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Geometry Exercise 2

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Geometry 2

  • This online quiz will test your knowledge of Geometry in Quantitative Aptitude.
  • This Online Test is useful for academic and competitive exams.
  • Multiple answer choices are given for each question in this test. You have to choose the best option.
  • After completing the test, you can see your result.
  • There is no negative marking for wrong answers.
  • There is no specified time to complete this test.

Two circles touch each other internally. Their radii are 2 cm and 3 cm. The biggest chord of the outer circle which is outside the inner circle is of length

geometry-38681.png
AB = geometry-38675.png
∴ AC = geometry-38671.pngcm

Two circles of radii 10 cm. 8 cm. intersect and length of the common chord is 12 cm. Find the distance between their centres.

Here, OP = 10 cm; O’P = 8 cm

geometry-38666.png

PQ = 12 cm

∴ PL = 1/2 PQ

⇒ PL = ½ × 12

⇒ PL = 6 cm

In rt. ∆ OLP, OP² = OL² + LP²

(using Pythagoras theorem)

⇒ 10² = OL² + 6²

⇒ OL² + 64; OL = 8

In ∆ O'LP, (O'L)² = O'P² - LP²

= 64 – 36 = 28

O’L² = 28 geometry-38638.png

O’L = 5.29 cm

∴ OO'= OL + O'L

OO'= 13.29 cm

Arc ADC is a semicircle and DB ⊥ AC. If AB = 9 and BC = 4, find DB.

m ∠ ADC = 90⁰ (Angle subtended by the diameter on a circle is 90°)

geometry-38626.png

∴ ∆ ADC is a right angled triangle.

∴ (DB)² = BA × BC

[since, DB is the perpendicular to the hypotenuse]

= 9 × 4 = 36

∴ DB = 6

A point P is 13 cm. from the centre of a circle. The length of the tangent drawn from P to the circle is 12 cm. Find the radius of the circle.
38791.png

Tangent at any point of a circle is ⊥ to the radius

In ∆OPT, OP² = PT² + OT²

(13)² = (12)² + OT²

⇒ 169 – 144 =OT²

⇒ 25 = OT²

⇒ 5 = OT

In the figure given below, O is the centre of the circle. If ∠OBC = 37°, then ∠BAC is equal to :
38785.png

We have, ∠OBC = ∠OCB = 37°

(equal angles of an isosceles triangle)

⇒ ∠COB = 180° – (37° + 37°) = 106°

Therefore, ∠BAC geometry-38921.png

PQRS is a square. SR is a tangent (at point S) to the circle with centre O and TR=OS. Then, the ratio of area of the circle to the area of the square is
39011.png

geometry-39115.png

In geometry-39108.png

geometry-39102.png geometry-39096.png

∴ geometry-39090.png

In the given figure, AB is chord of the circle with centre O, BT is tangent to the circle. The values of x and y are
38981.png

Given AB is a circle and BT is a tangent,

∠BAO = 32°

Here, ∠ OBT = 90°

[Since tangent is ⊥ to the radius at the point of contact]

OA = OB [ Radii of the same circle ]

∴ ∠ OBA = ∠OAB = 32° [ Angles opposite to equal side are equal]

∴ ∠ OBT = ∠ OBA + ∠ ABT = 90°

⇒ 32° + x = 90°.

⇒ ∠ x = 90° – 32° = 58° .

Also, ∠ AOB = 180° – ∠OAB – ∠OBA

= 180° – 32° – 32° = 116°

Now Y = ½ AOB [ Angle formed at the center of a circle is double the angle formed in the remaining part of the circle] = ½ × 116° = 58° .

In the figure given below, AB is a diameter of the semicircle APQB, centre O, ∠ POQ = 48° cuts BP at X, calculate ∠AXP.
38975.png

b = ½(48°) (∠at centre = 2 at circumference on same PQ) = 24°

∠AQB = 90° (∠In semi- circle)

∠QXB = 180° – 90° – 24° = 66°

OA is perpendicular to the chord PQ of a circle with centre O. If QR is a diameter, AQ = 4 cm, OQ = 5 cm, then PR is equal to
38969.png

AO =geometry-39083.png

= geometry-39077.png = geometry-39071.png= 3

Now, from similar ∆s QAO and QPR

OR = 2OA = 2× 3 = 6 cm.

In the given figure, m ∠ EDC = 54°. m ∠ DCA = 40°.Find x, y and z.
38963.png

m ∠ ACD = ½ M(are CXD) = m ∠ DEC

∴ m ∠ DEC = x = 40°

∴ m ∠ ECB = ½ m (are EYC) = m ∠ EDC

∴ m ∠ ECB = y = 54°

54 + x + z = 180° [Sum of all the angles of a triangle]

54 + 40 + z = 180°

∴ z = 86°.

