Circles Test

Result

Circles Test - 12

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

The length of tangent PQ, from an external point Q, is 24 cm. If the distance of the point Q from the center is 25 cm, then the diameter of the circle is

A
15 cm.

B
12 cm

C
7 cm

D
14 cm

Solution

Here \(\angle {OPQ}=90^{\circ}\) [Angle between tangent and radius through the point of contact]
\(\therefore {OQ}^{2}={OP}^{2}+{PQ}^{2} \Rightarrow(25)^{2}={OP}^{2}+(24)^{2}\)
\(\Rightarrow {OP}^{2}=625-576 \Rightarrow {OP}^{2}=49\)
\(\Rightarrow {OP}=7{cm}\)
Therefore, the diameter \(=2 \times {OP}=2 \times 7=14 {cm}\)
Hence, the correct option is (D).
Question 2
1 / -0

In the given figure, RQ is a tangent to the circle with center O. If SQ = 6 cm and QR = 4 cm, then OR is equal to

A
12 cm

B
5 cm

C
2.5 cm

D
10 cm

Solution

Here \(SQ=6 cm\), then \(OQ=OS=\frac62=3cm\) [Radii]
Now, in triangle \(OQR, OR^2=QR^2+OQ^2\)
\(\Rightarrow {OR}^{2}=(4)^{2}+(3)^{2}=16+9=25\)
\(\Rightarrow {OR}=5 {cm}\)
Hence, the correct option is (B).
Question 3
1 / -0

In the figure, PT is tangent to the circle with center O such that OP is 4 cm and ∠OPT=30^o length of the tangent is given by :

A
\(4\sqrt3\) cm

B
\(5\) cm

C
\(2\sqrt3\) cm

D
\(7\) cm

Solution

In right angled triangle \(OPT\),
\(cos30^{\circ}= \frac{PT}{PO} ⇒\frac{3}{\sqrt2} =\frac{ PT}{4}⇒ PT = 2\sqrt{3} cm\)
Hence, the correct option is (C).
Question 4
1 / -0

If PT is a tangent to the circle with center O, then x + y is equal to

A
90^∘

B
75^∘

C
60^∘

D
100^∘

Solution

Here ∠T = 90^∘ [Angle between tangent and radius through the point of contact]
Now, in triangle OPT, we know that
∠O + ∠P + ∠T = 180^∘
[Angle sum property of a triangle]
⇒x + y + 90^∘=180^∘
⇒x + y = 180^∘− 90^∘
⇒x + y = 90^∘
Hence, the correct option is (A).
Question 5
1 / -0

A tangent PQ at a point P o a circle of radius 5 cm meets a line through the center O at point Q, so that OQ = 12 cm. find the length PQ.

A
\(13\) cm

B
\(\sqrt{113}\) cm

C
\(\sqrt{119}\) cm

D
\(26\) cm

Solution

\(\angle {OPQ}=90^{\circ}\) [Angle between tangent and radius through the point of contact]
\(\therefore {OQ}^{2}={OP}^{2}+{PQ}^{2} \Rightarrow(12)^{2}=(5)^{2}+{PQ}^{2}\)
\(\Rightarrow {PQ}^{2}=144-25 \Rightarrow {PQ}^{2}=119\)
\(\Rightarrow {PQ}=\sqrt{119}\)
Hence, the correct option is (C).
Question 6
1 / -0

If the angle between two radii of a circle is 130^o, the angle between tangents at ends of radii is :

A
70^∘

B
90^∘

C
50^∘

D
60^∘

Solution

If angle between two radii of a circle is \(130^{\circ}\), the angle between tangents at ends of radii is \(\angle\) \(\mathrm{APB}=50^{\circ} .\) Because the angle between the two tangents drawn from an external point to a circle is supplementary of the angle between the radii of the circle through the point of contact.

Hence, the correct option is (C).
Question 7
1 / -0

In the given figure, the measure of ∠OQPis

A
40^∘

B
60^∘

C
90^∘

D
35^∘

Solution

Here ∠OPB = 90^∘ [Angle between tangent and radius through the point of contact]
⇒∠OPQ + ∠QPB = 90^∘
⇒∠OPQ + 50^∘ = 90^∘
⇒∠OPQ = 40^∘ But ∠OPQ = ∠OQP
[Angle opposite to equal radii]
∴∠OQP = 40^∘
Hence, the correct option is (A).
Question 8
1 / -0

In the adjoining figure, the measure Of PR is

A
10 cm

B
12 cm

C
18 cm

D
16 cm

Solution

Here ∠Q = 90^∘ [Angle between tangent and radius through the point of contact]
Now, in triangle OPQ, OP² = QO² + PQ²
⇒OP²= (6)² + (8)² = 36 + 64 = 100
⇒ OP = 10 cm
∴PR = OP + OR = OP + OQ [OR = OQ = Radii]
⇒ PR = 10 + 6 = 16 cm
Hence, the correct option is (D).
Question 9
1 / -0

PT is a tangent of the circle with center O and ∠TPO = 25^∘, then the measure of x is

A
100^∘

B
150^∘

C
115^∘

D
140^∘

Solution

Here ∠T = 90^∘ [Angle between tangent and radius through the point of contact]
Now, in triangle OPT, we know that
∠O + ∠P + ∠T = 180^∘ [Angle sum property of a triangle]
⇒∠O + 25^∘+ 90^∘= 180^∘
⇒∠O = 180^∘ − 115^∘ = 65^∘
Now, x + ∠TOP = 180^∘ [Linear pair]
⇒x + 65^∘ = 180^∘
⇒x = 180^∘−65^∘=115^∘
Hence, the correct option is (C).
Question 10
1 / -0

In the figure given alongside the length of PR is

A
20 cm

B
28 cm

C
26 cm

D
24 cm

Solution

Here ∠Q = 90^∘ and ∠S = 90^∘[Angle between tangent and radius through the point of contact]
Now, in triangle OPQ, OP² = OQ² + QP²
⇒OP² = (3)² + (4)²
⇒OP² = 16 + 9 = 25
⇒OP = 5 cm
Again in triangle RSO’,
O'R² = O'S² + RS²
⇒O'R² = (5)²+ (12)²
⇒O'R² = 25 + 144 = 169
⇒O’R = 13 cm
∴ PR = OP + OQ + O’S + O’R = 5 + 3 + 5 + 13 = 26 cm
Hence, the correct option is (C).