Theorem on Chords MCQ Quiz - Objective Question with Answer for Theorem on Chords - Download Free PDF

Given:

Parallel chords AB = 36 cm

Parallel chords CD = 48 cm

Distance between the chords = 42 cm

Formula Used:

The perpendicular from the center of a circle to a chord bisects the chord.

Pythagoras theorem: (radius)² = (half of chord length)² + (distance from center)²

Diameter = 2 × radius

Calculation:

Let the radius of the circle be r.

Let the distance of chord AB from the center be x cm.

Then the distance of chord CD from the center will be (42 - x) cm.

For chord AB:

r² = (36/2)² + x²

⇒ r² = 18² + x²

⇒ r² = 324 + x² (Equation 1)

For chord CD:

r² = (48/2)² + (42 - x)²

⇒ r² = 24² + (42 - x)²

⇒ r² = 576 + (1764 - 84x + x²)

⇒ r² = 576 + 1764 - 84x + x²

⇒ r² = 2340 - 84x + x² (Equation 2)

Equating Equation 1 and Equation 2:

324 + x² = 2340 - 84x + x²

⇒ 324 = 2340 - 84x

⇒ 84x = 2340 - 324

⇒ 84x = 2016

⇒ x = 2016 / 84

⇒ x = 24 cm

Substitute the value of x in Equation 1:

r² = 324 + (24)²

⇒ r² = 324 + 576

⇒ r² = 900

⇒ r = \(\sqrt{900}\)

⇒ r = 30 cm

Diameter = 2 × radius = 2 × 30 = 60 cm

∴ The diameter of the circle is 60 cm.

Given:

The angle subtended by arc ABC at the center of the circle = 138°.

Chord AB is produced to a point P, and we need to find ∠CBP.

Formula used:

The angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at the circle's circumference.

Calculation:

Let ∠AQC be the angle subtended by arc ABC at the circle's circumference.

⇒ ∠AOC = 2 × ∠AQC

⇒ ∠AQC = 1/2 × ∠AOC = 1/2 × 138° = 69°

Now, in the cyclic quadrilateral ABQC, the exterior angle ∠CBP is equal to the interior opposite angle ∠AQC.

⇒ ∠CBP = ∠AQC = 69°

∴ The measure of ∠CBP is 69°.

Given:

The two circles intersect at two points P and Q.

PR and PS are diameters of the two circles.

Formula Used:

In a circle, the angle subtended by the diameter on the circumference is 90º.

Calculation:

Since PR and PS are diameters of the two circles, they subtend a right angle at any point on the circumference.

Thus, ∠PQR is a right angle.

∠PQR = 90º

The correct answer is option 1.

Given:

Length of chord on both circles is 6 cm and 18 cm.

Concept used:

Pythagoras theorem.

Calculation:

Let the radii of the smaller circle and the bigger circle be r1 and r2.

In ΔOAC 

OC2 = OA2 + AC2

⇒ r12 - AC2 = OA2    ....(1)

In ΔOAB

OB2 = OA2 + AB2

⇒ r22 - AB2 = OA2    ....(2)

From the first equation and the second equation,

r12 - 9 = r22 - 81

⇒ r22-  r12 = 81 - 9 = 72

∴ The required difference in the square is 72 cm2.

Given:

Radii of two concentric circles are 26 cm and 10 cm.

The chord of the greater circle is a tangent to the smaller circle.

Formula Used:

Using Pythagoras theorem in the right triangle formed by the radius of the greater circle, the radius of the smaller circle, and half of the chord length:

\(R^2 = r^2 + \left(\frac{L}{2}\right)^2\)

Where:

R = radius of the greater circle = 26 cm

r = radius of the smaller circle = 10 cm

L = length of the chord

Calculation:

26² = 10² + \(\left(\frac{L}{2}\right)^2\)

676 = 100 + \(\left(\frac{L}{2}\right)^2\)

676 - 100 = \(\left(\frac{L}{2}\right)^2\)

576 = \(\left(\frac{L}{2}\right)^2\)

\(\left(\frac{L}{2}\right)\) = √576

\(\left(\frac{L}{2}\right)\) = 24

L = 24 × 2

L = 48 cm

The length of the chord is 48 cm.

Given:

ΔPQR is an equilateral triangle inscribed in a circle.

S is any point on the arc QR.

Concept used:

Each of the angles of an equilateral triangle is equal and 60° 

The angles subtended by an arc in the same segment of a circle are equal.

Calculation:

Both ∠PRQ and ∠PSQ are in the subtended by the arc PQ in the same segment, so

∠PRQ = ∠PSQ = 60°

 ∴ The correct option is 2

Given:

The radius of the circles is 8 cm

Calculation:

According to the diagram,

AD = DB

O1O2 = 8

Again O1A = O2A = 8 [Radius of the circle]

∠ADO1 = 90°

O1D = O2D = 4

AD = √(64 - 16)

⇒ √48 = 4√3

AB = 2 × 4√3 = 8√3

∴ The length of the common chord is 8√3 cm

Given:

Radii of circles are 13 cm and 15 cm 

Distance between centres = 14 cm 

Concept Used:

The line joining the centers of two circles is the perpendicular bisector of the common chord of the two circles.

