Maths M.C.Q

3. 7.5cm

Solution:

OA = OC

⇒ OA = OE + CE

⇒ OA = OE + 3

⇒ OE = OA - 3 ...(i)

$\text{AE}=\frac{1}{2}\text{AB}$ [Perpendicular drawn from the centre of a circle to the chord bisect the chord]

$=\frac{1}{2}(12)=6\text{cm}$

In right $\triangle\text{OEA},$

OA² = OE² + AE²

⇒ OA² = (OA - 3)² + AE² [From (i)]

⇒ OA² = OA² - 6OA + 9 + AE²

⇒ 6OA = 9 + 6²

⇒ 6OA = 9 + 36

$\Rightarrow\ \text{OA}=\frac{45}{6}=7.5\text{cm}$

So, the radius of the circle is 7.5cm.

2. $12\text{cm}$

Solution:

Let O be the centre of the circle with radius OA = 13cm.

AB is given to be 10cm.

Distance of a point to a line is always perpendicular to the line.

So, $\text{OL}\perp\text{AB}.$

We know that, the perpendicular drawn from the centre to the chord bisects the chord.

⇒ AL = LB = 5cm

In right $\triangle\text{OLA},$

OL² = AO² - AL² [By pythagoras theorem]

⇒ OL² = 13² - 5²

⇒ OL² = 169 - 25

⇒ OL² = 144

⇒ OL = 12cm

3. 60°

Solution:

Since BOC is a diameter, $\angle\text{BAC}=90^\circ.$

In $\triangle\text{BAC},$

$\angle\text{BAC}+\angle\text{ABC}+\angle\text{BCA}=180^\circ$ [Angle sum property]

$\Rightarrow\ 90^\circ+\angle\text{ABC}+30^\circ=180^\circ$

$\Rightarrow\ \angle\text{ABC}=60^\circ$

Since angles in the same segment of a circle are equal.

$\angle\text{CDA}=\angle\text{ABC}=60^\circ.$

4. 60°

Solution:

OA = OC [Radii of the same circle]

$\Rightarrow\ \angle\text{OBA}=\angle\text{OAB}=20^\circ$

In $\triangle\text{OAB,}$

$\angle\text{OBA}+\angle\text{OAB}+\angle\text{AOB}=180^\circ$ [Angle sum property]

$\Rightarrow\ 20^\circ+20^\circ+\angle\text{AOB}=180^\circ$

$\Rightarrow\ \angle\text{AOB}=140^\circ$

Now,

OB = OC [Radii of the same circle]

$\Rightarrow\ \angle\text{OBC}=\angle\text{OCB}=50^\circ$

In $\triangle\text{OCB},$

$\angle\text{OBC}+\angle\text{OCB}+\angle\text{COB}=180^\circ$ [Angle sum property]

$\Rightarrow\ 50^\circ+50^\circ+\angle\text{COB}=180^\circ$

$\Rightarrow\ \angle\text{COB}=80^\circ$

So,

$\angle\text{AOB}=\angle\text{AOC}+\angle\text{COB}$

$\Rightarrow\ \angle\text{AOC}=\angle\text{AOB}-\angle\text{COB}$

$\Rightarrow\ \angle\text{AOC}=140^\circ-80^\circ$

$\Rightarrow\ \angle\text{AOC}=60^\circ$

3. 60°

Solution:

We know that, the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

So,

$\angle\text{AOB}=2\angle\text{ACB}$

$=2(30^\circ)$

$=60^\circ$

1. 85°

Solution:

$\angle\text{BOA}+\angle\text{AOC}+\angle\text{BOC}=360^\circ$ [Angles around a point are 360°]

$\Rightarrow\ 100^\circ+90^\circ+\angle\text{BOC}=360^\circ$

$\Rightarrow\ \angle\text{BOC}=170^\circ$

Now,

$\angle\text{BAC}=\frac{1}{2}(\angle\text{BOC})=\frac{1}{2}(170^\circ)=85^\circ$

1. 8cm

Solution:

Construction: Join OC.

We know that, the perpendicular drawn from the centre to the chord bisects the chord.

So, $\text{CL}=\frac{1}{2}\text{CD}=\frac{1}{2}(30)=15\text{cm}$

AB is the diameter.

