 Important for :
1
In the figure given below, ABC is a triangle. BC is parallel to AE. If BC = AC, then what is the value of ∠CAE? » Explain it
D
Given that, BC || AE

∠CBA + ∠EAB = 180°

⇒ ∠EAB = 180° – 65° = 115°

∵   BC = AC

Hence, ΔABC is an isosceles triangle. ⇒   ∠CBA = ∠CAB = 65°

Now,  ∠EAB = ∠EAC + ∠CAB

⇒   115° = x + 65° ⇒ x = 50°.

Hence, option D is correct.

2
In the figure given below, AB is parallel to CD. ∠ABC = 65°, CDE = 15° and AB = AE. What is the value of AEF? » Explain it
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
The angles x°, a°, c° and (π – b)° are indicated in the figure given below. Which one of the following is correct? » Explain it
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 + ∠EBC = π

(∵ ∠PBC = ∠EBC)      (linear pair)

∠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°

∴   x° = a° + b° + c°.

Hence, option C is correct.

4
Consider the following statements
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.

Which of the above statements is/are correct?
» Explain it
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. Hence, option C is correct.

5
A wheel makes 12 revolutions per min. The angle in radian described by a spoke of the wheel in 1 s is:
» Explain it
B
= 12 × Its circumference

= 12 × 2πr

∴  In 1 s distance travelled by the wheel =

 12 × 2πr = 2 πr 60 5