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

Given:

The sum of their areas =  74 πsq cm

Distance between their centers = 12 cm.

Formula Used:

Area of circle = πr²

Calculation:

Let assume that radius of circle 1 = x

So, radius of circle 2 = 12 - x

Area of circle 1 = π(x)²

Area of circle 2 = π(12 - x)²

According to question ⇒ π(x)² + π(12 - x)² = 74π

⇒ x² + 144 - 24x + x² = 74 

⇒ 2x² - 24x + 70 = 0

⇒ x² - 12x + 35 = 0

⇒ (x - 7)(x - 5) = 0

⇒ x = 7 ⇒ x = 5 

∴ The radius of smaller circle is 5 cm

Given:

PQ is the diameter of a circle with center 'O'. Tangent drawn at P meets QR (produced) at S. ∠PSQ = 48º.

Formula Used:

Angle in the semicircle = 90º

Calculation:

In triangle PQR:

Since PQ is diameter ⇒ ∠PRQ = 90°

In triangle PSQ:

∠PSQ is the exterior angle to triangle PQR

∠PSQ = 48º

∠SPQ = 90° (angle made by tangent and chord)

∠PQS = 90 - 48 = 42° 

∠PQS = ∠PQR = 42° 

∴ The value of ∠PQR is 42°.

Given:

PQ is a chord in the minor segment of a circle.

R is a point on the minor arc PQ.

Tangents at points P and Q meet at point T.

∠PRQ = 102°

Formula used:

The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.

The sum of opposite angles in a cyclic quadrilateral is 180°.

The angle between the tangent and the chord through the point of contact is equal to the angle in the alternate segment (Tangent-Chord Theorem).

The sum of angles in a quadrilateral is 360°.

Calculation:

Let O be the center of the circle.

Consider the cyclic quadrilateral formed by P, R, Q and another point S on the major arc PQ.

The angle subtended by the chord PQ at any point in the major segment will be supplementary to ∠PRQ.

The angle subtended by the major arc PQ at the circumference is ∠PRQ = 102°.

The angle subtended by the minor arc PQ at the circumference (on the major segment) would be 180° - 102° = 78°.

∠PSQ = 180° - ∠PRQ (as PRQS is a cyclic quadrilateral if S is on the major arc).

So, ∠PSQ = 180° - 102° = 78°.

The angle subtended by the minor arc PQ at the center O, i.e., ∠POQ, is twice the angle subtended by it at the circumference in the major segment (∠PSQ).

⇒ ∠POQ = 2 × ∠PSQ

⇒ ∠POQ = 2 × 78°

⇒ ∠POQ = 156°

Now, consider the quadrilateral TP OQ. TP and TQ are tangents to the circle at P and Q, respectively.

We know that the radius is perpendicular to the tangent at the point of contact.

The sum of angles in the quadrilateral TP OQ is 360°.

⇒ ∠PTQ + ∠TPO + ∠POQ + ∠TQO = 360°

⇒ ∠PTQ + 90° + 156° + 90° = 360°

⇒ ∠PTQ + 336° = 360°

⇒ ∠PTQ = 360° - 336°

⇒ ∠PTQ = 24°

∴ The correct answer is option 2.

Given:-

(1) AT = 6 cm, AB = 10 cm & TB = ?

(2) TB ┴ AT

In right-angled, △ATB

Formula used:-

By Pythagoras theorem

(AB)² = (AT)² + (TB)²   

Calculation:-

⇒ (10)² = 6² + (TB)²

⇒ TB = 8

The radius (r) of the circle is TB = 8 cm

We know diameter is the largest chord of the circle

Concept:

we know that the line joining the centres of two circle perpendicularly bisects the common chord

Calculation:

radius of 1^st circle(r1) = 15/2 and that of 2^nd (r2) = 13/2

length of common chord = AB = 12 cm

let c = length of common chord/2 = 12/2 = 6

We can use the formula to get the distance between the centres

⇒ √((r1)² - c²) + √((r2)² – c²)

⇒ √{(7.5)² - 6²} + √{(6.5)² – 6²}

⇒ √20.25 + √6.25

⇒ 4.5 + 2.5 = 7 cm

Given:

AX = 24

XB = k

CX = (k + 2)

XD = 16

Formula Used:

If two chords AB and CD intersecting at point X.

Then, AX × XB = CX × XD

Calculation:

AX × XB = CX × XD

⇒ 24 × k = (k + 2) × 16

⇒ 3k = 2(k + 2)

⇒ 3k - 2k = 4

⇒ k = 4

So, the value of k is 4.

Given:

∠BAC = 60°

Concept used:

The angle subtended by an arc of a circle at the center is double the angle subtended by it any point on the remaining part of the circle.

