A3=2Rsin( 180 degrees /3)
=2Rsin60 degrees
= (root number 3) R,
A4=2Rsin( 180 degrees /4)
= (radical number 2)R,
So the perimeter of a regular triangle inscribed in a circle =(3 root numbers 3)R,
Area = [(root number 3)/4][ (root number 3) r] 2
=(3 radicals 3) r 2/4.
Circumference of a square inscribed in a circle = (radical number 2 of 4) r
Area = [(root number 2) r] 2
=2R^2.
(Note: area formula of regular triangle: area of regular triangle = [(root number 3)/4] times square of side length. )
7。 Because r/R=cos( 180 degrees /3)=cos60 degrees = 1/2.
So R=2r
And because r-r = 2.
So R=4.
So the side length of this regular triangle a3=4 root number 3.
Area = [(root number 3)/4] (root number 4 3) 2
= 12 root number 3.
8。 It is proved that: (1) Because all sides of a regular Pentagon are equal, all internal angles are equal to 108 degrees.
So angle AEB= angle ABE=36 degrees, angle BAC= angle BCA=36 degrees,
So angle EAP= angle EAB-angle BAC=72 degrees,
Angle EPA= angle ABE+ angle BAC=72 degrees,
So angle EAP= angle EPA,
So PE=AE=AB.
(2) Because angle AEB= angle BAC=36 degrees and angle ABE= angle PBA,
Triangle AEB is similar to triangle PAB,
So AB/BP=BE/AB
So ab 2 = be * BP
Because PE=AE=AB.
So PE 2 = be * BP.
Test analysi