⇒ AB = 2AM

Calculation:

From the figure, P and Q are the centers of two circles of radius 15 cm and 13cm respectively and AB is the common chord

In ΔAMP, PA = 15 cm

By Pythagoras theorem,

AM² = PA² - PM²

⇒ AM2 = 152 - PM2

⇒ AM2 = 225 - PM2     -----(1)

In ΔAMQ, QA = 13 cm

By Pythagoras theorem,

AM2 = QA2 - MQ2

⇒ AM2 = 132 - MQ2

⇒ AM2 = 169 - MQ2     -----(2)

From (1) and (2)

225 - PM2 = 169 - MQ2 

⇒ PM2  - MQ2 = 56

⇒ (PM + MQ) (PM - MQ) = 56

According to the question,

(PM + MQ) = PQ = Distance between centres = 14 cm ----(3)

⇒ 14 (PM - MQ) = 56

⇒ (PM - MQ) = 56/14 = 4                                               ----(4)

From (3) and (4)

PM = 9 cm

Putting value of PM in (1)

AM2 = 225 - 92

⇒ AM2  = 225 - 81

⇒ AM2  = 144

⇒ AM = 12 cm

Thus, length of common chord = AB = 2AM = 24 cm

∴ The length of the common chord is 24 cm.

Given:

AB = 28 cm and CD = 22 cm are two parallel chords on the same side of the centre of a circle.

The distance between them is 4 cm.

Calculation:

Let the OE is x, EF = 4 cm

As per the question,

OA = OC = radius of the circle

⇒ 14² + x² = (x+4)² + 11²

⇒ 196 + x² = x² + 16 + 8x + 121

⇒ 8x = 59

x = 59 / 8 

The radius will be,

√(142 + (\({59\over8}\))² = 15.82 cm

∴ The correct option is 3

Calculation

The perpendicular bisector of any chord passes through the centre of the circle,

OP ⊥ AB

In OPB, ∠OPB = 90⁰

PB² = OB² – OP²

⇒ PB² = 10² – 6²

⇒ PB² = 64

⇒ PB² = 8²

⇒ PB = 8 cm

OP bisects AB

⇒ AP = PB = 8 cm

∴ AB = AP + PB = 8 + 8 = 16 cm

The length of AB is 16 cm.

Concept used:

According to the given picture, ∠AOB = 2∠ACB

Again, ∠ACB + ∠APB = 180°

Calculation:

According to the given concept,

∠AOB = 2∠ACB

⇒ ∠ACB = 130/2 = 65° 

Again,  ∠ACB + ∠APB = 180°

⇒ ∠APB = 180° - 65° = 115°

∴ The correct answer is 115°

Given:

Radius of large circle (R) = 15 cm.

Radius of small circle (r) = 13 cm

Concept used:

A line drawn from the centre of the circle to a chord bisects the chord into two halves.

The tangent to the circle is perpendicular to the radius of the circle at the point of contact.

Pythagorean theorem:

H² = P² + B²

Where, H = hypotenuse ; P = perpendicular ; B = base

Calculation:

Here, O is the centre and PQ is the chord of the circle.

In △POR

⇒ OP² = OR² + PR²

⇒ (15)² = (13)² + PR²

⇒ PR² = 225 - 169 = 56

⇒ PR = √56 = 2√14 cm

PQ is tangent so, OR ⊥ PQ

PQ = PR + RQ

⇒ 2√14 + 2√14 (As PR = RQ)

⇒ 4√14 cm

∴ The correct answer is 4√14 cm. 

Given:

The angle subtended by a chord at the centre = 180°

Concept Used:

The angle subtended by a chord at the centre is twice the angle subtended on the circumference.

Calculation:

⇒ Angle subtended on the circumference = 180° / 2 = 90°

Therefore, the angle subtended by the same chord on the circumference is 90°.

Given:

The radius of the circles is 18 cm

Calculation:

According to the diagram,

AD = DB

O₁O₂ = 18

Again O₁A = O₂A = 18 [Radius of the circle]

∠ADO₁ = 90°

O₁D = O₂D = 9

AD = √(324 - 81)

⇒ (9√3) 

AB = 2 × (9√3) = 18√3 = 6√27

∴ The length of the common chord is 6√27 cm.

Given:

The diameter of the circle = 58 cm

The length of one chord = 42 cm

The length of another chord = 40 cm

Formula used:

H2 = B2 + P2

Calculation:

AB and PQ are two chords, and O is the center of the circle.

M is the midpoint of AB and N is the midpoint of PQ

OB = OQ = 29 cm [radius of the circle]

AB = 40 cm and OB = 29

Then,

AM = MB = 40/2 = 20 cm

In ΔMOB

(OB)2 = (OM)2 + (MB)2

⇒ (29)2 = (OM)2 + (20)2

⇒ 841 = (OM)2 + 400

⇒ (OM)2 = (841 – 400)

⇒ (OM)2 = 441

⇒ OM = 21 cm

Now,

PQ = 42 cm and OB = 29

NQ = PN = 42/2 = 21 cm

In ΔONQ

(OQ)2 = (ON)2 + (NQ)2

⇒ (29)2 = (ON)2 + (21)2

⇒ 841 = (ON)2 + 441

⇒ (ON)2 = (841 – 441)

⇒ (ON)2 = 400

⇒ ON = 20 cm

So according to the question.

⇒ 21 - 20 = 1

∴ The minimum distance between the two chords is 1 cm.