CRE 3 - Circles | Geometry, Mensuration & Trigonometry

Answer: Option C

Explanation :

Area of the circle, A = πr² = 36π

⇒ r² = 36

∴ Radius, r = 5 cm

∴ Diameter, d = 2 × 5 = 12 cm 

Hence, option (c).

Answer: Option B

Explanation :

Ratio of area and circumference of a circle = πr² : 2πr = r : 2

∴ 7/4 = r/2

∴ r = 7/2 cm

∴ Circumference of the circle = 2πr = 2 × 22/7 × 7/2 = 22 cm.

Hence, option (b).

Answer: Option A

Explanation :

A quarter circle can be considered as a sector with θ = 90° as shown below.

Hence, the area of the quarter circle has to be equal to the area of the sector and the perimeter has to be the same as the perimeter of the sector. 

Area of sector O-ACB = 1/4 × area of circle = 1/4 × πr² = 1/4 × 22/7 × 14 × 14 = 154 cm²

Perimeter of sector O-ACB = length(arc ACB) + length(OA) + length(OB) 

= $\left[\frac{1}{4}\times 2\mathrm{\pi r}\right]$ + r + r = $\left[\frac{1}{4}\times 2\times \frac{22}{7}\times 14\right]$ + 14 + 14 = 50 cm.

Hence, option (a).

Answer: Option B

Explanation :

A perpendicular drawn from the center of the circle to the chord bisects the chord. 

In the above figure, OD bisects chord AB  i.e. AD = DB

In ∆ODB, ∠ODB = 90°

Using Pythagoras theorem, we get,

OB² = DB² + OD² 

∴ 10² = DB² + 6²

∴ DB² = 100 – 36 = 64

∴ DB = 8 cm

Length of chord AB = 2 × DB = 2 × 8 = 16 cm

Hence, option (b).

Answer: Option C

Explanation :

The angle subtended by a chord on a point on the circumference (which lies in the major segment) of the circle measures half the angle subtended by the same chord at the center.

In the above figure, chord AC subtends an angle ∠ADB at point D lying on the major arc and an angle ∠AOB at the center of the circle.  

∴ m ∠AOB = 2 × m ∠ADB

m ∠AOB = 100°

∴ m ∠ADB = 50°

∴ The angle subtended by the chord AB in the major segment = 50°

Hence, option (c).

Answer: Option C

Explanation :

If two chords intersect internally as shown in the figure, 

DX × CX = AX × BX;

∴ AX = (6 × 8)/4 = 12 cm

AB = AX + BX = 12 + 4 = 16 cm

AB is the diameter of the circle. 

∴ Radius AO = 16/2 = 8 cm

Hence, option (c).

Answer: Option C

Explanation :

Circumference of a wheel = (22/7) × 1.4 = 4.4 m

∴ Total number of rotations = (1100/4.4) = 250 rotations

Hence, option (c).

Answer: Option D

Explanation :

Perimeter of the equilateral triangle = P

∴ Side of the equilateral triangle = P/3

∴ The area of the equilateral triangle = $\frac{\sqrt{3}}{4}\times {\left(\frac{P}{3}\right)}^2$

But the area of the equilateral triangle is equal to the area of the circle.

Area of the circle = πr² = $\frac{\sqrt{3}}{4}\times {\left(\frac{P}{3}\right)}^2$

∴ Radius of the circle, r = $\sqrt{\left(\frac{\sqrt{3}}{\pi }\right)}\times \frac{P}{6}$

∴ The circumference of the circle = 2πr = $\frac{\sqrt{\left(\pi \sqrt{3}\right)}}{3}P$

Hence, option (d).

Answer: Option A

Explanation :

The three semicircles in the figure can be named S₁, S₂ and S₃ as shown above. 

Since the two semicircles have equal radii, their area also has to be equal to each other. 

Area(S₂) = Area(S₃)

Hence, we can redraw the figure like this.

Area of the shaded portion is same as the area of S₁.

Radius of S₁ = 4.

∴ Area of shaded portion = ½ × π × 4 × 4 = 8π

Hence, option (a).

