Question

In the given figure, AED is an isosceles triangle with

AE = AD. ABCD is a parallelogram and EGF is a

line segment. If DCF = 65° and EFB = 100°, then

the number of diagonals of a regular polygon having

its each exterior angle equal to the measure of

AEG, is

Answer 1

The answer is fifteen degree (15°)

Answer 2

Answer: No. of diagonals = 54

Step-by-step explanation:

Step 1:  

∠EFB = 100°

∴ ∠CFE = 180° - 100° = 80° ….. [Linear Pair]

Step 2:

In ∆ ECF, using the angle sum property, we have

∠DCF + ∠CFE + ∠FEC = 180°

⇒ 65° + 80° + ∠FEC = 180°

⇒ ∠FEC = 180° - 145° = 35° …… (i)

Step 3:

ABCD is a parallelogram.

AD // BC and ∠DCF = 65° (given)

∴ ∠ADE = ∠DCF = 65° ….. [corresponding angles]

Also, ΔAED is given as an isosceles triangle with AE = AD

∴ ∠ADE = ∠AED = 65°  …… (ii)

Now,  

The Exterior angle, ∠AEG  

= ∠AED - ∠FEC  

= 65° - 35° ….. [from (i) & (ii)]

= 30°  

Step 4:

We know,

No. of sides in a regular polygon “n” with exterior angle 30°  

= 360° / exterior angle  

= 360° / 30°  

= 12

Thus,  

No. of diagonals in a regular polygon having n = 12 is,

= n(n-3)/2

= 12(12 - 3)/2

= 54