Prove that the interior angle of a regular five-sided polygon (pentagon) is three times the exterior angle of a regular decagon.

Open in App

Solution

One interior angle of a penagon $= (n - 2) \times \frac{180}{n}$

where $n = 5$

$= (5 - 2) \times \frac{180}{5} = 108$

Exterior angle of any polygon is 360.

One exterior angle of decagon $= \frac{360}{10} = 36$

$108 = 3 \times 36$

Therefore it's proved that one interior angle of pentagon is three times the exterior angle of decagon.

Suggest Corrections

Similar questions

Proove that interior angle of a regular Pentagon is 3 times the exterior angle of decagon.

The interior angle of a regular pentagon is four times the exterior angle of a regular decagon. State the whether the given statement is correct or not.

Q. The interior angle of a regular polygon with

n

sides is

6

times that of an exterior angle of a regular polygon with

\frac{3}{2} n

sides. Then

n

equals

Q. The measure of each interior angle of a regular polygon is five times the measure of its exterior angle. Find the number of sides in the polygon

Q. Fill in the blanks.
(i) Each interior angle of a regular octagon is (.........)°.
(ii) The sum of all interior angles of a regular hexagon is (.........)°.
(iii) Each exterior angle of a regular polygon is 60°. This polygon is a .........
(iv) Each interior angle of a regular polygon is 108°. This polygon is a .........
(v) A pentagon has ......... diagonals.

Join BYJU'S Learning Program

Prove that the interior angle of a regular five-sided polygon (pentagon) is three times the exterior angle of a regular decagon.

One interior angle of a penagon =(n−2)×180nwhere n=5=(5−2)×1805=108Exterior angle of any polygon is 360.One exterior angle of decagon =36010=36108=3×36Therefore it's proved that one interior angle of pentagon is three times the exterior angle of decagon.