Answer: Option C

Explanation :

Join OR as shown.

m ∠ORP = m ∠OPR = a  …(OP = OR; ∆ORP is isosceles)

Similarly, m ∠ORQ = m ∠OQR = b  …(OQ = OR, ∆ORQ is isosceles)

m ∠PRQ = a + b

Now, m ∠PRQ × 2 = m ∠POQ

∴ m ∠PRQ = 25°

∴ a + b = 25°

Hence, option (c).

Answer: Option B

Explanation :

Given QR is the diameter. Let QS be a chord.

Diameter subtends an angle 90° anywhere on the circle 

∴ ∠RPQ = 90°. 

Given, ∠RPS = 50°.

⇒ ∠SPQ = 90° - 50° = 40°.

Also, a chord subtends same angle in the same segment of the circle.

∴ ∠SRQ = ∠SPQ = 40°

Hence, option (b).

Answer: Option C

Explanation :

We can draw the following diagram from the infomation given.

​​​​​​​

The perpendicular drawn on a chord from the center of the circle bisects the chord
⇒ BX = 16/2 = 8 and
⇒ DY = 12/2 = 6 and

Now, in ∆BOX
⇒ BO² = BX² + OX²
⇒ 10² = 8² + OX²
⇒ OX = 6

Also, in ∆BOX
⇒ DO² = DY² + OY²
⇒ 10² = 6² + OY²
⇒ OY = 8

XY = OX + OY = 6 + 8 = 14

Hence, option (c).

Answer: Option B

Explanation :

∠COB = ∠AOD = 50°

⇒ ∠CDB = 25° [Angle subtended by a chord on circle in same segment is half of the angle it subtends on the center.]

Now, angle subtended by a chord in opposite segments are complementary. If we consider the chord CB, then

⇒ ∠CDB + ∠CEB = 180°

⇒ ∠CEB = 180° - 25° = 155°

Hence, option (b).

Answer: Option A

Explanation :

In the given figure if we join AB then ∠ABC = 90°. [Angle subtended by diameter AC on circle is right angle.]

∴ ∠ABE + ∠EBC = 90°

⇒ ∠ABE = 90° - 45° = 45°

Now, if we consider chord AE, angle it subtends at point B on circle will be same as the angle it subtends at point C on the circle.

⇒ ∠ABE = ∠ACE = 45°

Now, AC and ED are parallel

⇒ ∠ACE = ∠CED  = 45° [alternate interior angles]

Hence, option (a).

Answer: Option C

Explanation :

Let us join BC.

​​​​​​​

Now, ∠ABC = ½∠AOC = 25° (angle subtended by chord AC on circle is half of the angle it subtends at the center)

similalry, ∠DCB = ½∠DOB = 20° (angle subtended by chord AC on circle is half of the angle it subtends at the center)

In triangle, BPC, ∠BPD = ∠PCB + ∠PBC (exterior angle is sum of two opposite interior angles)

⇒ ∠BPD = 25° + 20° = 45°

Hence, option (c).

Answer: Option D

Explanation :

Let us join S and R with point X as shown in the figure.

​

​​​​​​∠SAR = 180° - ∠STR = 180° - 105° = 75°

∠SOR = 2∠SAR = 150°

Now, ∠OSR + ∠ORS + ∠SOR = 180°
⇒ ∠OSR + ∠OSR + 150° = 180° (∠OSR = ∠ORS )
⇒ ∠OSR = 15°   ...(1)

Also, RQRS is a cyclic quadratilateral, hence two opposite angles are supplementary
⇒ ∠PSR + ∠PQR = 180°
⇒ ∠PSR + 100° = 180°
⇒ ∠PSR = 80°   ...(2)

From (1) and (2)
∠PSO = 80° - 15° = 65°

In ∆POS
⇒ ∠PSO + ∠SPO + ∠POS = 180°
⇒ 65° + 65° + ∠POS = 180° (∠PSO = ∠SPO)
⇒ ∠POS = 50°

Hence, option (d).

Answer: Option A

Explanation :

For a circle, if two chords AB and CD intersect at a single point P outside the circle, then, 

PA × PB = PC × PD

∴ 6 × 12 = 3 × PD

∴ PD = 24 cm

∴ CD = PD – PC = 24 – 3 = 21 cm.

