Test: Circles- 1 - Grade 9 MCQ

If arc AB = arc BC = arc CD
then , angle suspended on circle is also equal
So, Sum of all angle is equal to angle APD
Angle APB+ BPC + CPD = angle APD
19 + 19 +19 = 57^o

Given, ABCD is a parallelogram and AEFD is a cyclic quadrilateral.
∠BEF=80^∘
Now, ∠ADC=∠BEF=80^∘ (Angle of a cyclic quadrilateral is equal to the opposite exterior angle )
Also, now in parallelogram ABCD,
∠ABC=∠ADC=80^∘ (Opposite angles of a parallelogram are equal)

Because the circles are congruent, therefore their radius are same

Length of arc AXB = [θ/(360°)] × 2πr = [(75°)/(360°)] × 2πr

Length of arc A’YB’ = [θ/(360°)] × 2πr = [(25°)/(360°)] × 2πr

Required Ratio = ([(75°)/(360°)] × 2πr) : ([(25°)/(360°)] × 2πr) = 3 : 1

Let AB be the chord of the circle with center O

.

Given that AB = Radius of the circle.

Also, AO = BO = Radius

∴ ΔOAB is an equilateral triangle.

Thus, ∠AOB = ∠OBA = ∠OAB = 60°

Also, angle subtended by an arc at the center of the circle is twice the angle subtended by it at any other point in the remaining part of the circle.

∴ ∠AOB = 2∠ACB

⇒ ∠ACB = 1/2 (∠AOB)

⇒ ∠ACB = 1/2 (60°) = 30°

Topic in NCERT: Angle Subtended by a Chord at a Point

Line in NCERT: "A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc."

In the given figure, the points A, B, C, and D form a cyclic quadrilateral because they lie on the circumferences of the two intersecting circles.

The opposite angles of a cyclic quadrilateral are supplementary, i.e.,

It is given that ∠ABC=75^∘

Using the property of cyclic quadrilaterals:

∠DBA = ∠DCA = 58° …(1)

[Angles in same segment]  ABCD is a cyclic quadrilateral : 

Sum of opposite angles = 180 degrees 

∠A +∠C = 180° 

75° + ∠C = 180° 

∠C = 105° 

Again, 

∠ACB + ∠ACD = 105° 

∠ACB + 58° = 105° 

or ∠ACB = 47° …(2) 

Now,  ∠ACB = ∠ADB = 47° [Angles in same segment] 

Also, ∠D = 77° (Given) 

Again From figure, ∠BDC + ∠ADB = 77° 

∠BDC + 47° = 77° 

∠BDC = 30° 

In triangle DPC 

∠PDC + ∠DCP + ∠DPC = 180° 

30° + 58° + ∠DPC = 180° 

or ∠DPC = 92° 

According to the question, we have a quadrilateral ABCD whose all vertices are lying on the circumference of the circle. The measure of the angles ∠B and ∠D is given.
The measure of ∠B = 81^∘ + x ………………………………(1)
The measure of ∠D=89^∘ …………………………………(2)
Since a cyclic quadrilateral is a quadrilateral whose all vertices lie on the circumference of the circle and the quadrilateral ABCD has its all vertices on the circumference of the circle so, the quadrilateral ABCD is cyclic.
In the given cyclic quadrilateral ABCD, we have ∠B and ∠D opposite to each other.
We know the property of a cyclic quadrilateral that the opposite angles are supplementary. In other words, the summation of the opposite angles is equal to 180^∘

 .Using this property, we can say that the summation of the angles, ∠B and ∠D is equal to 180^∘

So, ∠B+∠D=180^∘  ………………………..(3)
From equation (1) and equation (2), we have the measure of the angles ∠B and ∠D that is  ∠B=81^∘+x and ∠D=89^∘

Now, putting the measure of ∠B and ∠D in equation (3), we get

⇒ ∠B+∠D=180^∘

⇒ 81^∘+ x + 89^∘ = 180^∘ 

Solving the above equation further to get the value of x,

⇒ x + 170^∘ = 180^∘

⇒ x = 180^∘−170^∘

⇒ x = 10^∘

As we know that the greatest chord of a circle is called its diameter.
Hence, option (d) is correct.

Topic in NCERT: Equal Chords and their Distances from the Centre

Line in NCERT: "The distance of the diameter, which is the longest chord from the centre? Since the centre lies on it, the distance is zero."

We are given:

• ∠AOB=140^∘
• PQ and RS are equal chords.
• OA and OB are perpendiculars on the chords PQ and RS, respectively.
• We need to find ∠PAB.

The angle subtended by a chord at the center is twice the angle subtended on the remaining part of the circle.

• The chord AB subtends ∠AOB=140^∘ at the center.
• Therefore, the angle subtended by AB at any point on the circle (e.g., ∠PAB) is:

Substitute the given value of ∠AOB:

Property of Angles in a Triangle: In △ABC\triangle ABC△ABC, the sum of all angles is 180^∘

Using the Property of Angles at the Center and the Circumference: The angle subtended by an arc at the center is twice the angle subtended at any point on the circumference.

Here, the angle subtended by arc AC at the circumference is ∠ACB=30^∘. Therefore, the angle subtended by the same arc at the center (∠AOC) is:

Whenever a chord is drawn in a circle two segments are formed. One is called the minor segment while the other is called the major segment.
The angle formed by the chord in the minor segment will always be obtuse.

