^2=3r^2 \\ AP=\sqrt{3} r \mathrm{units} \end{gathered}$$

(1)

In right ΔABC, we have

$$\begin{gathered} A{C}^2=A{B}^2+B{C}^2 \\ A{C}^2=(4{)}^2+(3{)}^2=16+9 \\ A{C}^2=25 \\ AC=5 \mathrm{cm} \\ \mathrm{Now}, \mathrm{AD} = \mathrm{AB} – \mathrm{BD} = (4 – \mathrm{x}) \mathrm{cm} \\ \mathrm{AF}=\mathrm{AD}=\left(4-\mathrm{x}\right) \mathrm{cm}(\mathrm{Tangents} \mathrm{drawn} \mathrm{from} \mathrm{an} \mathrm{external} \mathrm{point} \mathrm{A}) \\ \mathrm{Also}, \\ \mathrm{EC}=\mathrm{BC}\mathrm{\{}–\mathrm{BE}=3-\mathrm{xcm} \\ \mathrm{CF}=\mathrm{EC}=\left(3-\mathrm{x}\right) \mathrm{cm}(\mathrm{Tangents} \mathrm{drawn} \mathrm{from} \mathrm{an} \mathrm{external} \mathrm{point} \mathrm{C}) \\ \mathrm{AC}=\mathrm{AF}+\mathrm{CF}=\left(4-\mathrm{x}\right)+\left(3-\mathrm{x}\right) \\ 5=7-2\mathrm{x} \\ 2\mathrm{x}=7-5=2 \\ \mathrm{x}=1 \mathrm{cm} \end{gathered}$$

(3)

We know that the angle subtended by a chord at the centre is twice the angle subtended by it on the circumference.

$$\begin{gathered} \angle AOC=2\angle ADC \\ 110°=2\angle ADC\Rightarrow \angle ADC=55°\Rightarrow \angle D=55° \end{gathered}$$

$$\begin{gathered} \therefore \angle AOC = 110° \\ \Rightarrow Reflex \angle AOC = 360° – 110° = 250° (As shown in Fig.) \end{gathered}$$

$$\begin{gathered} Also, reflex \angle AOC = 2\angle ABC \\ 250°=2\angle ABC\Rightarrow \angle ABC=125° \end{gathered}$$

$\Rightarrow \angle B = 125° \Rightarrow Statements \left(i\right) and \left(iv\right) are correct.$

(4)

In a cyclic quadrilateral, the sum of opposite angles is 180°. Only rectangle satisfies the condition of cyclic quadrilateral.