Question

Prove that the interior angle of a regular polygon is 3 times the exterior angle of regular decagon

Answer 1

Answer:

n=10

180*10=i*10+360 (i is interior angle)

i=144 degree

Exterior angle=180-144=36

i/ext=4

Answer 2

Answer:

$\huge\fbox\red{Solution} \:$

Sum of all interior angles of a

regular polygon = ( n - 2) × 180

= (5 - 2) × 180

= 3 × 180

= 540

$each \: interior \: angle \: = \frac{540}{5} \\ \\ = {108}^{0}$

Interior Angle . Regular polygon = 108° -----› (1)

$each \: exterior \: angle = \frac{360° \: }{n \: }$

$exterior \: angle \: of \: r.d \: = \frac{360°}{10} \\ \\ = 36$

E.A.R.D = 36° -------› (2)

Divide eq (1) by eq (2) / Taking ratio of EQ (1) & EQ (2)

$\frac{ I.A.R.P}{ E.A.R.D} = \frac{108}{36}$

I.A.R.P = 3 ( E.A.R.D )

interior Angle of regular pentagen = 3 ( exterior angles of regular decagon )

Step-by-step explanation:

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