Important for :

1

D

Given that, BC || AE∠CBA + ∠EAB = 180°

⇒ ∠EAB = 180° – 65° = 115°

Hence, ΔABC is an isosceles triangle.

⇒ ∠CBA = ∠CAB = 65°

Now, ∠EAB = ∠EAC + ∠CAB

⇒ 115° = x + 65° ⇒ x = 50°.

Hence, option D is correct.

2

B

Given that,∠ABC = 65° and ∠CDE = 15°

Here, ∠ABC + ∠TCB = 180° (

∠TCB = 180° – ∠ABC

Now, in ΔCDE

∠CED = 180° – (∠ECD + ∠EDC) (

= 180° – (– 65° + 15° ) = 100°

⇒ ∠FEC = 180° – 100° = 80°

Given that, AB = AE.

i.e. ΔABE an isosceles triangle.

⇒ 65° + x° + 80° = 180°

Hence, option B is correct.

3

C

∠PCT + ∠PCB = π (Linear pair)∠PCB = π – (π – b°) = b° ..... (i)

In ΔBPC,

∠PCB + ∠BPC + ∠PBC = π

∠PBC = π – ∠PCB – ∠BPC = π – b° – a° ..... (ii)

∠ABE = π – ∠PBC = π – (π – b° – a°) = a° + b° ...(iii)

Now, in ΔABE

Sum of two interior angles = Exterior angle

∠EAB + ∠ABE = ∠BES ⇒ c° + b° + a° = x°

Hence, option C is correct.

4

C

Statements I and II are both true, because the locus of points which are equidistant from two parallel lines is a line parallel to both of them and draw mid way between them.

Also, it is true that the perpendicular distances of any point on this locus line from two original parallel lines are equal. Further, no point outside this locus line has this property.

Also, it is true that the perpendicular distances of any point on this locus line from two original parallel lines are equal. Further, no point outside this locus line has this property.

Hence, option C is correct.

5

B

In 1 min = 60 s distance travelled by the wheel= 12 × Its circumference

= 12 × 2πr

∴ In 1 s distance travelled by the wheel = |
12 × 2πr | = | 2 | πr |

60 | 5 |

∵ Angle = |
Arc | = | 2/5 πr | = | 2π |

Radius | r | 5 |

Which is the required angle.

Hence, option B is correct.