Aregular pentagon is inscribed in a circle with a radius of 5 cm. what is the perimeter of the pentagon?

P(p)  = 29.4 cm

Step-by-step explanation:

A regular pentagon (has all five sides equal)

Let call O center of the pentagon (which is the center of the circle.

We can get isosceles triangles with segment of lines between O and each of the corners of the pentagon. As is easy to see in such triangles two sides are radius of the circle and the other one is side of the pentagon.

In these triangles ∠ O  are 72 °  as you have to divide 360 ÷ 5 in order to get five equal sides. Then if we get a triangle for instance triangle AOB with:

∠ AOB   72 °  then the sides and angles

sides OB = OA = radius of circle = 5 cm and

∠OAB  = ∠OBA  =  54°  

Remember that sum of internal angles in a triangle is 180°

then   180° - 72°  =  108°  two equal angles  108/2   =  54°

We are able to apply sines law

Then      OB/ sin 54° =  L / sin 72°     where L is side of the pentagon

From tables:

sin 72°  = 0.9510

sin 54°  = 0.8090

By subtitution

L / sin 72°    =    5 / sin 54°    ⇒    L  =   5 *  sin 72° / sin 54°

L  =  5*  0.9510/ 0.8090    ⇒   L  =  5.88 cm

Then the perimeter of pentagon is P(p)

P(p)  =  5 * 5.88

P(p)  = 29.4 cm

hi there!

question 1 answer: p/q can be terminating when q = 5^m * 2^n

question 2 answer: -1 and -1

question 3 answer: -4

question 4 answer: 5/12

hope this : )

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