Expert: Sherman D. - 3/17/2005

Question
1) the sides of an isosceles triangle have lengths 5, 10, and 10. what are the measures of its angles?
2) a regular pentagon is inscribed in a circle of radius 4 in. find the area of the pentagon.

Answer
1.)
2x + (1/2)x = 180
4x + x = 360
5x = 360
x = 72

the angles are 36°,72°, and 72°

here is what i did, since 2 sides are the same, i just put 2x and since 5 is half of 10, i put (1/2)x. The sides that are 10 would have the same length, and the side that is 5 would have half the length as the sides that are 10.

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2.)
If the pentagon is inscribed in a circle, then that would make the circle circumscribed of the pentagon.

You would use this formula

K = 5R^2 * sqrt(10 + 2sqrt[5])/8
K = 5(4)^2 * sqrt(10 + 2sqrt(5))/8
K = 5(16) * sqrt(10 + 2sqrt(5))/8
K = 80 * sqrt(10 + 2sqrt(5))/8
K = 10sqrt(10 + 2sqrt(5))
K = 38.04226065
K = about 38 in.^2

for more info on this, go to http://mathforum.org/dr.math/faq/formulas/faq.regpoly.html