Question

Please answer this question -

Answer 1

||✪✪ QUESTION ✪✪||

Find The sum of ∠A + ∠B + ∠C + ∠D + ∠E + ∠F + ∠G ?

|| ★★ FORMULA USED ★★ ||

• Sum of interior Angle of Polygon with n sides = (n-2)*180°.
• Sum of all Three angles of a ∆ is 180°..

|| ✰✰ ANSWER ✰✰ ||

❁❁ Refer To Image First .. ❁❁

As we can see In The Fig. The star So formed as 7 sides .

So,

➼ Sum of interior Angles of Polygon with 7 sides is = (7-2)*180° = 5*180° = 900°..

So,

➪ Each Angle of Polygon is = (900/7)°.

Now,

➻ One Angle of a ∆ = 180° - (900/7) = (360/7)° .

Than,

➺ Another Angle of ∆ = 900° - (2*360/7)° = (540/7)°

Hence,

☛ All the Angles of polygon ( ∠A + ∠B + ∠C + ∠D + ∠E + ∠F + ∠G ) = 7 * (540/7)° = 540° ..

•°• Sum of ∠A + ∠B + ∠C + ∠D + ∠E + ∠F + ∠G is 540°.

꧁____________________꧂

Answer 2

REFER TO THE ATTACHMENT FIRSTLY

in the figure MNOPQRS is a seven sided regular polygon

so, ∠1+∠2+∠3+∠4+∠5+∠6+∠7 =  (7-2)*180 °

∠1+∠2+∠3+∠4+∠5+∠6+∠7 = 900°

and

∵ MNOPQRS is a regular heptagon

∴∠1 =∠2 =∠3 =∠4 =∠5 =∠6 =∠7 =  (900/7 )°

NOW,

since, AMS is line segment therefore

∠ AMN = 180° - ∠1   ( being linear pair )

∠AMN = 180° - (900/7) = (360/7)°

similarly ,

∠ ANM = (360 / 7)°

NOW,

in Δ AMN

by angle sum property of triangle

∠A = 180° - 2(360/7)°

∠A =  (540/7)°

SO,

∠A+∠B+∠C+∠D+∠E+∠F+∠G = 7 ( 540 /7 ) °

∠A+∠B+∠C+∠D+∠E+∠F+ ∠G= 540 °

HENCE THE CORRECT OPTION IS (D) 540°