So, $\text{AO}=\frac{1}{2}\text{AB}=\frac{1}{2}(34)=17\text{cm}.$

In $\triangle\text{OLC},$

OL² = OC² - CL²

⇒ OL² = 17² - 15²

⇒ OL² = 289 - 225

⇒ OL² = 64

⇒ OL = 8cm

2. 75°

Solution:

OB = BC [Given]

$\Rightarrow\ \angle\text{OBC}=\angle\text{BCO}=25^\circ$ [Angles opposite equal sides are equal]

Now,

$\angle\text{OBC}=\angle\text{BOC}+\angle\text{BCO}=25^\circ+25^\circ=50^\circ$

OA = OB [Radii of the same circle]

$\Rightarrow\ \angle\text{OAB}=\angle\text{OBA}=50^\circ$

In $\triangle\text{AOC},$

$\angle\text{AOD}=\angle\text{OAC}+\angle\text{ACO}$

$=\angle\text{OAB}+\angle\text{BCO}$

$=50^\circ+25^\circ$

$=75^\circ$

3. 30cm

Solution:

Let O be the centre of the circle with radius OA = 17cm.

Since $\text{OC}\perp\text{AB}.$

In right $\triangle\text{OCA},$

OA²= OC² + AC² [By pythagoras theorem]

AC² = OA² - OC²

⇒ AC² = 17² - 8²

⇒ AC² = 289 - 64

⇒ AC² = 225

⇒ AC = 15cm

We know that, the perpendicular drawn from the centre to the chord bisects the chord.

⇒ AB = 2AC = 2(15) = 30cm.

1. 30°

Solution:

Since AB || CD, $\angle\text{BAD}=\angle\text{CDA}=30^\circ$ [Alternate angles]

Since AOB is a diameter, $\angle\text{ADB}=90^\circ$

$\angle\text{CDB}=\angle\text{CDA}+\angle\text{ADB}=30^\circ$

$\Rightarrow\ \angle\text{CDB}=30^\circ+90^\circ$

$\Rightarrow\ \angle\text{CDB}=120^\circ$

We know that the opposite angles of a quadrilateral are supplementary.

$\angle\text{CAB}+\angle\text{CDB}=180^\circ$

$\Rightarrow\ \angle\text{CAD}+\angle\text{DAB}+\angle\text{CDB}=180^\circ$

$\Rightarrow\ \angle\text{CAD}+30^\circ+120^\circ=180^\circ$

$\Rightarrow\ \angle\text{CAD}=30^\circ$

2. 45°

Solution:

Since BOC is a diameter of a circle, $\angle\text{BAC}$ is 90°.

Given that AB = AC.

$\Rightarrow\ \angle\text{ABC}=\angle\text{ACB}$

In $\triangle\text{BAC},$

$\angle\text{ABC}+\angle\text{ACB}+\angle\text{BAC}=180^\circ$ [Angle sum property]

$\Rightarrow\ \angle\text{ABC}+\angle\text{ABC}+90^\circ=180^\circ$

$\Rightarrow\ 2\angle\text{ABC}=90^\circ$

$\Rightarrow\ \angle\text{ABC}=45^\circ$

1. 10cm

Solution:

Construction: Join OD.

OB = OD

⇒ OB = OE + EB

⇒ OB = OE + 4

⇒ OE = OB - 4 ...(i)

In right $\triangle\text{OED},$

OD² = OE² + DE²

⇒ OD² = (OB - 4)² + DE² [From (i)]

⇒ OD² = OB² - 80A + 16 + 8²

⇒ 8OD = 16 + 64

⇒ 8OD = 80

$\Rightarrow\ \text{OD}=\frac{80}{8}=10\text{cm}$

So, the radius of the circle is 10cm.

3. 60°

Solution:

$\angle\text{AOB}=2\angle\text{ACB}$

$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}\angle\text{AOB}$

$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}(90^\circ)$

$\Rightarrow\ \angle\text{ACB}=45^\circ$

$\angle\text{COA}=2\angle\text{CBA}=2(30^\circ)=60^\circ$

Since AOD is a straight line,

$\therefore\ \angle\text{COD}+\angle\text{AOC}=180^\circ$

$\therefore\ \angle\text{COD}+60^\circ=180^\circ$

$\therefore\ \angle\text{COD}=120^\circ$

$\Rightarrow\ \angle\text{CAO}=\frac{1}{2}\angle\text{COD}=\frac{1}{2}\times120^\circ=60^\circ$

3. 120°

Solution:

Since $\triangle\text{BDC}$ is an equilateral traingle, $\angle\text{BAC}=60^\circ.$

Since ABCD is a cyclic equilateral,

$\angle\text{BAC}+\angle\text{BDC}=180^\circ$

$\Rightarrow\ 60^\circ+\angle\text{BDC}=180^\circ$

$\Rightarrow\ \angle\text{BDC}=120^\circ$

2. 30°

Solution:

We know that the opposite angles of a quadrilateral are supplementary.

