Test: Circles- 2 - Class 9 MCQ

The region between a chord and either of its arcs is called a segment of the circular region or simply a segment of the circle. The segment formed by minor arc along with chord, is called minor segment and the segment formed by major arc, is called the major segment.

AP+PB=AB [diameter of circle]
4+16=20[diameter]
so r=AO = CO=10
so clearly PO= OA-AP=10-4=6
PO=6 OC=10 So just apply pythagoreus theorem in triangle PCO
so CP will come 8 So chord CD = 2x8 = 16

The correct answer is 50°

GIVEN: Chords AB and CD intersect at right angle. ∠BAC = 40°

TO FIND: ∠ABD

SOLUTION

We can simply solve the above problem as follows;

Let the point of intersection of the chord be 'O'.

∠AOC = 90°

Therefore,

In ΔAOC

∠AOC + ∠OAC + ∠ACO = 180°

90 + 40 + ∠ACO = 180

∠ACO = 180- 130 = 50°

Now,

AO :CO = 5:4 [since larger angle denotes shorter length]

Therefore,

Ratio of-

OB:OD=4:5 [ since they intersect and ratio gets changed]

or,

∠OBD:∠ODB=5:4 [larger length has smaller ratio]

Therefore,

∠OBD+∠ODB+90=180

5x + 4x + 90 = 180

9x + 90 = 180

x + 10 = 20

x = 10

∠OBD = 5 × 10 = 50°

Since,

∠OBD = ∠ABD

Therefore,

∠ABD = 50°

(Angle at the centre is double the angle at the circumference subtended by the same chord)

For a regular n-sided polygon inscribed in a circle, the central angle subtended by each side is:

Central Angle = 360^∘ / n

For the pentagon (n=5), the central angle subtended by AB is:

360^∘ / 5 =72^∘

For the hexagon (n=6), the central angle subtended by AC is:

360^∘ / 6 = 60^∘

The angle ∠APB between the two intersecting chords AB and AC can be calculated using the property of angles formed by intersecting chords: ∠APB = 1 /2 × (Sum of the arcs subtended by the opposite segments)

• Arc subtended by AB: 72^∘
• Arc subtended by AC: 60^∘
• The sum of the opposite arcs involved is: 72^∘+120^∘=192^∘

2ACB=AOB
ACB=40
CAB+ACB+ABC=180
CAB=180-70
CAB=110
since,OA=OB(radius)
OAB=OBA

AOB+OAB+OBA=180
2OBA=100
OBA=50

CAO=CAB-OBA=110-5O
CAO=60^o

The angle subtended by a chord at the center of the circle is twice the angle subtended by the same chord at any point on the circumference.

This property is expressed as:

∠ACB = 1 / 2 × ∠AOB

∠ AOB = 80^∘ (angle subtended by the arc AB at the center).

Using the property: ∠ACB = 1 / 2 × ∠AOB

Substituting ∠AOB=80^∘

∠ACB = 1 / 2 ​× 80^∘= 40^∘

We know that, the perpendicular from the centre of a circle to a chord bisects the chord.
AC=CB=(1/2)AB=(1/2)×8=4cm
OA = 5cm (given)
AO²=AC²+OC²
(5²)=(4)²+OC²
OC = 3 cm
[Take positive value as length can't negative]
OA = OD = 5cm
CD = OD - OC = 5 - 3 = 2cm.

Given:
∠OAB = 40° ...(1)


In ∆OAB,
OA = OB
∴ ∠OAB = ∠OBA = 40° (angles opposite to equal sides are equal) ...(2)

Now,
∠OAB + ∠OBA + ∠AOB = 180° (angle sum property)
⇒ 40° + 40° + ∠AOB = 180° (From (1) and (2))
⇒ ∠AOB = 180° − 80°
⇒ ∠AOB = 100° ...(3)


We know, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
Thus, ∠AOB = 2∠ACB
⇒ 100° = 2∠ACB (From (3))
⇒ ∠ACB = 50°


Hence, ∠ACB = 50°.

Two chords AD and BC intersect each other at right angles at P and ∠DAB=35o.

In ΔABP, Ext. ∠APC=∠B+∠A

⇒90o=∠B+35^o

∠B=90^o − 35^o =55^o

∵∠ABC and ∠ADC are in the same segment.

