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

Concept used:

Center angle is double of the major angle

In cyclic quadrilateral sum of opposite angle is 180° 

Calculation:

∠POR = 2 × ∠PSR

⇒ ∠PSR = 150°/2 = 75°

⇒ ∠PQR + ∠PSR = 180° 

⇒ ∠PQR + 75° = 180°

⇒ ∠PQR = 105°

∴ The correct answer is 105°

Concept used:

Center angle is double of the major angle

In cyclic quadrilateral sum of opposite angle is 180° 

Calculation:

∠POR = 2 × ∠PSR

⇒ ∠PSR = 150°/2 = 75°

⇒ ∠PQR + ∠PSR = 180° 

⇒ ∠PQR + 75° = 180°

⇒ ∠PQR = 105°

∴ The correct answer is 105°

Concept used:

PA × PB = PC × PD

Calculation:

PA × PB = PC × PD

⇒ 10 × 22 = 11 × PD

PD = \(\frac{10 \times 22}{11}\) = 20

∴ The correct answer is 20

Given:

∠BAT = 40° 

Concept used:

Alternate segment theorem:

If a chord is drawn from the point of contact of the tangent then the angles made by the chord with the tangent are equal to the angles formed in the corresponding alternate segments.

∠BAT = ∠BCA

∠BAP = ∠BDA

Calculation: 

According to question,

∠BAT = 40° 

By alternate segment theorem,

∠BAT = ∠BCA = 40° 

∠AOB  = 2∠BCA

⇒ 2 × 40° 

⇒ 80° 

Since, OA = OB 

∴ ∠AOB + ∠OBA + ∠OAB = 180° 

⇒ 80° + 2∠OAB = 180° 

⇒ 2∠OAB = 100° 

⇒ ∠OAB = 50° 

Ratio = ∠AOB : ∠OAB

⇒ 80° : 50° 

⇒ 8 : 5

∴ The correct answer is 8 : 5.

Given:

A = (x₁, y₁) = (3, 2)

B = (x2, y2) = (6, 5)

C = (x3, y3) = (9, 8)

Formula used:

Distance between two points = \(\sqrt{(x_2 - x_1)^2 + (y_2-y_1)^2}\)

Calculation:

AB = \(\sqrt{(6-3)^2 + (5-2)^2} = \sqrt{9+9} =3\sqrt2cm\)

BC = \(\sqrt{(9-6)^2 + (8-5)^2} = \sqrt{9+9} = 3\sqrt2cm\)

CA = \(\sqrt{(9-3)^2 + (8-2)^2} = \sqrt{36 + 36} =6\sqrt2cm\)

the distance between A and C = AB + BC

Hence, A, B, and C forms a line segment.

Hence, the correct option is 2.

Using the theorem, angle in the semi-circle is a right angle,

⇒ ∠BOA = 90°

Theorem: The alternate segment theorem states that an angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment.

⇒ ∠BOQ = ∠BAO = 60°

In ΔABO,

Sum of the angles of the triangle is 180°

⇒ ∠ABO = 180° – ∠BOA – ∠BAO = 180° – 90° – 60° = 30°

∠OPT = 90° [∵ Radius is perpendicular to tangent]

∠OPQ = 90° - ∠QPT = 90° - α

∠OQP = 90° - α [∵ OQ = OP]

In triangle OQP

∠O + ∠Q + ∠P = 180°

∠O + 90 - α + 90 - α = 180

∴ ∠O = 2α

Given:

∠AOC = 110°

∠AOB = 90°

Concept used:  

The angle subtended by an arc at the centre is double the angle subtended by it at the circumference of the circle.

Angles around a point will always add up to 360°

Calculation:

∠AOC = 110°

∠AOB = 90°

Angles around a point will always add up to 360°

∠BOC = 360° - (∠AOC + ∠AOB)

⇒ ∠BOC = 360° - (110° + 90°)

⇒ ∠BOC = 160°

Now,

∠BAC = ∠BOC/2

∠BAC = 160°/2

∠BAC = 80°

∴ The value x is 80°.