Hence, option (a).

Answer: Option E

Explanation :

In ∆PAB,

PA = PB    …(tangents from an external point to a circle are congruent)

∴ m ∠PBA = m ∠PAB; …(angles opposite to congruent sides)

∴ m ∠PBA = 60° …(sum of angles in a triangle is 180°)

Similarly, in ∆QBC 

m ∠CBQ = m ∠BCQ = 50°

∵ m ∠PBA + m ∠CBQ + x = 180° …(linear angles)

∴ x = 180° – 60° – 50° = 70°

Now, m ∠AOC = 2 × x

∴ m ∠AOC = 140°

Hence, option (e).

Answer: 114

Explanation :

Draw two lines joining A and C with center O.

Now in quadrilateral AOCP, sum of all four angles should be 360°.

Also, ∠OAP = ∠OCP = 90° (radius is line joining center and point of tangency is perpendicular to the tangent)

∴ 90° + ∠AOC + 90° + 48° = 360°

⇒ ∠AOC = 132°

Angle subtended by chord AC in the major segment is half of the angle it subtends at the center.

∴ ∠ADC = ½ × ∠AOC = 66°

Now, angles subtended by a chord in different segments are supplementary.

∴ ∠ADC + ∠ABC = 180°

⇒ ∠ABC = 180° - 66° = 114°.

Hence, 114.

Answer: 44

Explanation :

We know 2 tangents drawn from an external point to a circle are equal in length.

⇒ AP = AQ = 22

∴ If we consider point B as the external point, we have two tangents i.e., BP and BD.

⇒ BP = BD

Similarly, if we consider point C as the external point, we have two tangents i.e., CD and CQ.

⇒ CD = CQ

Now, Perimeter of ∆ABC = AB + BD + CD + AC

= AB + BP + CQ + AC

= AP + AQ

 = 22 + 22 = 44

Hence, 44.

Answer: 13

Explanation :

When two secants are drawn from an external point to a circle

⇒ PC × PD = PA × PB

∴ 6 × 15 = 5 × (5 + AB)

⇒ AB = 13

Hence, 13.

Answer: 2.5

Explanation :

If we consider the smaller circle, two secants are drawn from P.

∴ PC × PB = PF × PE

⇒ 9 × 12 = 8 × PE

⇒ PE = 27/2

Now, if we consider the bigger circle, two secants are drawn from P.

∴ PB × PA = PE × PD

⇒ 12 × 18 = 27/2 × PD

⇒ PD = 16

∴ DE = PD – PE = 16 – 27/2 = 16 – 13.5 = 2.5

Hence, 2.5.

Answer: 9

Explanation :

Let PA = x

When a secant and a tangent are drawn from an external point to a circle

⇒ PA × PB = PT²

∴ x × (x + 5) = 6²

⇒ x² + 5x – 36 = 0

⇒ (x + 9)(x - 4) = 0

∴ x = -9 or 4. (x = -9 is rejected)

∴ PB = x + 5 = 9.

Hence, 9.

Answer: Option D

Explanation :

We can draw the following figure from the given information.

​​​​​​​

Since OR = OP = OQ = OS (radius)

⇒ ∠OPQ = ∠OQP & ∠QSO = ∠QOS

⇒ ∠POQ = 60°

⇒ ∠POR = 180 - (60° + 30°) = 90°

Hence, option (d).

Answer: 100

Explanation :

Given, ∠ATC = 30° and ∠ACT = 50°

In ∆ACT, external angle ,

∠CAB = ∠ACT + ∠ATC = 50° + 30° = 80°   …(1)

Also, chord CA and tangent CT make an angle of 50°, hence ∠CBA = 50°.

[Angle made by a chord and tangent is same as the angle subtended by the chord in opposite segment.]

Now, in ∆ABC, sum of all angles = 180°

⇒ ∠CBA + ∠BAC + ∠ACB = 180°

⇒ 50° + 30° + ∠ACB = 180°

⇒ ∠ACB = 100°

Now, angle make by a chord in major segment is half the angle it subtends at the center.

∴ ∠AOB = 2 × ∠ACB

⇒ ∠AOB 2 × 100° = 200°

Hence, 100.