∴∠ADC =∠ABC = 55^o

Angle BPA = 90°

In triangle OAB

OA = OB (radius of a circle)

∠OAB = ∠OBA

∠OBA = 40º (angles opposite to equal sides are equal)

Using the angle sum property

∠AOB + ∠OBA + ∠BAO = 180º

Substituting the values

∠AOB + 40º + 40º = 180º

By further calculation

∠AOB + 80º = 180º

∠AOB = 180º - 80º

∠AOB = 100º

As the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle

∠AOB = 2 ∠ACB

Substituting the values

100º = 2 ∠ACB

Dividing both sides by 2

∠ACB = 50º

Therefore, ∠ACB is equal to 50º.

Given, ∠BAO = 60^∘
In ΔAOB
OA = OB
⇒ ∠OBA=∠BAO [angles opposite to equal sides are equal]
⇒ ∠BAO=∠OBA=60^∘

We know that, ∠ABC=∠ADC [ angles in the same segment AC are equal]
∴ ∠ADC=60^∘

1. The central angle subtended by the minor arc RP is:

   Minor arc RP =∠POQ =120^∘

2. The fraction of the circle represented by the minor arc RP is:

   Fraction=Central Angle / Total Angle of the Circle

   Substituting the values:

   Fraction = 120^∘ / 360^∘= 1 / 3

Since AD is the diameter, the angle subtended by the diameter at any point on the circle is always 90^∘. Therefore:

∠ACD = 90^∘

Given AE = DE, triangle ADE is isosceles. Therefore:

∠DAE =∠DEA

Since ABCD is a cyclic quadrilateral, opposite angles of a cyclic quadrilateral sum to 180^∘:

∠ABC + ∠ADC = 180^∘

Substituting ∠ABC=115^∘

∠ADC =180^∘−115^∘= 65^∘

In triangle ADE, since AE = DE, we have:

∠DAE=∠DEA

As ∠ADC=65^∘(external angle of triangle ADE), we know that:

∠DAE = 1 / 2 × 65^∘= 32.5^∘

Finding ∠CAE: Since ∠CAE=∠DAE, we find:

∠CAE = 70^∘

It is given that

∠DAB = 60º

∠ABD = 50º

As ∠ADB = ∠ACB … (1) [angles in same segment of a circle are equal]

In triangle ABD

Using the angle sum property

∠ABD + ∠ADB + ∠DAB = 180º

Substituting the values

50 + ∠ADB + 60 = 180º

By further calculation

∠ADB + 110 = 180

So we get

∠ADB = ∠ACB = 70º

Therefore, ∠ACB is equal to 70º.

In the figure, ABCD is a trapezium in which AD || BC and ∠B=70^o.

∵ AD || BC

∴∠A+∠B = 180^o (Sum of cointerior angles)

⇒ ∠A+70^o=180^o

⇒∠A = 180^∘−70^∘=110^∘

∴∠A=110^∘

But ∠A+∠C =180^∘ and ∠B+∠D=180^o

(Sum of opposite angles of a cyclic quadrilateral)

∴110^o+∠C = 180^o

⇒∠C = 180^o−110^o=70^o

and 70^o+∠D = 180^o

⇒∠D = 180^o−70^o=110^o

∴∠A=110^o, ∠C=70^o and ∠D=110^o

360^o - 130^o =230^o

Therefore, angle ABC =1/2reflex angle AOC

Angle ABC =1/2 230^o

So, angle ABC =115^o

In ∆ BDC and ∆ BAC
Angle BAC = BDC
(angle made on same segment BC)
Since ABC is making right angle (90)
So,
In ∆ABC
ABC +BAC+ACB=180
(angle sum property of triangle)
90+42+ACB=180
ACB=180-132
ACB=48^o

Use the Property for Angles Subtended at the Center: The total angle subtended by the arc AB at the center O is given by:

∠AOB = 2 × ∠CAB + 2 × ∠CBA

Substitute the Given Values:

∠AOB = 2 × 20^∘ + 2 × 35^∘ 

Simplify: ∠AOB = 40^∘+ 70^∘= 110^∘