$\angle\text{ADC}+\angle\text{ABC}=180^\circ$

$\Rightarrow\ 120^\circ+\angle\text{ABC}=180^\circ$

$\Rightarrow\ \angle\text{ABC}=60^\circ$

 Since BOC is a diameter $\angle\text{ACB}=90^\circ.$

In $\triangle\text{CAB},$

$\angle\text{ABC}+\angle\text{BAC}+\angle\text{ACB}=180^\circ$ [Angle sum property]

$\Rightarrow\ 60^\circ+\angle\text{BAC}+90^\circ=180^\circ$

$\Rightarrow\ \angle\text{BAC}=30^\circ$

2. 50°

Solution:

In $\triangle\text{OAB},$

$\text{OA}=\text{OB}$ [Radii of the same circle]

$\Rightarrow\ \angle\text{OAB}=\angle\text{OBA}$ [Angle opposite equal sides are equal]

$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$ [Angle sum property]

$\Rightarrow\ 40^\circ+40^\circ+\angle\text{AOB}=180^\circ$

$\Rightarrow\ \angle\text{AOB}=100^\circ$

We know that, the angle subtended by an arc of a circle at the centre is double the angle subtended by it at any point on the remaining part of the circle.

So, $\angle\text{ACB}=\frac{1}{2}\angle\text{AOB}$

$=\frac{1}{2}(100^\circ)$

$=50^\circ$

2. 60°

Solution:

Since angles in the same segment are equal,

$\angle\text{BDC}=\angle\text{BAC}=40^\circ$

In $\triangle\text{BDC},$

$\angle\text{BDC}+\angle\text{BCD}+\angle\text{CBD}=180^\circ$ [Angle sum property]

$\Rightarrow\ 40^\circ+80^\circ+\angle\text{CBD}=180^\circ$

$\Rightarrow\ \angle\text{CBD}=60^\circ$

3. 105°

Solution:

Since CF || AB, $\angle\text{ABC}=\angle\text{BCF}=85^\circ$

$\angle\text{BAD}=\angle\text{BCE}$

$\Rightarrow\ \angle\text{BAD}=\angle\text{BCF}+\angle\text{ECF}$

$\Rightarrow\ \angle\text{BAD}=105^\circ$

2. 80°

Solution:

Since ABCD is a cyclic equilateral,

$\angle\text{CBE}=\angle\text{ADC}=100^\circ$

Since ADF is a straight line,

$\angle\text{CDF}+\angle\text{ADC}=180^\circ$

$\Rightarrow\ \angle\text{CDF}+100^\circ=180^\circ$

$\Rightarrow\ \angle\text{CDF}=80^\circ$

3. 100°

Solution:

OA = OB [Radii of the same circle]

$\Rightarrow\ \angle\text{OAB}=\angle\text{OBA}=50^\circ$ [Angle opposite equal sides are equal]

$\angle\text{BOD}=\angle\text{OAB}+\angle\text{OBA}$

$\Rightarrow\ \angle\text{BOD}=50^\circ+50^\circ$

$\Rightarrow\ \angle\text{BOD}=100^\circ$

4. 110°

Solution:

Since ABCD is a cyclic qyadrilateral, we have:

$\angle\text{BAD}+\angle\text{BCD}=180^\circ$

$\Rightarrow\ \angle\text{BAD}+110^\circ=180^\circ$

$\Rightarrow\ \angle\text{BAD}=70^\circ$

Since ABEF is a cyclic qyadrilateral, we have:

$\angle\text{BAD}+\angle\text{BEF}=180^\circ$

$\Rightarrow\ 70^\circ+\angle\text{BEF}=180^\circ$

$\Rightarrow\ \angle\text{BEF}=110^\circ$

3. 115°

Solution:

Minor $\angle\text{AOC}=130^\circ$

Major $\angle\text{AOC}=360^\circ-130^\circ$

⇒ Major $\angle\text{AOC}=230^\circ$

Since $\angle\text{ABC}=\frac{1}{2}\text{major}\angle\text{AOC}$

$\Rightarrow\ \angle\text{ABC}=\frac{1}{2}(230^\circ)$

$\Rightarrow\ \angle\text{ABC}=115^\circ$

3. 100°

Solution:

In $\triangle\text{OAB},$

OA = OB [Radii of the same circle]

$\Rightarrow\ \angle\text{OBA}=\angle\text{OAB}=20^\circ$ [Angle opposite equal sides are equal]

In $\triangle\text{OAC},$

OA = OC [Radii of the same circle]

