Theorem on Tangents MCQ Quiz in मल्याळम - Objective Question with Answer for Theorem on Tangents - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Calculation

By the theorem,

LQ × LP = LY × LX

Let the length of XY be x.

⇒ 5 × 15 = 3 × (3 + x)

⇒ 25 = x + 3

⇒ x = 22

The length of XY is 22.

Given:

The radius of smaller circle = 3 cm

The radius of large circle = 6 cm

Formula used:

Direct common tangent = 2 × √(R × r)

Where, R = radius of large circle;

and r = radius of the small circle

Calculation:

Direct common tangent = 2 × √(R × r)

⇒ 2 × √(3 × 6)

⇒ 2 × 3 × √2 = 6√2 cm

∴ The correct answer is 6√2 cm.

Given:

PA and PB are tangents to the circle with center O such that ∠APB = 54°

Calculation:

We know that the radius and tangent are perpendicular at their point of contact.

So,

∠PAO = ∠PBO = 90°

So,

∠AOB = 360 - 54 - 180

⇒ 126°

Also AO = OB = Radius

So, ∠OAB = ∠OBA = 54/2

⇒ 27°

∴ The required answer is 27°

Given:

PT is a tangent to a circle, whose centre is O.

OP = 17 cm

OT = 8 cm

Formula used:

Pythagoras Theorem: In a right-angle triangle, the hypotenuse is equal to - 

OP² = OT² + TP²

Calculation:

This, the length of the tangent TP is equal to:

TP2 = OP2 - OT2  

TP2 = (17)2 - (8)2  

TP2 = 289 - 64

TP2 = 225 

TP = √225 = 15

∴ The length of the tangent segment PT is equal to 15 cm.

Given:

∠APB = 60º

Formula used:

In ΔOAP and ΔOBP:

OA = OB (radii of the circle)

AP = BP (length of tangents from an external point)

OP = OP (common side)

By SSS congruence, ΔOAP ≅ ΔOBP

∠POA = ∠POB

∠OPA = ∠OPB

Therefore, OP is the angle bisector of ∠APB and ∠AOB

Hence, ∠OPA = ∠OPB = 1/2 (∠APB )

= 1/2 × 60°

= 30°

Calculation:

By angle sum property of a triangle,

In ΔOAP

∠A + ∠POA + ∠OPA = 180°

OA ⊥ AP     (The tangent at any point of a circle is perpendicular to the radius through the point of contact.)

Therefore, ∠A = 90°

90° + ∠POA + 30° = 180°

120° + ∠POA = 180°

∠POA = 180° - 120°

∠POA = 60°

Therefore, Option (2) i.e 60° is the correct answer.

Given:

The diameters of the two circles are 12 cm and 20 cm

The distance between their centres is 16 cm

Calculation:

According to the diagram,

The two circles are touching each other externally

So, the number of common tangents will be 3

∴ The required answer is 3.

Given:

Radius of circle with center P = 7 cm.

Radius of circle with center Q = 3 cm.

Formula Used:

Ratio of division of line segment by external tangents = Ratio of radii of circles

Calculation:

Let the direct common tangents to the circles meet PQ extended at A.

Since the tangents are direct common tangents, they will divide the line segment PQ externally in the ratio of the radii of the circles.

Ratio of radii of the circles = 7:3

Therefore, A divides PQ externally in the ratio 7:3

The correct answer is option 3.

Concept used:

Explanation:

The largest chord in a circle is the diameter. This meets the circle at two points. The points of tangency lie on the two points where the diameter meets the circle.

∴ A circle can have 2 parallel tangents at most.

So, only one parallel tangent is possible for a given tangent.

Important points:

This diameter meets the circle in exactly two points. So there is no more space for a third or more parallel tangents

Calculation : 

∴ The correct answer is 3.