# CRE 1 - Basics | Geometry - Circles

**CRE 1 - Basics | Geometry - Circles**

What is the diameter (in cm) of a circle with area 36π cm^{2}?

- A.
20 cm

- B.
12.5 cm

- C.
12 cm

- D.
6 cm

Answer: Option C

**Explanation** :

Area of the circle, A = πr^{2} = 36π

⇒ r^{2} = 36

∴ Radius, r = 5 cm

∴ Diameter, d = 2 × 5 = 12 cm

Hence, option (c).

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**CRE 1 - Basics | Geometry - Circles**

What is the circumference (in cm^{2}) of a circle for which the ratio of the area and the circumference is 7 : 4? [Take π = 22/7]

- A.
44 cm

- B.
22 cm

- C.
15π cm

- D.
7π/2 cm

Answer: Option B

**Explanation** :

Ratio of area and circumference of a circle = πr^{2} : 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).

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**CRE 1 - Basics | Geometry - Circles**

What is the area (in cm^{2}) and the perimeter (in cm) of a quarter circle with radius 14 cm? [Take π = 22/7]

- A.
154, 50

- B.
98π, 50

- C.
49π, 22π

- D.
49π, 2π

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 × πr^{2 }= 1/4 × 22/7 × 14 × 14 = 154 cm^{2}

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).

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**CRE 1 - Basics | Geometry - Circles**

What is the length of a chord which is at a distance of 6 cm from the center of a circle with radius 10 cm?

- A.
12 cm

- B.
16 cm

- C.
24 cm

- D.
18 cm

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,

OB^{2} = DB^{2} + OD^{2}

∴ 10^{2} = DB^{2} + 6^{2}

∴ DB^{2} = 100 – 36 = 64

∴ DB = 8 cm

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

Hence, option (b).

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**CRE 1 - Basics | Geometry - Circles**

A chord subtends an angle of 100° at the center of a circle. What is the angle subtended by the chord in the major segment of the circle?

- A.
80°

- B.
160°

- C.
50°

- D.
260°

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).

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**CRE 1 - Basics | Geometry - Circles**

In the given circle with center O, AB and CD are two chords that intersect at point X. DX = 6 cm, CX = 8 cm, and BX = 4 cm. What is the radius of the circle?

- A.
12 cm

- B.
10 cm

- C.
8 cm

- D.
Can't be determined

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).

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**CRE 1 - Basics | Geometry - Circles**

What is the circumference of a circle whose area is equal to that of an equilateral triangle with perimeter P?

- A.
$\sqrt{\left(\pi \sqrt{3}\right)}P$

- B.
$\sqrt{\sqrt{3}}P$

- C.
√3 P

- D.
$\frac{\sqrt{\left(\pi \sqrt{3}\right)}}{3}P$

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 = πr^{2} = $\frac{\sqrt{3}}{4}\times {\left(\frac{P}{3}\right)}^{2}$

∴ 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).

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**CRE 1 - Basics | Geometry - Circles**

Radius of two smaller semicircles shown in the figure is 2 cm. What is the area (in cm^{2}) of the shaded portion bounded by three semicircles as shown in the figure?

- A.
8π

- B.
16π - 1

- C.
8π - 1

- D.
None of these

Answer: Option A

**Explanation** :

The three semicircles in the figure can be named S_{1}, S_{2} and S_{3} as shown above.

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

Area(S_{2}) = Area(S_{3})

Hence, we can redraw the figure like this.

Area of the shaded portion is same as the area of S_{1}.

Radius of S_{1} = 4.

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

Hence, option (a).

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**CRE 1 - Basics | Geometry - Circles**

In the given figure, O is the center of the circle. The sum of angles a and b is:

- A.
50°

- B.
100°

- C.
25°

- D.
Can't be determined

- E.
None of these

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).

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**CRE 1 - Basics | Geometry - Circles**

In the given figure, QR is the diameter of the circle. P and S are points on the circle such that ∠RPS = 50°. Find ∠SRQ.

- A.
30°

- B.
40°

- C.
50°

- D.
60°

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).

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**CRE 1 - Basics | Geometry - Circles**

What is the distance in cm between two parallel chords of length 16 cm and 12 cm in a circle of radius 10 cm?

- A.
15

- B.
18

- C.
14

- D.
None of these

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

⇒ BO^{2} = BX^{2} + OX^{2}

⇒ 10^{2} = 8^{2} + OX^{2}

⇒ OX = 6

Also, in ∆BOX

⇒ DO^{2} = DY^{2} + OY^{2}

⇒ 10^{2} = 6^{2} + OY^{2}

⇒ OY = 8

XY = OX + OY = 6 + 8 = 14

Hence, option (c).

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**CRE 1 - Basics | Geometry - Circles**

In the figure AB and CD are two diameters of the circle intersecting such that ∠COB = 50°. E is any point on arc CB. Find ∠CEB?

- A.
165°

- B.
155°

- C.
145°

- D.
100°

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).

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**CRE 1 - Basics | Geometry - Circles**

In the adjoining figure, chord ED is parallel to the diameter AC of the circle. If angle CBE = 45°, then what is the value of angle DEC?

- A.
45°

- B.
65°

- C.
55°

- D.
30°

- E.
None of these

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).

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**CRE 1 - Basics | Geometry - Circles**

A truck has wheels of diameter 1.4 m. How many times must a wheel of the truck rotate to cover a distance of 1.1 km?

- A.
210

- B.
200

- C.
250

- D.
None of these

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).

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**CRE 1 - Basics | Geometry - Circles**

Two chords AB and CD of circle whose centre is O, meet at the point P inside the circle such that ∠AOC = 50°, ∠BOD = 40°. Then the value of ∠BPD is

- A.
60°

- B.
40°

- C.
45°

- D.
75°

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).

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**CRE 1 - Basics | Geometry - Circles**

In the figure given below, O is the center of the circle and ∠PQR = 100° and ∠STR = 105°. What is the value (in degrees) of ∠SOP?

- A.
80

- B.
45

- C.
65

- D.
50

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).

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