Calculation:

∠BIC = 2 × ∠BAC = 2 × 60° = 120° 

Given:

∠ABO = 40°

Concept:

The angle subtended by an arc at the center is double than subtended by it on other parts of the circle. In the figure below, ∠ AOB = 2∠ACB

Calculation:

In the figure ∠ABO = ∠BAO (angles formed by the radius are equal)

In Δ AOB, ∠ABO + ∠BAO + ∠BOA = 180°

⇒ ∠BOA = 180° - (40 + 40) = 100°

⇒ ∠BOA (external) = 360° - 100° = 260°.

∠ACB = \(1\over 2\) ∠BOA (external) (since angle formed by the arc at the center is double compared to the angle formed by the arc at other points on the circumference).

∴ ∠ACB = \(1\over 2\) × 260° = 130°

∴ The angle formed by ∠ACB is 130°.

Given:

∠APB = 122° 

Concept used:

Sum of opposite sides of a cyclic quadrilateral are 180°

The angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle’s circumference.

Calculation:

In the given figure,

APBT is a cyclic quadrilateral.

∠ATB + ∠APB = 180° [sum of opposite angles are 180°]

⇒ x° + 122° = 180° 

⇒ x = (180° – 122°)

⇒ x = 58° 

∠ATB = 58° 

And we know that the angle subtended by an arc of a circle at its center is twice the angle it subtends anywhere on the circle’s circumference.

⇒ ∠AOB = 2 × ∠ATB

⇒ ∠AOB = 2 × 58° 

⇒ ∠AOB = 116° 

Now,

OA = OB [radius of the circle]

Then,

∠OAB + ∠AOB + ∠OBA = 180° 

⇒ θ + 116° + θ = 180° 

⇒ 2θ + 116° = 180° 

⇒ 2θ = (180° – 116°)

⇒ 2θ = 64° 

⇒ θ = 32° 

so,

∠OAB = θ = 32° 

∴ The required value of ∠OAB is 32°.

Shortcut Trick 

From the above given diagram we have

⇒ θ = P - 90°

⇒ θ = 122° - 90° = 32°

∴ The correct answer is 32°.

from the following figure.

∠AOB = 110° and ∠AOC = 130°

As we know,

∠AOB + ∠AOC + ∠BOC = 360°

∠BOC = 360° - 110° - 130° = 120°

∠BAC = ∠BOC/2

∴ ∠BAC = 120°/2 = 60°

Calculation:

RP = RQ + PQ = 14.4 + 11.2 = 25.6

As we know,

RP × RQ = RT × RS

⇒ 25.6 × 14.4 = RT × 12.8

⇒ RT = 28.8 cm

Given:

Radii of two circles are 12 cm and 5 cm. Distance between their centres is 25 cm.

Formula used:

Length of the direct common tangent of two circles = \(\sqrt {D^2 - (r_1 - r_2)^2}\) (D = Distance between their centres, r₁ = radius of the bigger circle, and r₂ = radius of the smaller 

Calculation:

Let the centres be at P and Q.

QN = Radius of the bigger circle = 12 cm

PM = Radius of the smaller circle = 5 cm

Let MN be the direct common tangent.

According to the formula,

Length of MN

⇒ \(\sqrt {25^2 - (12 - 5)^2}\)

⇒ \(\sqrt {576}\)

⇒ 24 cm

∴ The length of the direct common tangent is 24 cm.

∠QRP = 28°

As we know,

∠ORP = 90° [tangents at R]

⇒ ∠ORQ = ∠ORP – ∠QRP = 90° – 28° = 62°

⇒ ∠OQR = ∠ORQ = 62° [OQ = OR = radius]

As we know,

∠OQR = ∠ QRP + ∠QPR

⇒ 62 = 28 + ∠QPR

⇒ ∠QPR = 62 – 28 = 34

⇒ ∠SPR = ∠QPR = 34

Given:

Chord = 21cm 

Diameter = 25 cm 

Concept:

The perpendicular drawn from the center on the chord, bisects the chord in two equal parts. 

Pythagoras Theorem:

OA² = OD² + AD²

Calculation:

Let AB be the chord of length 21 cm. 

⇒ OD be the perpendicular distance 

⇒ AO be the radius of the circle 

OA2 = OD2 + AD2

⇒ (25/2)² = OD2 + (21/2)²

⇒ 625/4 = OD2 + 441/4 

⇒ OD2 = 625/4 - 441/4 = 184/4 = 46 

∴ OD = √46 

Given:

Radius = 10 cm

Concept used:

Concept of similarity

AAA similarity → If all three angles in one triangle are the same as the corresponding angles in the other, then the triangles are similar.

Calculation:

In, the given figure,

ΔDAO ∼ ΔAPO [∵ ∠OAP = ∠ODA = 90° and ∠AOD is common in two triangles]

So,

AD/AP = DO/AO

⇒ 6/AP = 8/10 [6, 8, 10 pythagoras triplets]

⇒ AP = 60/8

⇒ AP = 7.5 cm

∴ Length of tangent AP is 7.5 cm