Important for :

1

2

B

Given that,

∠ABC = 65° and ∠CDE = 15°

Here, ∠ABC + ∠TCB = 180°

(∵ AB || CD)

∠TCB = 180° – ∠ABC

∴ ∠TCB = 180° – 65° = 115°

∵ ∠TCB + ∠DCB = 180°

(Linear pair)

∴ ∠DCB = 65°

Now, in ΔCDE

∠CED = 180° – (∠ECD + ∠EDC)

(∵ ∠ECD = ∠BCD)

= 180° – (– 65° + 15° ) = 100°

∵ ∠DEC + ∠FEC = 180°

⇒ ∠FEC = 180° – 100° = 80°

Given that, AB = AE.

i.e. ΔABE an isosceles triangle.

∴ ∠ABE = ∠AEB = 65°

∵ ∠AEB + ∠AEF + ∠FEC = 180°

(straight line)

⇒ 65° + x° + 80° = 180°

∴ x° = 180° – 145° = 35°.

Hence, option B is correct.

3

C

∠PCT + ∠PCB = π

(Linear pair)

(Linear pair)

∠PCB = π – (π – b°) = b°

..... (i)

..... (i)

In ΔBPC,

∠PCB + ∠BPC + ∠PBC = π

∠PBC = π – ∠PCB – ∠BPC = π – b° – a°

..... (ii)

..... (ii)

∵ ∠ABE + ∠EBC = π

(∵ ∠PBC = ∠EBC) (linear pair)

(∵ ∠PBC = ∠EBC) (linear pair)

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

...(iii)

...(iii)

Now, in ΔABE

Sum of two interior angles = Exterior angle

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

∴ x° = a° + b° + c°.

Hence, option C is correct.

4

I. The locus of points which are equidistant from two parallel lines is a line parallel to both of them and drawn mid way between them.

II. The perpendicular distance of any point on this locus line from two original parallel lines are equal. Further, no point outside this locus line has this property.

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