ABCD is a cyclic quadrilateral in which BC || AD, ∠ADC = 110° and ∠BAC = 50° find ∠DAC

∠ABC + ∠ADC = 180° (sum of opposites angles of cyclic quadrilateral is 180°)

geometry-39065.png

⇒ ∠ABC + 110° = 180° [ABCD is a cyclic quadrilateral]

⇒ ∠ABC = 180 – 110

⇒ ∠ABC = 70°

Since AD || BC

∴ ∠ABC + ∠BAD = 180° [Sum of the interior angles on the same side of transversal is 180°]

70° + ∠ BAD = 180°

⇒ ∠BAD = 180° – 70° = 110°

⇒ ∠BAC + ∠DAC = 110°

⇒ 50° + ∠DAC = 110°

⇒ ∠DAC = 110° – 50 ° = 60°

In a circle of radius 17 cm, two parallel chords are drawn on opposite sides of a diameter. The distance between the chords is 23 cm. If length of one chord is 16 cm, then the length of the other one is :

Let PQ and RS be two parallel chords of the circle on the opposite sides of the diameter AB and PQ = 16 cm

geometry-39561.png

Now, PN = 8 (Since ON is the perpendicular bisector)

In ∆ PON, ON² = OP² - PN²

= (17)² – (8)² = 289 – 64 = 225

⇒ ON = 15

∴ OM = 23 – 15 = 8

In ∆ ORM, RM² = OR² - OM²

= (17)² – (8)² = 289 – 64 = 225

⇒ RM = 15

⇒ RS = 15 × 2 = 30 cm

In given fig, if ∠BAC = 60° and ∠BCA = 20° find ∠ADC
39261.png

In ∆ABC, ∠B = 180° – ( 60° + 20°) [By ASP]

geometry-39523.png

⇒ ∠B = 100°

But ∠B + ∠D = 180°

[Since ABCD is a cyclic quadrilateral; Sum of opposite is 180°]

100° + ∠D = 180°

⇒ ∠ADC = 80°

A point P is 26 cm. away form the centre O of a circle and the length PT of the tangent draw from P to the circle is 10cm. Find radius of the circle

24 cm

In a circle of radius 10 cm, a chord is drawn 6 cm from its centre. If an another chord, half the length of the original chord were drawn, its distance in cm from the centre would be:

geometry-39516.png
In a triangle ∆AMO,

geometry-39510.png= 8

Therefore, the length of the another chord A’B′ = 8 cm.

Now, A’N = 4

In ∆ OA'N, ON² = (OA')² - A'N²

= 10² – 4² = 100 – 16 = 84

⇒ ON = geometry-39492.png

AB and CD two chords of a circle such that AB = 6 cm, CD = 12 cm. And AB || CD. The distance between AB and CD is 3 cm. Find the radius of the circle.

Draw OE ⊥ CD and OF ⊥ AB

geometry-39473.png
AB||CD [Given]

Let ‘r’ be the radius of the circle

Now in rt. ∆ OED, (OD)² = (OE)² + (ED)²

[using Pythagoras theorem]
geometry-39448.png
⇒ r² = x² + 36 ...(1)

In rt. ∆ OFB, (OB)² = (OF)² + (EB)²

⇒ r² = (x + 3)² + (3)²

⇒ r² = x² + 6x + 9 + 9

⇒ r² = x² + 6x + 18 ...(2)

From (1) and (2), we get

x² + 36 = x² + 6x + 18

⇒ 36 = 6x + 18

⇒ 36 - 18 = 6x

⇒ 6x = 18

⇒ x = 3

For (1), r² = (3)² + (6)²

r² = 9 + 36

⇒ r² = 45

geometry-39374.pngcm

A circle with radius 2 is placed against a right angle. Another smaller circle is also placed as shown in the adjoining figure. What is the radius of the smaller circle? 39225.png

geometry-39368.png
OABC is square with side = 2

geometry-39362.png
OB = 2geometry-39356.png = OD + r + O’B = 2 + r + rgeometry-39350.png
⇒ r(geometry-39344.png+1) = 2(geometry-39338.png–1)
geometry-39332.png
geometry-39326.png

In the figure below, which of the following is the relationship between ‘x’ and ‘y’ if the equal circles shown are tangents to each other and to the sides of the rectangle
39194.png

Diameter of circle = x

∴ y = 4x

∴ x = ¼ y

In the cyclic quadrilateral ABCD BCD =120° , m(arc DZC) = 7°, find DAB and m (arc CXB).
39188.png

m ∠ DAB + 180° – 120° =60°

[Opposite angles of a cyclic quadrilateral]

m (arc BCD) = 2m ∠ DAB = 120°.

geometry-39320.png

∴ m (arc CXB) = m (BCD) – m (arc DZC) = 120° – 70° = 50° .

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