Circles Test

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Circles Test - 22

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Weekly Quiz Competition

Question 1
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CP and CQ are tangents to a circle with centre O. ARB is another tangent touching the circle at R. If$$ CP = 11 cm, BC =7 cm$$, then the length BR is :

A
$$11cm$$

B
$$7cm$$

C
$$3cm$$

D
$$4cm$$

Solution

CQ and CP are the tangents to a circle
So, $$CQ = CP$$ (Tangents to a circle from a common point (C) are equal in length)
$$CP = CQ = 11$$

Also, $$BC + BQ = 11$$

$$\Rightarrow BC+7=11$$

$$\Rightarrow BQ = 4$$

So, $$ BR = BQ = 4$$. (Tangents to a circle from a common point (B) are equal in length)

Thus, option D will be the answer.
Question 2
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In Fig. 3, PQ and PR are the tangents to the circle with center O such that $$\angle QPR = 50^{\circ}.$$
Then $$\angle OQR$$ is equal to:

A
$$25^{\circ}$$

B
$$30^{\circ}$$

C
$$40^{\circ}$$

D
$$50^{\circ}$$

Solution

In quadrilateral PQOR, $$\angle P + \angle PQO +\angle PRO + \angle QOR = {360}^{\circ}$$

But $$\angle PQO = \angle PRO = {90}^{\circ}$$ (Tangent is perpendicular to radius at point of contact)
Thus, $${50}^{\circ} + 90^{\circ} + {90}^{\circ} + \angle QOR = {360}^{\circ}$$

So, $$\angle QOR = {130}^{\circ}$$

In triangle OQR, $$\angle OQR + \angle ORQ + \angle QOR = {180}^{\circ}$$

Hence, $$2. \angle OQR + \angle QOR = {180}^{\circ}$$
(because, $$\angle OQR = \angle ORQ$$; (since $$OQ = OR$$, radii ))

Thus, $$2.\angle OQR + {130}^{\circ} = \angle {180}^{\circ}$$

So, $$\angle{OQR}=25^{\circ}$$

Therefore option A is the answer.
Question 3
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In the given figure, the length of PR is :

A
20 cm

B
26 cm

C
24 cm

D
28 cm

Solution

Tangent from an external point makes right angle with radius at the point of contact.

$$\angle PQO' = 90^o \Rightarrow PO' = \sqrt{QO'^2 + PQ^2} $$ (by Pythagoras theorm)
$$PO'=\sqrt{3^2+4^2}$$
$$=\sqrt{9+16}$$
$$=\sqrt{25}=5\ cm$$

$$\angle OSR = 90^o \Rightarrow OR = \sqrt{OS^2 + SR^2} $$ (by Pythagoras theorm)
$$OR =\sqrt{5^2+12^2}$$
$$=\sqrt{25+144}$$
$$=\sqrt{169}=13\ cm$$

Distance between centers of circles touching externally is equal to the sum of their radii.

$$O'O = 3+5 = 8cm$$ (sum of radii)

$$\Rightarrow PR = PO' + O'O + OR = 5+8+13 = 26 cm$$.

Hence, option B is the right answer.
Question 4
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In the given figure, if PT is a tangent of the circle with center O and $$\angle TPO = 25^{\circ} $$, then the
the measure of x is :

A
$$120^{\circ}$$

B
$$125^{\circ}$$

C
$$110^{\circ}$$

D
$$115^{\circ}$$

Solution

Since tangent is perpendicular to radius at the point of contact, we have:
$$\angle PTO = 90^o$$.
Hence by the exterior angle formula applied to triangle $$OTP$$, we get:
$$x = 90 + 25 = 115^o$$.
So option D is the correct answer.
Question 5
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In above Fig. 2, $$AB$$, $$AC$$ and $$ PQ$$ are tangents. If $$AB = 5 \, \mathrm{cm}$$ and $$PQ= PB+QC$$, then perimeter of $$\Delta APQ$$ is:

A
$$10\ \mathrm{cm}$$

B
$$15\ \mathrm{cm}$$

C
$$12.5\ \mathrm{cm}$$

D
$$20\ \mathrm{cm}$$

Solution

Tangents drawn from the same external point are equal in length.
So, $$AB = AC = 5 \ \mathrm{cm}$$ and similarly
$$PQ = PB + QC$$
$$\Rightarrow AP + PQ + AQ = (AP + PB) + (AQ + QC) = AB + AC = 10 \ \mathrm{cm}$$
So the perimeter of $$ \triangle APQ$$ is $$10\ \mathrm{cm}$$
So option A is the right answer.
Question 6
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If the angle between two radii of a circle is $$140^{\circ}$$, then the angle between the tangents at the ends of the radii is :

