Geometry Exercise 2

AB =
∴ AC = cm

Here, OP = 10 cm; O’P = 8 cm

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

O’L = 5.29 cm

∴ OO'= OL + O'L

OO'= 13.29 cm

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

∴ ∆ ADC is a right angled triangle.

∴ (DB)² = BA × BC

[since, DB is the perpendicular to the hypotenuse]

= 9 × 4 = 36

∴ DB = 6

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

We have, ∠OBC = ∠OCB = 37°

(equal angles of an isosceles triangle)

⇒ ∠COB = 180° – (37° + 37°) = 106°

Therefore, ∠BAC

In

∴

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° .

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

∠AQB = 90° (∠In semi- circle)

∠QXB = 180° – 90° – 24° = 66°

AO =

= = = 3

Now, from similar ∆s QAO and QPR

OR = 2OA = 2× 3 = 6 cm.

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°.

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

⇒ ∠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°

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

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

⇒ ∠B = 100°

But ∠B + ∠D = 180°

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

100° + ∠D = 180°

⇒ ∠ADC = 80°

24 cm

In a triangle ∆AMO,

= 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 =

Draw OE ⊥ CD and OF ⊥ AB

AB||CD [Given]

Let ‘r’ be the radius of the circle

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

[using Pythagoras theorem]

⇒ 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

cm

OABC is square with side = 2

OB = 2 = OD + r + O’B = 2 + r + r
⇒ r(+1) = 2(–1)

Diameter of circle = x

∴ y = 4x

∴ x = ¼ y

m ∠ DAB + 180° – 120° =60°

[Opposite angles of a cyclic quadrilateral]

m (arc BCD) = 2m ∠ DAB = 120°.

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