$\Rightarrow\ \angle\text{OCA}=\angle\text{OAC}=30^\circ$ [Angle opposite equal sides are equal]

Now, $\angle\text{BAC}=\angle\text{BAO}+\angle\text{CAO}$

$=20^\circ+30^\circ$

$=50^\circ$

$\angle\text{BOC}=2\angle\text{BAC}=2(50^\circ)=100^\circ.$

2. 70°

Solution:

BC = BD [Given]

$\angle\text{BDC}=\angle\text{CBD}=35^\circ$ [Angle opposite equal sides are equal]

In $\triangle\text{BCD},$

$\angle\text{BCD}+\angle\text{BDC}+\angle\text{CBD}=180^\circ$ [Angle sum property]

$\Rightarrow\ \angle\text{BCD}+35^\circ+35^\circ=180^\circ$

$\Rightarrow\ \angle\text{BCD}=110^\circ$

Since ABCD is a cyclic quadrilateral,

$\angle\text{BAD}+\angle\text{BCD}=180^\circ$

$\Rightarrow\ \angle\text{BAD}+110^\circ=180^\circ$

$\Rightarrow\ \angle\text{BAD}=70^\circ$

2. 80°

Solution:

Given that AB = CD.

Since equal chord, subtend equal angles at the centre,

$\angle\text{COD}=\angle\text{AOB}=80^\circ.$

2. 50°

Solution:

OA = OB [Radii of the same circle]

$\Rightarrow\ \angle\text{OBA}=\angle\text{OAB}=50^\circ$

Since angles in the same segment are equal, $\angle\text{ABC}=\angle\text{CDA}.$

That is, $\angle\text{ABO}=\angle\text{CDA}=50^\circ$

4. 10cm

Solution:

In $\triangle\text{BEO}$ and $\triangle\text{CFO},$

OB = OC [Radii of the same circle]

$\angle\text{OBE}=\angle\text{OCF}$ [Alternate angles since AB || CD]

$\angle\text{BOE}=\angle\text{COF}$ [Vertically angles]

$\Rightarrow\ \triangle\text{BEO}\cong\triangle\text{CFO}$ [ASA congruence criterion]

⇒ OE = OF [C.P.C.T.]

Since chord are equidistant from the centre are equal, AB = CD = 10 cm.

2. 12cm

Solution:

In $\triangle\text{BOD}$ and $\triangle\text{CAB},$

Since BOC is the diameter, $\angle\text{CAB}=90^\circ.$

Also, $\angle\text{ODB}=90^\circ.$

So, $\angle\text{DBO}=\angle\text{ABC}$ [Common angles]

$\Rightarrow\ \triangle\text{BOD}\sim\triangle\text{BCA}$ [AA congruence criterion]

$\Rightarrow\ \frac{\text{OD}}{\text{CA}}=\frac{\text{BO}}{\text{BA}}$

$\Rightarrow\ \frac{\text{OD}}{\text{CA}}=\frac{1}{2}$ [Since radius = 2 diameter]

$\Rightarrow\ \frac{6}{\text{CA}}=\frac{1}{2}$

⇒ CA = 12cm that is, AC = 12cm.

3. 30°

Solution:

Since BOA is a diameter.

$\angle\text{AOC}+\angle\text{BOC}=180^\circ$

$\Rightarrow\ 120^\circ+\angle\text{BOC}=180^\circ$

$\Rightarrow\ \angle\text{BOC}=60^\circ$

So, $\angle\text{BDC}=\frac{1}{2}\angle\text{BOC}=\frac{1}{2}(60^\circ)=30^\circ$

2. 6cm

Solution:

We know that, the line joining centres is the perpendicular bisector of the common chord.

Then,

AP = 5cm, BP = 3cm and AB = 4cm

AP² = 5² = 25

BP² + AB² = 3² + 4² = 25

In $\triangle\text{ABP},$

Since AP² = BP² + AB²

$\triangle\text{ABP},$ is a right-angled triangle and PQ = 2BP = 2(3) = 6cm.

4. 65°

Solution:

OA = OB [Radii of the same circle]

$\Rightarrow\ \angle\text{OAB}=\angle\text{OBA}$

In $\triangle\text{OAB},$

$\angle\text{BOA}+\angle\text{OAB}+\angle\text{OBA}=180^\circ$ [Angle sum property]

$\Rightarrow\ 50^\circ+\angle\text{OAB}+\angle\text{OAB}=180^\circ$

$\Rightarrow\ 2\angle\text{OAB}=130^\circ$

$\Rightarrow\ \angle\text{OAB}=65^\circ$

3. $3\sqrt{3}\text{cm}$

Solution:

Let $\triangle\text{ABC}$ be an equilateral triangle.

