7 2 Circles Quiz Review

Circles Questions are given here to help the students understand all basic concepts related to circles, and improve their problem-solving skills on applications of unlike theorems on circles. In this article, y'all will detect questions on circles, along with solutions for each of these questions. Students can also practice the additional questions provided at the stop of the article.

What is a circle?

In geometry, a circumvolve is the collection of all the points in a plane, which are at a stock-still altitude from a stock-still bespeak in the plane. The fixed point is called the centre, and the fixed distance is called the radius of the circle.

Larn more about circles here.

Circles Questions and Answers

1. If arcs AXB and CYD of a circle are congruent, find the ratio of AB and CD.

Solution:

Given that arcs AXB and CYD of a circle are congruent,

i.e. arc AXB ≅ arc CYD.

Circles Questions Q1

Nosotros know that if two arcs of a circumvolve are coinciding, their corresponding chords are also equal.

i.eastward. chord AB = chord CD

Thus, AB/CD = ane

AB/CD = 1/one

AB : CD = 1 : one

ii. In the figure, AOC is the bore of the circle and arc AXB = (ane/two)arc BYC. Find ∠BOC.

Solution:

Given,

arc AXB = (1/two) arc BYC

∠AOB = (ane/2) ∠BOC

Also, ∠AOB + ∠BOC = 180º

Therefore, (i/2) ∠BOC + ∠BOC = 180º {linear pair since AOC is the diameter}

(3/2) ∠BOC 180º

∠BOC = (two/three) × 180º = 120º

three. A circular park of a radius of xx m is situated in a colony. Three boys Ankur, Syed and David are sitting at an equal distance on its boundary, each having a toy telephone in his hands to talk to each other. Find the length of the string of each telephone.

Solution:

Let A, B, and C be the positions of Ankur, Syed and David, respectively.

As they are sitting at equal distances, the triangle is equilateral

Ad ⊥ BC is drawn.

Advertising is the median of ΔABC, and it passes through the centre O.

O is the centroid of the ΔABC. OA is the radius of the triangle.

OA = 2/3 AD

Let a m be the side of the triangle.

So, BD = a/2 k

In ΔABD,

Past Pythagoras theorem,

ABⁱⁱ = BD² + Ad²

⇒ AD^two = AB² – BD²

⇒ AD² = a² – (a/2)²

⇒ ADⁱⁱ = 3a²/4

⇒ AD = √3a/two

OA = 2/three AD

20 1000 = (⅔) × (√3a/2)

⇒ a = 20√3 g

Therefore, the length of the string of the toy is 20√3 m.

iv. In the effigy, ABCD is a circadian quadrilateral in which Air conditioning and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.

Solution:

Given,

ABCD is a circadian quadrilateral in which Air-conditioning and BD are its diagonals.

∠DBC = 55° and ∠BAC = 45°

∠CAD = ∠DBC = 55° (Angles in the same segment)

Thus, ∠DAB = ∠CAD + ∠BAC

= 55° + 45°

= 100°

We know that the opposite angles of a circadian quadrilateral are supplementary.

∠DAB + ∠BCD = 180°

∠BCD = 180° – 100° = fourscore°

Tangent to a Circle:

The tangent to a circle is perpendicular to the radius through the point of contact.
The lengths of the two tangents from an external point to a circumvolve are equal.

v. The length of a tangent from point A at a altitude of 5 cm from the heart of the circumvolve is iv cm. Observe the radius of the circle.

Solution:

Let O be the eye of the circle and AB is the tangent.

Circles Questions A5

OB⊥ AB and then ∆OAB is a right-angled triangle

OA² = AB² + OB² (by Pythagoras Theorem)

v^two = 4^two + OBⁱⁱ

OBⁱⁱ = 5^two – 4ⁱⁱ

= 25 – 16

OB² = ix

OB = √9 = 3

Hence, the radius of the circle = three cm.

6. Out of the two concentric circles, the radius of the outer circle is 5 cm and the chord Air-conditioning of length eight cm is a tangent to the inner circle. Observe the radius of the inner circle.

Solution:

Allow C_i and C₂ be the two circles having the same middle O.

AC is the chord that touches the C₁ at indicate D.

Join OD.

Circles Questions A6

Also, OD is perpendicular to Air-conditioning

∴ AD = DC = four cm [ Perpendicular line OD bisects the chord]

In correct ΔAOD,

OA² = Advertⁱⁱ + Do^two {by Pythagoras theorem}

⇒ DO² = 5² − 4²

= 25 − 16

= 9

⇒ Do = three

Therefore, the radius of the inner circle = iii cm.

7. AB is a bore, and Air-conditioning is a chord of a circumvolve with center O such that ∠BAC = xxx°. The tangent at C intersects extended AB at a indicate D. Testify that BC = BD.

Solution:

Given:

O is the middle of a circle, AB is the diameter, and AC is the chord Ac such that ∠BAC=30°.

A tangent from C meets AB at D on producing.

Join BC and OC.

Circles Questions A7

∠BCD = ∠BAC = 30° (Angles in the alternate segment)

Arc BC subtends ∠Doc at the centre of the circumvolve and ∠BAC at the remaining part of the circumvolve.

∴∠BOC. =two∠BAC = 2 × xxx° = 60°

In triangle OCD,

∠BOC or ∠DOC = threescore° (Proved above)

∠OCD = 90° (∵ OC ⊥ CD)

∴∠Doc +∠ODC = 90°

⇒ 60° + ∠ODC = 90°

∴ ∠ODC = 90° − 60° = 30∘

In triangle BCD,

∠ODC or ∠BDC =∠BCD = 30°

∴BC = BD

Hence proved.

eight. If an isosceles triangle ABC, in which AB = Air conditioning = 6 cm, is inscribed in a circle of radius 9 cm, notice the area of the triangle.

