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Math Olympiad Test: Circles- 3 - Class 10 MCQ


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10 Questions MCQ Test Olympiad Preparation for Class 10 - Math Olympiad Test: Circles- 3

Math Olympiad Test: Circles- 3 for Class 10 2024 is part of Olympiad Preparation for Class 10 preparation. The Math Olympiad Test: Circles- 3 questions and answers have been prepared according to the Class 10 exam syllabus.The Math Olympiad Test: Circles- 3 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- 3 below.
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Math Olympiad Test: Circles- 3 - Question 1

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 for Math Olympiad Test: Circles- 3 - 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

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 for Math Olympiad Test: Circles- 3 - 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

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 for Math Olympiad Test: Circles- 3 - 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

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 for Math Olympiad Test: Circles- 3 - 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

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 for Math Olympiad Test: Circles- 3 - 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

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 for Math Olympiad Test: Circles- 3 - 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

In the given figure, AB and PQ intersect at M. If A and B are centres of circles then _______.

Detailed Solution for Math Olympiad Test: Circles- 3 - 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

In the following figure, PT is of length 8 cm. OP is 10 cm. Then the radius of the circle is___.

Detailed Solution for Math Olympiad Test: Circles- 3 - Question 8

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

∴ r = 6 cm

Math Olympiad Test: Circles- 3 - Question 9

In the given figure, O is the centre of the circle, then ∠OZ is______.

Detailed Solution for Math Olympiad Test: Circles- 3 - 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

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 for Math Olympiad Test: Circles- 3 - 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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