ML Aggarwal Solution Class 9 Chapter 15 Circle MCQs

 MCQs

Choose the correct answer from the given four options (1 to 6) :

Question 1

If P and Q are any two points on a circle, then the line segment PQ is called a

(a) radius of the circle

(b) diameter of the circle

(c) chord of the circle

(d) secant of the circle

Sol :

chord of the circle (c)


Question 2

If P is a point in the interior of a circle with centre O and radius r, then

(a) OP = r

(b) OP > r

(c) OP ≥ r

(d) OP < r

Sol :

OP > r (b)


Question 3

The circumference of a circle must be

(a) a positive real number

(b) a whole number

(c) a natural number

(d) an integer

Sol :

A positive real number (a)


Question 4

AD is a diameter of a circle and AB is a chord. If AD = 34 cm and AB = 30 cm, then the distance of AB from the centre of circle is

(a) 17 cm

(b) 15 cm

(c) 4 cm

(d) 8 cm

Sol :

AD is the diameter of the circle whose length is AD=34 cm

AB is the chord of the circle whose length is AB=30 cm







Distance of the chord from the centre is OM

Since the line through the centre of the chord of the circle is the perpendicular bisector, 

We have ∠OMA=90°

and AM=BM

Thus, ΔAMO is a right angled triangle


Now, by applying Pythagoras theorem

⇒OA2=AM2+OM2...(i)

Since the diameter AD=34 cm, radius of the circle is 17 cm

∴OA=17 cm

Since AM=BM and AB=30 cm

∴We have AM=BM=15 cm

We have, OA2=AM2+OM2

⇒(17)2=(15)2+OM2

⇒OM2=289-225

⇒OM2-64

⇒OM=√64=8

Ans (d)


Question 5

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is

(a) 6 cm

(b) 8 cm

(c) 10 cm

(d) 12 cm

Sol :
⇒Given that AB=12 cm and BC=16 cm and ∠ABC=90°







⇒Every angle inscribed in a semicircle is a right angle

⇒Since the inscribed angle 

⇒∠ABC=90° , the arc ABC is a semicircle

⇒Thus , AC is the diameter of the circle passing through the centre

⇒Now by Pythagoras theorem

⇒AC2=AB2+BC2

=122+162

=14+256=400

⇒AC=∠400=20 cm

∴Diameter of the circle is 20 cm

Thus, the radius of the circle passing through 

A,B and C is 10 cm

Ans (c)


Question 6

In the given figure, O is the centre of the circle. If OA = 5 cm, AB = 8 cm and OD ⊥ AB, then length of CD is equal to

(a) 2 cm

(b) 3 cm

(c) 4 cm

(d) 5 cm

Sol :











$\mathrm{OC}=\sqrt{\mathrm{AO}^{2}-\mathrm{AC}^{2}}$

$=\sqrt{25-16}=\sqrt{9}$ cm

=3 cm

Sicne , OD=OA=5 cm

∴CD=OD-OC=5-3 cm=2 cm

Ans (a)

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