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

Given:

Two circles centered at P and Q intersect at two points C and D.

Line AB is tangent to both circles at A and B.

∠ADB = 68°

Calculations:

Drew segment CD,

∠CAB = ∠ADC ---- (1) Tangent secant angle theorem.

∠CBA = ∠CDB ---- (1) Tangent secant angle theorem.

From (1) and (2),

∠CAB + ∠CBA = ∠ADC + ∠CDB

Now, ∠ADB = ∠ADC + ∠CDB --- (3)

In, ΔACB,

∠CAB + ∠CBA + ∠ACB = 180^∘  (Sum of angle of a triangle)

∠ADC + ∠CDB + ∠ADB = 180^∘   from (1) and (2),

∠ADB + ∠ACB = 180^∘   from equation (3)

68^∘ + ∠ACB = 180∘

∠ACB = 180∘ - 68^∘

∴ ∠ACB = 112°

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:

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:

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:

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

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 data:

AB = 14 cm

OP = 24 cm

Concept used:

Chord theorem circle states that if the radius of the circle is perpendicular to the chord of the circle, it bisects it and vice-versa.

In the circle, OA is the radius that bisects the chord AB perpendicularly.

Calculations:

In a right-angled triangle OPA,

AP² + OP² = OA²  (From Pythagoras theorem)

Substituting the value of OP and AP

⇒ OA² = 7² + 24² 

⇒ OA2 = 625

⇒ OA = 25

The radius is 25 cm.

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.