Let AD be one of its medians.

Then, $\text{AD}\perp\text{BC}$ and BD = 4.5cm

In right $\triangle\text{ADB},$

$\therefore\ \text{AD}=\sqrt{\text{AB}^2-\text{BD}^2}$ [By pythagoras theorem]

$=\sqrt{9^2-\frac{9^2}{2}}$

$=\sqrt{81-\frac{81}{2}}$

$=\sqrt{\frac{243}{2}}$

$=\frac{9\sqrt{3}}{2}\text{cm}$

Let G be the centroid of $\triangle\text{ABC}.$

Then, AG : GD = 2 : 1

$\therefore$ radius $=\text{AG}=\frac{2}{3}\text{AD}=\Big(\frac{2}{3}\times\frac{9\sqrt{3}}{2}\Big)=3\sqrt{3}\text{cm}$

3. 110°

Solution:

Minor $\angle\text{AOB}=140^\circ$

Major $\angle\text{AOB}=360^\circ-140^\circ$

⇒ Major $\angle\text{AOB}=220^\circ$

Since $\angle\text{ACB}=\frac{1}{2}\text{major}\angle\text{AOB}$

$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}(220^\circ)$

$\Rightarrow\ \angle\text{ACB}=110^\circ$

2. 100°

Solution:

Since AB || DC,

$\angle\text{ADC}+\angle\text{BAD}=180^\circ$

$\Rightarrow\ 100^\circ+\angle\text{BAD}=180^\circ$

$\Rightarrow\ \angle\text{BAD}=80^\circ$

We know that the opposite angles of a quadrilateral are supplementary.

$\angle\text{BAD}+\angle\text{ABC}=180^\circ$

$\Rightarrow\ 80^\circ+\angle\text{ABC}=180^\circ$

$\Rightarrow\ \angle\text{ABC}=100^\circ$

3. 80°

Solution:

$\angle\text{AED}=\angle\text{ECB}+\angle\text{EBC}$

$\Rightarrow\ 110^\circ=\angle\text{ECB}+30^\circ$

$\Rightarrow\ \angle\text{ECB}=80^\circ$

Since angles in the same segment are equal,

$\angle\text{ADB}=\angle\text{ECB}=80^\circ$

3. 8.5cm

Solution:

Construction: Join AC.

$\frac{\text{AE}}{\text{CE}}=\frac{\text{DE}}{\text{BE}}$

⇒ AE × BE = DE × CE ...(i)

Then,

AE = AB + BE = 11 + 3 = 14cm, BE = 3cm, CE = (x + 3.5)cm and DE = 3.5cm

So, from (i), we get

14 × 3 = 3.5 × (CD + 3.5)

$\Rightarrow\ \frac{14\times3}{3.5}=\text{CD}+3.5$

⇒ 12 = CD + 3.5

⇒ CD = 8.5cm

3. 115°

Solution:

Minor $\angle\text{AOB}=130^\circ$

Major $\angle\text{AOB}=360^\circ-130^\circ$

⇒ Major $\angle\text{AOB}=230^\circ$

Since $\angle\text{ACB}=\frac{1}{2}\text{major}\angle\text{AOB}$

$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}(230^\circ)$

$\Rightarrow\ \angle\text{ACB}=115^\circ$

1. 50°

Solution:

Construction: Let E be a point on the reamaining part of the circumference of the circle.

Join AE and CE.

$\angle\text{AEC}=\frac{1}{2}\angle\text{AOC}$

$\Rightarrow\ \angle\text{AEC}=\frac{1}{2}(100^\circ)=50^\circ$

Since AECB forms a cyclic quadrilateral.

$\Rightarrow\ \angle\text{CBD}=\angle\text{AEC}=50^\circ$

2. 50°

Solution:

Since angles in the same segment of a circle are equal.

$\angle\text{CDB}=\angle\text{BAC}$

That is , $\angle\text{ODB}=\angle\text{OAC}=50^\circ.$

3. 70°

Solution:

Since angles in the same segment of a circle are equal.

$\angle\text{BAC}=\angle\text{BDC}=60^\circ.$

In $\triangle\text{BDC},$

$\angle\text{BDC}+\angle\text{DBC}+\angle\text{BCD}=180^\circ.$

$\Rightarrow\ 60^\circ+50^\circ+\angle\text{BCD}=180^\circ$

$\Rightarrow\ \angle\text{BCD}=70^\circ$