Circle MCQ Quiz in मल्याळम - Objective Question with Answer for Circle - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

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°.

Calculation:

As the expression given is C₁C₂ = r₁ + r₂

C1C2 represents the distance between the centers of the circles.

r₁ represents the radius of the first circle

r₂ represents the radius of the second circle.

C1C2 = r1 + r2 represents that the distance between the centers of the circle is equal to the sum of the radius, which means the circle touches each other externally.

When the circle touches each other externally then there are only three tangents are possible.

i.e. two direct common tangents and one transverse tangent.

∴ The correct answer is Option 3.

Given:

DE is the chord of a circle.

KD = 9 cm ; DE = 7 cm

Concept used:

If a tangent and a secant are drawn from a common point outside the circle then,

Tengent² = line of secant × (line of sacant - chord) 

Calculation:

From tangent and secant relation:

⇒ KH² = KD × KE

⇒ KH² = 9 × 16

⇒ KH = √144 = 12 cm

∴ The correct answer is 12 cm. 

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:

O is the centre of this circle. Tangent drawn from a point P, touches the circle at Q.

PQ = 24 cm and OQ = 10 cm

Concept used:

1. If a tangent is drawn on a circle from an external point, then at the point of tangency, it is perpendicular to the radius.

2. In a right-angled triangle, Hypotenuse² = Base² + Height²

Calculation:

​According to the concept,

∠OQP = 90°

Hence, OP is the hypotenuse of ΔOQP which is a right-angled triangle.

Now, OP = \(\sqrt {24^2 + 10^2}\) = 26 cm

∴ The value of OP is 26 cm.

Given:

AB = 10 cm

CD = 24 cm

PQ = 7 cm

Formula used:

Pythagorean formula : 

H² = P² + B²

H = Hypotaneous, P = perpendicular, B = base

Calculations:

In Δ COQ,

CO² = CQ² + OQ²

⇒ r2 = 12² + OQ²  

⇒ r2 = 144 + OQ²     ---(1) 

In Δ AOP,

OP = OQ + PQ = OQ + 7

AO² = AP² + OP²

⇒ r2 = 5² + (7 + OQ)²

⇒ r2 = 25 + (49 + OQ² + 14OQ)     ---(2)

From equation (1) and (2):

144 + OQ² = 25 + (49 + OQ2 + 14OQ)

⇒ 144 + OQ2 = 25 + 49 + OQ² + 14OQ

⇒ 144 = 74 + 14OQ

⇒ 144 - 74 = 14OQ

⇒ 70 = 14 OQ

⇒ OQ = 70/14 = 5 cm

So, from equation (1),

⇒ r2 = 144 + OQ2 = 144 + 5² = 144 + 25 = 169

⇒ r = √169 = 13 cm

Hence, diameter of the circle = 2r = 2 × 13 = 26 cm

∴ The correct answer is option (1).

Given:-

Two chords AB and CD intersect each other at point P.

AB = 18 cm, PB = 6 cm, and CP = 4 cm

Formula used:-

If two chords AB and CD intersect each other at point P inside the circle,
then AP × PB = CP × PD

Calculation:-

AP = AB - PB

⇒ AP = 18 - 6 = 12 cm

According to the formula

AP × PB = CP × PD

⇒ 12 × 6 = 4 × PD

⇒ PD = 72/4

∴ The required answer is 18.

Given:

PQ is a diameter.

∠QOR = 100°

Concept used:

the angle subtended by the chord at the centre of the circle is twice

the angle subtended at any point on the circle's circumference. 

Calculation:

⇒ ∠QOR + ∠ROP = 180°

⇒ ∠ROP = 180 - 100 = 80°

∠ROP = 2 × ∠PSR

⇒ ∠PSR = 80/2 = 40°

∴ The correct answer is 40°.

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.

Given:

Radius of large circle = 25 cm

Radius of small circle = 4 cm

Formula used:

Direct common tangent = 2 × √(R × r)

Calculation:

Direct common tangent (AB) = 2 × √(R × r)

⇒ 2 × √(25 × 4)

⇒ 2 × 10 = 20 cm

∴ The correct answer is 20 cm.