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Math Olympiad Test: Circles- 4 - Class 10 MCQ


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10 Questions MCQ Test Olympiad Preparation for Class 10 - Math Olympiad Test: Circles- 4

Math Olympiad Test: Circles- 4 for Class 10 2024 is part of Olympiad Preparation for Class 10 preparation. The Math Olympiad Test: Circles- 4 questions and answers have been prepared according to the Class 10 exam syllabus.The Math Olympiad Test: Circles- 4 MCQs are made for Class 10 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Math Olympiad Test: Circles- 4 below.
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Math Olympiad Test: Circles- 4 - Question 1

Let s denote the semiperimeter of a triangle ABC in which BC = a, CA = b, AB = c. If a circle touches the sides BC, CA, AB at D, E, F, respectively, find BD.

Detailed Solution for Math Olympiad Test: Circles- 4 - Question 1

According to question,
s = (a + b + c/2) ⇒ 2s = a + b + c
B is external point and BD and BF are tangents and from an external point the tangents drawn to a circle are equal in length
∴ BD = BF;
Similarly, AF = AE; CD = CE
s = Semi perimeter = 
⇒ 2s = AB + AC + BC
⇒ 2s = AF + FB + AE + EC + BD + DC
⇒ 2s = 2AE + 2CE + 2BD
⇒ s = AE + CE + BD
⇒ s = AC + BD ⇒ s - b = BD.

Math Olympiad Test: Circles- 4 - Question 2

Two concentric circles of radii a and b, where a > b, are given. The length of a chord of the larger circle which touches the other circle is

Detailed Solution for Math Olympiad Test: Circles- 4 - Question 2

In figure, AB is a chord of circle C1 which is a tangent to C2.
Since, tangent is perpendicular to radius through point of contact
∴ ∠OCA = 90° ⇒ OA = a, OC = b
In ∆OCA, (OA)2 = (OC)2 + (AC)2
⇒ a2 = b2 + (AC)2 ⇒ AC = 
∴ Length of chord AB = 2AC = 2 

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Math Olympiad Test: Circles- 4 - Question 3

How many tangents can a circle have?

Detailed Solution for Math Olympiad Test: Circles- 4 - Question 3

A tangent to a circle is a line that intersects the circle at only one point. On every point on the circle, one tangent can be drawn as shown in the figure below.

As per the above diagram, we see that a circle can have infinitely many tangents.

Math Olympiad Test: Circles- 4 - Question 4

In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If ∠PBT = 30°, then BA : AT is

Detailed Solution for Math Olympiad Test: Circles- 4 - Question 4

∠BPA = 90° (Angle in semicircle)
In ∆ BPA, ∠ABP + ∠BPA + ∠PAB = 180°
⇒ 30° + 90° + ∠PAB = 180°
⇒ ∠PAB = 60°
Also, ∠POA = 2∠PBA
⇒ ∠POA = 2 × 30° = 60°
⇒ OP = AP ...(i)
(side opposite to equal angles)

In ∆OPT, ∠OPT = 90°
∠POT = 60° and ∠PTO = 30° [angle sum property of a D]
Also ∠APT + ∠ATP = ∠PAO [exterior angle property]
∴ ∠APT + 30°= 60° ⇒ ∠APT = 30°
∴ AP = AT ...(ii) (side opposite to equal angles)
From (i) and (ii), AT = OP = radius of the circle; and AB = 2r
⇒ AB = 2AT ⇒ AB/AT = 2 ⇒ AB : AT = 2 : 1

Math Olympiad Test: Circles- 4 - Question 5

In the given figure, A and B are the centres of two circles that intersect at X and Y. PXQ is a straight line. If reflex angle QBY = 210°, find obtuse angle PAY.

Detailed Solution for Math Olympiad Test: Circles- 4 - Question 5

In quadrilateral YBQX
∠YXQ = 1/2 reflex ∠QBY = 1/2 x 210° = 105°
PQ is a straight line
⇒ ∠PXY + ∠YXQ = 180° (linear pair)
⇒ ∠PXY = 180° - 105° = 75°
Now, in quadrilateral PAYX
∠PXY = 1/2∠PAY
∴ ∠PAY = 2∠PXY = 150°

Math Olympiad Test: Circles- 4 - Question 6

If O is the centre of a circle, AOC is its diameter and B is a point on the circle such that ∠ACB = 50°. If AT is the tangent to the circle at the point A, then ∠BAT =

Detailed Solution for Math Olympiad Test: Circles- 4 - Question 6

∠ABC = 90° [Angle in a semicircle ]
In ∆ ABC, we have
∠ACB + ∠CAB + ∠ABC = 180°

⇒ 50° + ∠CAB + 90° = 180°
⇒ ∠CAB = 40°
Now, ∠CAT = 90° ⇒ ∠CAB + ∠BAT = 90°
⇒ 40° + ∠BAT = 90°⇒ ∠BAT = 50°

Math Olympiad Test: Circles- 4 - Question 7

Match the columns.