A
$$90^{\circ}$$

B
$$40^{\circ}$$

C
$$70^{\circ}$$

D
$$60^{\circ}$$

Solution

Since tangents and radii are perpendicular at the point of contact, in the quadrilateral formed by the two radii and the tangents at their ends, we have two right angles at the two points of contacts.
Let the angle between the tangents be $$x$$. Then
$$140^{\circ} + 90^{\circ} + 90^{\circ} + x = 360 $$ (Sum of angles in a quadrilateral is $$360^{\circ}$$)
$$320^{\circ} + x = 360 $$
$$\Rightarrow x = 40^{\circ}$$
So option B is the right answer.
Question 7
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In the figure, if the semiperimeter of $$\triangle ABC = 23$$ cm, then AF + BD + CE is:

A
46 cm

B
11.5 cm

C
23 cm

D
34.5 cm

Solution

We know that, tangents drawn from an external point are equal in length. So
$$AF = AE, \ BF = BD, \ CD = CE$$.
Given that the semiperimeter of the triangle is $$23\ cm$$.
Therefore, the perimeter of $$\triangle ABC=AB + BC + CA = 2\times 23 = 46 \ cm$$.

$$AB + BC + CA = (AF + BF) + (BD + CD) + (AE + CE)$$

$$= (AF + BD + CE) + (AE + BF+ CD) $$

$$= 2(AF + BD + CE)$$

$$\Rightarrow AF + BD + CE = \dfrac{1}{2}\times 46 = 23 \ cm$$

Hence, option C is correct.
Question 8
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In the given quadrilateral $$OQPR,$$ if $$\angle QPR=50^\circ $$ then $$\angle QOR $$ equals

A
$$120^{\circ}$$

B
$$130^{\circ}$$

C
$$145^{\circ}$$

D
$$110^{\circ}$$

Solution

Tangent is perpendicular to the radius at the point of contact.
Therefore, $$\angle{OQP}=\angle{ORP}=90^{\circ}$$

Sum of all the four interior angles of a quadrilateral is $$360^{\circ}.$$ So,
$$\angle{PQO}+\angle{QOR}+\angle{ORP}+\angle{RPQ}=360^{\circ}$$
$$\Rightarrow 90^{\circ}+\angle{QOR}+90^{\circ}+50^{\circ}=360^{\circ}$$ [Given $$\angle{RPQ}=50^{\circ}$$]
$$\Rightarrow \angle{QOR}=130^{\circ}$$

Hence, option $$B$$ is correct.
Question 9
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In the given figure, a circle touches the side $$BC$$ of $$\triangle ABC$$ at $$P$$ and $$AB$$ and $$AC$$ produced at $$Q$$ and $$R$$ respectively. If $$AQ=15cm$$, find the perimeter of $$\triangle ABC$$.

A
$$30cm$$

B
$$24cm$$

C
$$15cm$$

D
$$18cm$$

Solution

$$ The\quad lengths\quad of\quad the\quad tangents,\quad from\quad a\quad point\quad to\quad a\quad circle,\\ are\quad equal.\\$$
Therefore AB = AP, BQ = BP, BP = CP
Perimeter = AC + AB + BC = 2(AB) + BP + PC = 2(AB) + 2(BQ) = 2(AQ) = 30 cm
Question 10
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To draw a pair of tangents to a circle which are inclined to each other at an angle of $$ 35^{\circ}$$, it is required to draw tangents at the end point of those two radii of the circle, the angle between them is:

A
$$ 70^{\circ}$$

B
$$ 110^{\circ}$$

C
$$ 140^{\circ}$$

D
$$ 145^{\circ}$$

Solution

PA and PB are tangents drawn from an external point P to the circle.

$$\angle{OAP} = \angle{OBP} = 90^{\circ}$$ (Radius is perpendicular to the tangent at point of contact)

In quadrilateral OAPB,

$$\angle{APB} + \angle{OAB} + \angle{AOB} + \angle{OBP} = 360^{\circ}$$

$$35^{\circ} + 90^{\circ} + \angle{AOB} + 90^{\circ} = 360^{\circ}$$

$$215^{\circ}$$ +$$ \angle{AOB}$$ = $$360^{\circ}$$

$$\angle{AOB} = 360^{\circ} – 280^{\circ} = 145^{\circ}$$

Thus, the angle between the two radii, OA and OB is $$145^{\circ}$$.