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CBSE Class 10  >  Class 10 Test  >  Olympiad Preparation  >  Math Olympiad Test: Circles- 3 - Class 10 MCQ

Math Olympiad Test: Circles- 3 - Free MCQ with solutions Class 10


MCQ Practice Test & Solutions: Math Olympiad Test: Circles- 3 (10 Questions)

You can prepare effectively for Class 10 Olympiad Preparation for Class 10 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Math Olympiad Test: Circles- 3". These 10 questions have been designed by the experts with the latest curriculum of Class 10 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 10 minutes
  • - Number of Questions: 10

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Math Olympiad Test: Circles- 3 - Question 1
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In the given figure, PT touches the circle at R whose centre is O. Diameter SQ when produced meets PT at P. Given ∠SPR = x° and ∠QRP = y°. Then,

Detailed Solution: Question 1

Consider chord QR

∴∠QRP = ∠QSR

⇒ ∠QSR = y° [∴ QRP = y°]

⇒ ∠PSR = y°

Since angle in a semicircle is a right angle. 

∴∠QRS = 90°

Now, ∠PRS = ∠QRP + ∠QRS

⇒ ∠PRS = y° + 90°

In ΔPRS, we have

∠SPR + ∠PRS + ∠PSR = 180°

⇒ x° + y°+ 90° + y° = 180° 

⇒ x° + 2y° = 90°

Math Olympiad Test: Circles- 3 - Question 2
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In the given figure, a circle touches the side BC of ΔABC at P and touches AB and AC produced at Q and R respectively. If AQ = 5 cm, find the perimeter of ΔABC.

Detailed Solution: Question 2

Perimeter of ΔABC = AB + BC + AC
= AB + (BP + PC) + AC
= (AB + BQ) + (CR + AC)
[∵ BP = BQ, PC = CR]
= AQ + AR
(∵AQ = AR, tangents of circle from same point A)
= 2 AQ = 2 x 5 = 10 cm.

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Math Olympiad Test: Circles- 3 - Question 3
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Two circles of radii 10 cm and 8 cm intersect each other and the length of common chord is 12 cm. The distance between their centres is.

Detailed Solution: Question 3

M is the mid-point of AB,

∴ AM = 6 cm

AO (r1) = 10 cm,

AO' (r2) = 8 cm

AB is perpendicular to OO' , then

In ΔAOM, 100 = 36 + OM2 [using pythagoras theorem]

⇒ OM = 8 cm; In ΔAMO', 64 = 36 + MO'2

⇒ √28 = MO' = 2√7 = MO'

∴ OO' = (2√7 + 8) cm

Math Olympiad Test: Circles- 3 - Question 4
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A circle inscribed in ΔABC having AB = 10 cm, BC = 12 cm, CA = 28 cm touching sides at D, E, F (respectively). Then AD + BE + CF is ______.

Detailed Solution: Question 4

x + y = 10 cm ... (i)
y + z = 12 cm    ... (ii)
x + z = 28cm    ... (iii)
Adding (i), (ii) and (iii), we get
2(x + y + z) = 50
⇒ x + y+ z = 25

Math Olympiad Test: Circles- 3 - Question 5
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In the given figure, O is the centre and SAT is a tangent to the circle at A. If ∠BAT = 30°, find ∠AOB and ∠AQB.

Detailed Solution: Question 5

∠BAT = 30° (given)
Since angle made by chord with a tangent is equal to angle made by it in alternate segment
∴ ∠APB = 30°
Now, ∠APB + ∠AQB = 180°
[Opposite angles of cyclic quadrilateral APBQ]
⇒ 30° + ∠AQB = 180° ⇒ ∠AQB = 150°
Also, ∠AOB = 60° [Angle subtended by an arc at centre is double the angle subtended by it on remaining part of circle] 
∴ ∠AOB = 60°, ∠AQB = 150°

Math Olympiad Test: Circles- 3 - Question 6
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In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6 cm, BC = 7 cm, CD = 4 cm, then AD equals _________.

Detailed Solution: Question 6

x + y = 6 cm ... (i)

y + z = 7 cm ... (ii)

z + w = 4 cm ... (iii)

Using (ii) and (iii),

y - w = 3 cm

Now using (i) x + (3 + w) = 6 cm

⇒ x + w = 3 cm

⇒ AD = 3 cm

Math Olympiad Test: Circles- 3 - Question 7
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In the given figure, AB and PQ intersect at M. If A and B are centres of circles then _______.

Detailed Solution: Question 7

In the given figure, PQ is the common chord of the two circles.
⇒ AB bisects the common chord PQ at M.
∴ PM = MQ
Moreover, PQ is perpendicular to AB.
∴ Option (C) is correct.

Math Olympiad Test: Circles- 3 - Question 8
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In the following figure, PT is of length 8 cm. OP is 10 cm. Then the radius of the circle is___.

Detailed Solution: Question 8

Since OT is perpendicular to FT, Using pythagoras theorem,
100 = 64 + r2

∴ r = 6 cm

Math Olympiad Test: Circles- 3 - Question 9
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In the given figure, O is the centre of the circle, then ∠OZ is______.

Detailed Solution: Question 9

OX= OY = OZ = r    (radius of circle)
Taking ∠X = θ and ∠Z = α
Then in ΔYXO, OX = OY ⇒ ∠X = ∠OYX = θ
Now, in ΔOZY OZ= OY⇒ ∠Z= ∠OYZ = α
Now, XOZY is a quadrilateral,
⇒ ∠X+∠Y+ ∠Z+∠O = 360°
⇒ θ + θ + α + α +∠O = 360°
⇒ ∠O = 360° - 2(α + θ)
⇒ ∠O = 360° - 2 (∠OZX + ∠XZY + ∠OXZ + ∠ZXY)
⇒ ∠O = 360° - 2 (∠OZX + ∠OXZ) - 2(∠XZY + ∠ZXY)
⇒ ∠O = 360° - 2(180° - ∠O) - 2(∠XZY + ∠ZXY)
⇒ ∠O = 360° - 360° + 2∠O - 2(∠XZY + ∠ZXY)

Math Olympiad Test: Circles- 3 - Question 10
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In the given figure, O is the centre of the circle. If PA and PB are tangents, then the value of ∠AQB is

Detailed Solution: Question 10

Since PA and PB are tangents.
Also, tangent is ⊥ r to radius through point of contact,

∴ ∠PAO = 90° and ∠PBO = 90°

In quadrilateral APBO

∠APB + ∠PAO + ∠PBO + ∠AOB = 360°
80° + 90° + 90° + ∠AOB = 360°
⇒ ∠AOB = 100° ⇒ ∠AQB = 1/2∠AOB = 50° 

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About this Class 10 Practice Test

This test contains 10 MCQs on Math Olympiad Test: Circles- 3 with detailed solutions for each question. It is part of the Olympiad Preparation for Class 10 course on EduRev, designed to align with the current Class 10 syllabus. Each solution explains not just the correct answer but also gives you an understanding of the topic — not just to attempt the question but also help you prepare for the exam with practice.

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