Solution:

Given:

In a circle, ΔABC is inscribed in which AB = AC = 6 cm and the radius of the circle is nine cm

To Notice: Expanse of triangle ABC

Construction: Join OB and OC

Circles Questions 8A

Proof:

In △AOB and △AOC,

OB = OC [Radii of same circle]

AO = AO [Common]

AB = AC = 6 cm [Given]

△AOB ≅ △AOC [By SSS congruence criterion]

∠OAB = ∠OBC [Corresponding parts of congruent triangles are equal ]

⇒ ∠MAB = ∠MBC

In △ABM and △AMC,

AB = Air conditioning = half dozen cm [Given]

AM = AM [Mutual]

∠MAB = ∠MBC [Proved in a higher place]

△ABM ≅ △AMC [By SAS congruence criterion]

∠AMB = ∠AMC [Corresponding parts of congruent triangles are equal ]

Now,

∠AMB + ∠ AMC = 180°

∠AMB + ∠AMB = 180°

2 ∠AMB = 180

∠AMB = ninety°

Nosotros know that a perpendicular from the heart of the circle bisects the chord.

So, OA is the perpendicular bisector of BC.

Let OM = 10

AM = OA – OM = 9 – x [ As OA = radius = ix cm]

In right angled ΔAMC,

(AM)² + (MC)² = (Air-conditioning)ⁱⁱ {By Pythagoras theorem}

(nine – x)² + (MC)² = (6)^two

81 + 10²– 18x + (MC)^two = 36

(MC)² = 18x – x² – 45….(i)

In right △OMC ,

(MC)² + (OM)² = (OC)ⁱⁱ {By Pythagoras Theorem}

18x – ten² – 45 + (x)ⁱⁱ = (ix)²

18x – 45 = 81

18x = 126

x = vii

AM = 9 – 10 = 9 – 7 = 2 cm

In right △AMC,

(AM)² + (MC)ⁱⁱ = (Air conditioning)² {Past Pythagoras Theorem}

(2)² + (MC)² = (6)ⁱⁱ

four + (MC)² = 36

(MC)² = 32

MC = four√two cm

Likewise, MC = BM

BC = 2MC = 2(four√2) = 8√two cm

Area of ΔABC = (1/2) × BC × AM

= (i/2) × 8√2 × ii

= 8√2 cm²

9. Prove that the tangents fatigued at the ends of a diameter of a circle are parallel.

Solution:

Showtime, draw a circle and connect 2 points A and B such that AB becomes the bore of the circle.

Now, depict two tangents PQ and RS at points A and B, respectively.

Circles Questions 9A

Now, both radii i.east. AO and OB are perpendicular to the tangents.

And so, OB is perpendicular to RS and OA perpendicular to PQ

And then, ∠OAP = ∠OAQ = ∠OBR = ∠OBS = ninety°

From the to a higher place figure, angles OBR and OAQ are alternating interior angles.

As well, ∠OBR = ∠OAQ and ∠OBS = ∠OAP {since they are also alternate interior angles}

So, we can say that the line PQ and the line RS will exist parallel to each other.

Hence proved.

10. If two circles intersect at two points, prove that their centres prevarication on the perpendicular bisector of the common chord.

Solution:

Given

Permit circle C₁ have centre O and circumvolve C_two have centre O'.

PQ is the common chord.

Circles Questions 10A

To prove:

OO' is the perpendicular bisector of PQ i.e.

1. PR = RQ

2. ∠PRO = ∠PRO' = ∠QRO = ∠QRO' = ninety°

Construction:

Join PO, PO', QO, QO'

Proof

In △POO' and △QOO',

OP = OQ (Radius of circle C1)

O'P = O'Q (Radius of circumvolve C2)

OO' = OO' (Common)

∴ △POO' ≅ △QOO' (by SSS Congruence rule)

∠POO' = ∠QOO' (CPCT) —-(1)

Besides,

In △POR and △QOR,

OP = OQ (Radius of circle C1)

∠POR = ∠QOR ( From (one))

OR = OR (Common)

∴ △OPO' ≅ △OQO' (by SAS Congruence Dominion)

PR = QR (CPCT)

& ∠PRO = ∠QRO (CPCT) —-(2)

Since PQ is a line

∠PRO + ∠QRO = 180° (linear Pair)

∠PRO + ∠PRO= 180° ( From (2))

2∠PRO = 180°

∠PRO = 180° / ii

∠PRO = 90°

Therefore,

∠QRO = ∠PRO = ninety°

Also,

∠PRO' = ∠QRO = xc° (vertically reverse angles)

∠QRO' = ∠PRO = 90° (vertically opposite angles)

Since, ∠PRO = ∠PRO' = ∠QRO = ∠QRO' = 90°

∴ OO' is the perpendicular bisector of PQ.

Hence proved.

Video Lessons on Circles

Introduction to Circles

Parts of a Circle

Surface area of a Circumvolve

All about Circles

Practice Questions on Circles

ABCD is a circadian quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Farther, if AB = BC, find ∠ECD.
Two concentric circles are of radii v cm and three cm. Notice the length of the chord of the larger circumvolve which touches the smaller circle.
Ii circles with centres O and O′ intersect at two points A and B. A line PQ is drawn parallel to OO′ through A(or B) intersecting the circles at P and Q. Prove that PQ = 2 OO′.
A is a point at a distance of 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circumvolve at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, discover the perimeter of the ∆ABC.
A chord of a circle is equal to the radius of the circle. Detect the angle subtended past the chord at a point on the small-scale arc, and also at a point on the major arc.