MENU
Question -

The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the angles.



Answer -

Let the four angles of a quadrilateral which are in A.P., be
a – 3d, a – d, a + d, a + 3d
Common difference = 10°
Now sum of angles of a quadrilateral = 360°
a – 3d + a – d + a + d + a + 3d = 360°
=> 4a = 360°
=> a = 90°
and common difference = (a – d) – (a – 3d) = a – d – a + 3d = 2d
2d = 10°
=> d = 5°
Angles will be
a – 3d = 90° – 3 x 5° = 90° – 15° = 75°
a – d= 90° – 5° = 85°
a + d = 90° + 5° = 95°
and a + 3d = 90° + 3 x 5° = 90° + 15°= 105°
Hence the angles of the quadrilateral will be
75°, 85°, 95° and 105°

Comment(S)

Show all Coment

Leave a Comment

Free - Previous Years Question Papers
Any questions? Ask us!
×