Detailed Solution for Math Olympiad Test: Circles- 4 - Question 7

(a) Draw OC perpendicular on AB,
In Δ OBC & Δ OAC
∠OCB = ∠OCA (each 90°)
OB = OA (radii of same circle)

OC is common side in both triangles
∴ ΔOBC ≅ ΔOAC (RHS congruence)

Now, in ΔOCA (OA)2 = (OC)2 + (CA)2

Hence, radius of circle = 5 cm
(b) In ΔOPT,
(OP)2 = (OT)2 + (PT)2 
⇒ (x + 4)2 = x2 + (8)2
⇒ x2 + 16 + 8x = x2 + 64
⇒ 8x = 48 ⇒ x = 6 cm
(c) PQ = 10 cm
We know, length of tangents drawn from an external point to a circle are equal.
⇒ PQ = PR ...(i)
Also, SQ = SU ...(ii)
and TU = TR ...(iii)
Now, perimeter of DPST
= PS + ST + PT = PS + SU + UT + PT
= PS + SQ + TR + PT (Using (ii) & (iii))
= PQ + PR = PQ + PQ (Using (i))
= 2 PQ = 2 × 10 = 20 cm.

Math Olympiad Test: Circles- 4 - Question 8

AB is a chord of length 24 cm of a circle of radius 13 cm. The tangents at A and B intersect at a point C. Find the length AC.

Detailed Solution for Math Olympiad Test: Circles- 4 - Question 8

Given, Chord AB = 24 cm, Radius OB = OA = 13 cm
Draw OP ⊥ AB
In D OPB, OP ⊥ AB ⇒ AP = PB
[Perpendicular from centre on chord bisect the chord] =(1/2)AB= 12
Also, OB2 = OP2 + PB2
⇒ (13)2 = OP2 + PB2 ⇒ 169 = OP2 + (12)2
⇒ OP2 = 169 - 144 = 25 ⇒ OP = 5 cm
In Δ BPC, BC2 = x2 + BP2 [By pythagoras theorem]
BC2 = x2 + 144 ...(i)
In ΔOBC, OC2 = OB2 + BC2
⇒ (x + 5)2 = (13)2 + BC2 ⇒ x = 288/10=28.8cm

Put the value of x in (i), we get
BC2 = x2 + 144 = ((144)2/25) + 144⇒ BC = 31.2
⇒ AC = BC = 31.2 cm

Math Olympiad Test: Circles- 4 - Question 9

In the given figure, O is the centre of a circle; PQL and PRM are the tangents at the points Q and R respectively and S is a point on the circle such that ∠SQL = 50° and ∠SRM = 60°. Then, find ∠QSR and ∠RPQ.

Detailed Solution for Math Olympiad Test: Circles- 4 - Question 9

Since, PQL is a tangent and OQ is a radius, so ∠OQL = 90°
∴ ∠OQS = (90°- 50°) = 40°
Now, OQ = OS ⇒ ∠OSQ = ∠OQS = 40°
Similarly, ∠ORS = (90°- 60°) = 30°
And, OR = OS ⇒ ∠OSR = ∠ORS = 30°
∴ ∠QSR = ∠OSQ + ∠OSR = (40°+ 30°) = 70°
Now, ∠ROQ = 2 ∠QSR = 140°
∠ROQ + ∠ORP + ∠OQP + ∠RPQ = 360° (Angle sum property of quadrilateral QORP)
⇒ 140° + 90° + 90° + ∠RPQ = 360°
⇒ ∠RPQ = 40°

Math Olympiad Test: Circles- 4 - Question 10

The radii of two concentric circles are 13 cm and 8 cm. AB is a diameter of the the bigger circle. BD is a tangent to the  smaller circle touching it at D. Find the length AD.

Detailed Solution for Math Olympiad Test: Circles- 4 - Question 10

Produce BD to meet the bigger circles at E. Join AE. Then
∠AEB = 90° [Angle in a semicircle]
OD ⊥ BE
[∵ BE is tangent to the smaller circle at D and OD is its radius] BD = DE [∵ BE is a chord of the circle and OD ^ BE]
∴ OD || AE [∵ ∠AEB = ∠ODB = 90°]
In ΔAEB, O and D are mid-points of AB and BE. Therefore, by mid-point theorem, we have

OD = 1/2AE
⇒ AE = 2 × 8 = 16 cm
In ΔODB, we have
OB2 = OD2 + BD2 [By Pythagoras Theorem]
⇒ 169 = 82 + BD2
⇒ BD2 = 169 – 64 = 105 ⇒ BD = 105 cm
⇒ DE = √105 cm [∵ BD = DE]
In ΔAED, we have
AD2 = AE2 + ED2 [By Pythagoras Theorem]
⇒ AD2 = 162 + (√105)2 = 256 + 105 = 361
⇒ AD = 19 cm

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