AE is a straight line so,

angle ABD + angle EBD = 180°

angle ABD + 95° = 180°

angle ABD = 180° -  95°

angle ABD = 85°

so, it is interior angle of circle

So, angle AOD is double of angle suspended on the circle

angle AOD = 2ABD

angle AOD = 2(85°)

angle AOD = 170°

As we are given that, both the triangle are congruent which means their corresponding angles are equal.  Therefore, ∠AOB = AO’B = 50° Now, by degree measure theorem, we have  ∠APB = ∠AOB/2 = 25° 

Topic in NCERT: Angle in a semicircle is a right angle.

Line in NCERT: "Angle in a semicircle is a right angle."

1. Understanding Non-Collinear Points:
- Non-collinear points are points that do not all lie on the same straight line. For example, if we have three points A, B, and C, they form a triangle if they are non-collinear.

2. Circle through Two Points:
- If we take any two points, say A and B, an infinite number of circles can be drawn through these two points. This is because circles can be drawn with different radii and centers that still pass through points A and B.

3. Adding the Third Point:
- When we add a third point C, which is not on the line formed by A and B, we can only draw one unique circle that passes through all three points A, B, and C. This is because a circle is uniquely defined by three non-collinear points.

4. Conclusion:
- Therefore, the number of circles that can be drawn through three non-collinear points is exactly one.

Final Answer: The answer is (a) 1.

Understand the Geometry:

• ∠ABC and ∠BCA are angles of triangle △ABC.
• ∠BDC is an exterior angle of △ABC, subtending the same arc BC.

Angle Subtended by the Same Arc: In a cyclic quadrilateral, the angles subtended by the same arc on the opposite sides of the circle are related. Here:

• ∠ABC and ∠ADC are subtended by the arc AC.
• ∠ABC+∠ADC=180^∘ (cyclic quadrilateral property).

However, we focus on triangle ABC for simplicity.

Exterior Angle Property: The exterior angle ∠BDC of a triangle is equal to the sum of the two non-adjacent interior angles. Therefore, in △ABC:

∠BDC=∠CAB+∠BCA

Substituting the given values:

40^∘=50^∘+∠BCA

Solve for ∠BCA: Rearrange to find ∠BCA:

∠BCA=90^∘

In the circle, AB and CD are two chords which intersect each other at P at right angle i.e. ∠CPB=90^o.

∠CAB and ∠CDB are in the same segment.

∴ ∠CDB=∠CAB=x

Now, in ΔPDB, Ext. ∠CPB=∠D+∠DBP

⇒90^o = x+y (∵CD⊥AB)

Hence, x+y = 90^o

Angle ADC = Angle CBA(: AC is the same point that makes the angle ADC and angle CBA.)
43° = Angle CBA
In triangle ACB
Angle BAC + Angle ACB + Angle ABC = 180°
49° + Angle ACB +43° = 180°
180° - 92° = ∠ACB
88° = Angle ACB

The angle subtended by a chord (or arc) at the center of the circle is twice the angle subtended at the circumference of the circle.

Mathematically:

∠QOR = 2 × ∠QPR

Substituting ∠QPR=50^∘ into the formula:

∠QOR = 2 × 50^∘ = 100^∘

As per the given question:

We can see, OP is the radius of the circle to the tangent PT and OQ is the radius to the tangents TQ.

So, OP ⊥ PT and TQ ⊥ OQ

∴ ∠OPT = ∠OQT = 90°

Now, in the quadrilateral POQT, we know that the sum of the interior angles is 360°

So, ∠PTQ + ∠POQ + ∠OPT + ∠OQT = 360°

Now, by putting the respective values, we get,

⇒ ∠PTQ + 90° + 110° + 90° = 360°

⇒ ∠PTQ = 70°

(b) 50°
OA = OB
⇒ ∠OBA = ∠OAB = 40°
Now, ∠AOB = 180° - (40° + 40°) = 100°
∴ ∠ACB=1/2∠AOB=1/2×100°=50°

Step 1: Draw a diagram:

- Draw a circle and label it as O.

- Draw two chords AB and CD intersecting at point P inside the circle.

- Label the angles ∠DPB and ∠DAP.

Step 2: Identify the relationships:

- The angles formed by intersecting chords are related to each other.

- The angle subtended by an arc on one side of an intersecting chord is equal to the sum of the opposite angles formed by the intersecting chord.

Step 3: Identify the angles:

- ∠DPB = 88° (given)

- ∠DAP = 46° (given)

Step 4: Identify the opposite angles:

- ∠ABC and ∠CDP are opposite angles formed by the intersecting chord AB.

- ∠DPC and ∠BAP are opposite angles formed by the intersecting chord CD.

Step 5: Write the relationship equation:

- According to the relationship mentioned in Step 2, we can write the equation: ∠DPB + ∠DPC = ∠ABC + ∠BAP

Step 6: Substitute the given values:

- ∠DPB = 88° - ∠DPC = ∠DAP (opposite angles are equal)

- ∠DAP = 46°

- ∠ABC = ? (to be found)

- ∠BAP = ∠ABC (opposite angles are equal)

Substituting these values in the equation from Step 5, we get: 88° + ∠DAP = ∠ABC + ∠ABC

Step 7: Simplify the equation:

88° + 46° = 2∠ABC

134° = 2∠ABC

Step 8:

Solve for ∠ABC:

Divide both sides of the equation by 2:

134°/2 = ∠ABC

67° = ∠ABC