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

Calculation: 

As per the question,

PB = AB - AP

⇒ 20 - 12 = 8cm

As we know

PA × PB = PD ×  PC

⇒ 12 × 8 = PD × 16

⇒ PD = 6

Then CD = PD + PC 

⇒ 16 + 6 = 22 cm

∴ The correct option is 1

Given:

MO = 9 cm

ON = 5 cm

OQ = 6 cm

Formula Used:

In a circle, if two chords MN and PQ intersect at O, then: MO × ON = OQ × OP

Calculation:

Using the formula, MO × ON = OQ × OP:

⇒ 9 × 5 = 6 × OP

⇒ 45 = 6 × OP

⇒ OP = 45 / 6

⇒ OP = 7.5 cm

The value of OP is 7.5 cm.

Given:

Radius of the circle = 5√13 cm

Distance of the chord from the centre = 10 cm

Formula Used:

Length of the chord = 2 × √(radius² - distance from centre²)

Calculation:

Radius = 5√13 cm

Distance from centre = 10 cm

⇒ Length of the chord = 2 × √((5√13)² - 10²)

⇒ Length of the chord = 2 × √(325 - 100)

⇒ Length of the chord = 2 × √225

⇒ Length of the chord = 2 × 15

⇒ Length of the chord = 30 cm

The length of the chord is 30 cm.

Given:

A pair of straight lines from an external point F intersects a circle at A and B (FA

∠ACF = 50º

∠AFC = 30º

Calculation:

Given ∠ACF = 50º

In \(\triangle \)ACF:

∠ACF = 50º, ∠AFC = 30º

So, ∠CAF = 180º - (50º + 30º)   --- Sum of all angles of a triangle is 180º

In \(\triangle \)OAC:

∠OCA = 90º - ∠ACF ------ (Radius is normal to tangent)

∠OCA = 90º - 50º = 40º

∠OCA = ∠OAC   ---(Angle opposite to radius are equal in a triangle)

So, ∠OAC = 40º

∠AOC = 180º - (40º + 40º) = 100º

∠OAB = 180º - (∠OAC + ∠CAF) ---- (Sum of all angle on a straight line is 180º)

∠OAB = 180º - (40º + 100º)

∠OAB = 40º

In \(\triangle \)OAB:

∠OAB = ∠OBA    ---  (Angle opposite to radius are equal in a triangle)

So, ∠OBA = 40º

∠AOB = 180º - (∠OAB + ∠OBA)  --- (Sum of all angles of a triangle is 180º)

∠AOB = 180º - (40º + 40º) = 100º

The correct answer is option 3.

Given :

O is the centre of the circle

∠AOC =140°

Calculation:

O is the centre of the circle

Take a point D in remaining arc

∠AOC = 140°

∠AOC = 2∠ADC ( Angle formed by chord at centre = twice angle formed in same arc segment)

⇒ 140° =  2∠ADC

⇒ ∠ADC  = 70°

ABCD will be  a cyclic quadrilateral

Sum of opposite angles of cyclic quadrilateral = 180°

⇒ ∠ABC + ∠ADC= 180°

⇒ 70° + ∠ABC = 180°

⇒ ∠ABC = 110°

∴ Option 2 is the correct answer.

Given:

The angle subtended by arc ABC at the center of the circle = 138°.

Chord AB is produced to a point P, and we need to find ∠CBP.

Formula used:

The angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at the circle's circumference.

Calculation:

Let ∠AQC be the angle subtended by arc ABC at the circle's circumference.

⇒ ∠AOC = 2 × ∠AQC

⇒ ∠AQC = 1/2 × ∠AOC = 1/2 × 138° = 69°

Now, in the cyclic quadrilateral ABQC, the exterior angle ∠CBP is equal to the interior opposite angle ∠AQC.

⇒ ∠CBP = ∠AQC = 69°

∴ The measure of ∠CBP is 69°.

Given:

AB = 24 cm & CD = 18 cm

Distance between chords = 21 cm

Calculations:

Let AO and CO be the radius of circle,

In Δ APO,

AO² = AP² + OP² = 12² + x² = 144 + x²

In Δ CQO,

CO² = CQ² + OQ² = 9² + (21 - x)² = 81 + (21 - x)²

Since, AO and CO are the radius of the circle, So

AO = CO

⇒ AO² = CO²

⇒ 144 + x² =  81 + (21 - x)2

⇒ 144 + x2 =  81 + 21² + x² - 2 × 21 × x

⇒ 144 = 81 + 441 - 42x

⇒ 42x = 522 - 144

⇒ x = 378 / 42 = 9 cm

So, AO² = 144 + x² = 144 + 9² = 144 + 81 = 225

⇒ AO = √225 = 15 cm

Hence, the radius of circle = AO = 15 cm

∴ The correct answer is option (1).

Given:

The angle subtended by a chord on the major arc of a circle is 132°.

Concept used:

The angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.

Calculation:

The major arc for the AB chord is ACB.

∠AGB = 132° 

According to the concept,

∠AGB  = 2 × ∠ACB

⇒ ∠ACB = 132°/2

⇒ ∠ACB = 66°

The angle ACB is 66°

Given:

Length of chord on both circles is 6 cm and 18 cm.

Concept used:

Pythagoras theorem.

Calculation:

Let the radii of the smaller circle and the bigger circle be r1 and r2.

In ΔOAC 

OC2 = OA2 + AC2

⇒ r12 - AC2 = OA2    ....(1)

In ΔOAB

OB2 = OA2 + AB2

⇒ r22 - AB2 = OA2    ....(2)

From the first equation and the second equation,

r12 - 9 = r22 - 81

⇒ r22-  r12 = 81 - 9 = 72

∴ The required difference in the square is 72 cm2.

Given:

Bigger circle radius(R) = 20 cm

Smaller circle radius(r) = 16 cm

Formula Used:

(Hypotenuse)² = (Base)² + (Perpendicular)²

Calculations:

As BOC is an right-angled triangle at C

by Pythagoras theorem,

⇒ (BO)2 = (CO)2 + (BC)²

⇒ (20)² = (16)² + (BC)²

⇒ 400 = 256 + BC²

⇒ BC = \(\sqrt{400 - 256}\)

⇒ BC = \(\sqrt{144}\)

⇒ BC = 12 cm

When a line from centre fall on chord, It divide it into 2 equal parts

⇒ AB = 2BC = 2(12) = 24 cm

⇒ Hence, The length of the chord is 24 cm