SOLUTION: In a circle with diameter of 20 cm, a regular five-pointed star touching its circumference is inscribed. Find the area of the star.

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Question 1180960: In a circle with diameter of 20 cm, a regular five-pointed star touching its circumference is inscribed. Find the area of the star.
Found 2 solutions by MathLover1, Edwin McCravy:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

there are 5 central angles
each angle is 360%2F5+=+72 degrees
label a point of the star on the circumference A, center of the circle O and the intersection of the side of A and the side of the point of the star adjacent to A going clockwise B.
angle AOB is 72%2F2+=+36+degrees and angle AOB is 36%2F2+=+18 degrees
for triangle AOB, we have
36+%2B18+%2Bbeta+=+180
beta+=+180+-36+-18+=+126 degrees
radius of circle = 20%2F2+=+10+cm
Now use the law of sines
%28OB%2Fsin%2818%29%29+=+%2810%2Fsin%28126%29%29
OB+=+10%2Asin%2818%29%2Fsin%28126%29
OB+=+3.8197
Area of triangle+AOB+=+%281%2F2%29%2A3.8197%2A10%2Asin%2836%29+=+11.2258
Area of the five pointed star = 10%2A11.2258+=+112.258cm%5E2


Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!


By sketching in a few other lines, it shouldn't take you long
to figure out that the star is made up of 10 triangles
all congruent to ΔOPQ.  Also, you can easily get:

∠POQ = 36o, OQ = 20, ∠OQP = 18o, ∠OPQ = 126o

By the law of sines,

OQ%2Fsin%28OPQ%29=PQ%2Fsin%28POQ%29

20%2Fsin%28126%5Eo%29=PQ%2Fsin%2836%5Eo%29

Solve that and get

PQ = 14.53085056

Then to find the area of that triangle, we use the
formula:

Area%22%22=%22%22expr%281%2F2%29OQ%2APQ%2Asin%28OQP%29

Area%22%22=%22%22expr%281%2F2%2920%2A14.53085056%2Asin%2818%5Eo%29

Area%22%22=%22%2244.90279766cm%5E2

We multiply that by 10 and get

449.0279766 cm2    <---answer

Edwin