Alternate MethodIn triangle AOC,

OC = OA (radius of the circle)

Therefore, the angles opposite to these sides will be equal. (∠OCA = ∠OAC)

Let ∠OCA and ∠OAC be y

110 + 2y = 180

2y = 70

y = 35

In triangle AOB, 

OA = OB (radius of the circle)

Therefore, the angles opposite to these sides will be equal.  (∠OAB = ∠OBA)

Let ∠OAB and ∠OBA be z

90 + 2z = 180

2z = 90

z = 45

Thus, the value of 'x' will be (y + z) = 35 + 45 = 80

Given

AB = 2 cm, OB = 5 cm and OD = 3.5 cm

Property:

When two chords AB and CD intersect outside a circle at a point O, then,

OA × OB = OC × OD

Applying the same

⇒ (AB + OB) × OB = OC × OD

⇒ 7 × 5 = OC × 3.5

⇒ OC = 35/3.5 = 10

⇒ CD = OC – OD = 10 – 3.5 = 6.5 cm

Given, AB ∶ BC = 4 ∶ 5

⇒ BC = (5/4) × AB

∵ AC = AB + BC

⇒ AC = AB + (5/4) × AB = (9/4) × AB

Applying the tangent-secant theorem, the tangent and the secant are related as,

⇒ AD² = AB × AC

⇒ AD² = (9/4) × AB²

⇒ AD/AB = √(9/4) = 3/2

In this diagram

let PB = x

According to the property of secant

PB × PA = PD × PC               [PC = PD + CD]

x × 15 = 6 × 10 

x =  4 = PB 

Given:

AB = 12, PB = 8, and OP = 18

⇒ PA = AB + PB

⇒ PA = 8 + 12

⇒ PA = 20

⇒ PD = 18 - r

⇒ PC = 18 + r

Where, r = radius of circle

Formula Used:

In two chords AB and CD intersect at outer point of P then,

PB × PA = PD × PC

Calculation:

PB × PA = PD × PC

⇒ 8 × 20 = (18 - r)(18 + r)

⇒ 160 = 18² - r²

^⇒ r² = 324 - 160

⇒ r² = 164

⇒ r = \(\sqrt {164} \)

⇒ r = \(2\sqrt {41} \)

∴ The radius of circle is \(2\sqrt {41} \)

We know that,

AO × OB = OC × OD

⇒ (9x – 2) × (2x + 2) = (4x) × (7x – 2)

⇒ 18x² – 4x + 18x – 4 = 28x² – 8x

⇒ 10x² – 22x + 4 = 0

⇒ 10x² – 20x – 2x + 4 = 0

⇒ 10x(x – 2) – 2(x – 2) = 0

⇒ (x – 2)(10x – 2)  = 0

⇒ x = 2 or x = 0.2

⇒ x = 2

AO = (9x – 2) = (18 – 2) = 16 cm

∴ The value of AO is 16cm.

Radius of the circle is 5 cm

∴ Side of the square AB = BC = 5√2 cm

Suppose PB = x cm

In ΔPBC

⇒ PC² = [(5√2)² + x²]

We know that∶

PC² = PB × PA

⇒ [(5√2)² + x²] = x × (x + 5√2)

⇒ 50 + x² = x² + 5√2x

⇒ x = 5√2 cm

Given:

DE is a tangent to the circle with centre O

∠ACD = 50°

∠AOB = 140°

Concept:

Theorem∶ The angle subtended by an arc at the centre of the circle is double the angle subtended by it at the circumference of the circle.

Theorem∶ According to the alternate segment theorem, the angle between a tangent and a chord is equal to the angle made by the chord in the alternate segment of the circle.

Calculation:

According to the question:

∵ OA = OB = radius of circle

⇒ ΔAOB is an isosceles triangle.

⇒ ∠OAB = ∠OBA

∵ Sum of angles of triangle = 180°

⇒ ∠OAB + ∠OBA + ∠AOB = 180°

⇒ ∠OAB = ∠OBA = (180° – 140°)/2 = 20°

According to the concept,

⇒ ∠ACB = 1/2 × ∠AOB = 140°/2 = 70°

⇒ ∠BCE = 180° – ∠ACB – ∠ACD = 180° – 70° – 50° = 60°

Again, according to the concept,

⇒ ∠BCE = ∠BAC = 60°

⇒ ∠OAC = ∠BAC – ∠OAB = 60° – 20° = 40°

∴ The required result will be 40°.