David Allen

Posted on Sunday, 12 January, 2003 - 08:57 pm:

Here is my question, its been puzzling me for days and has two parts to it.

A regular pentagon ABCDE is inscribed in a circle with centre O and radius 5cm.

(Then I have a nice wee diagram of just that).

Calculate the perimeter of the pentagon, correct to 2 decimal places. (In triangle OAB you may assume that the line from O to the midpoint of AB is at right angles to AB)

ALSO:

What is the percentage of the area of the circle lies outside the pentagon?

Now I know I should know the formuli to get these answers but I am completely stumped. Please can you help?

Anuj Shah

Posted on Sunday, 12 January, 2003 - 09:27 pm:

Take triangle OAB, you know OA is 5cm and angle OAB is 54

^{\circ}

(Half the interior angle of a regular pentagon). The length AM (where M is the midpoint of AB) is then 5.cos(54

^{\circ}

). So the perimeter is 50.cos(54

^{\circ}

)

The area of triangle OAM is (1/2)(5)(5.cos(54 $^{\circ}$ ))=25.cos(54 $^{\circ}$ )/2. The area of the pentagon is then 125.cos(54 $^{\circ}$ ). So the area of the circle outside the pentagon is $π . 5^{2} - 125 . \cos (54^{\circ}) = 25 (π - 5 . \cos (54^{\circ}))$ . You can then find the percentage.