five pointed star
Geometry
Level
3
What is the area of regular five pointed star with side 16?!!
The answer is 79.43.
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Note: Pentagon= PG , Isosceles Triangle= IT and " * " for Multiplication
Since it is regular shape, so all parts are symmetrical.
Let the side of PG is X , and two sides of IT is Y
We know that interior angles of pentagon are 108
By vertically opposite angle : angle between sides of two IT will be 108 Then,
180 - 108 =72 _ will be the _ A & B internal angles of IT. Sum of all angles of Triangle is 180 . So, Third angle i.e.
C = 180 -2*72 = 36.
From 16 cm line we find that:
_16=2*Y+X _
Y=(16-X)/2 now by applying sine law to IT we get
Y/Sin(72) = X/Sin(36)
X=0.617571531* Y put value of Y in terms of X
we get X= 3.774928164 cm _ and _ Y= 6.112535918 cm
Area of PG = 1/4x sqrt(5 (5+2 sqrt(5)))x X^2
Solving this we get ....................................................... PG Area= 24.51694514 cm^2 --------(i)
Area of IT = Base * Height /2 = X * H/2
H can be find out by applying Pythagoras H= 5.813826166 cm
Solving this we get Area of IT = 10.97 cm^2
We have 5 IT So, ..............................................................5* Area of IT = 54.85 cm^2 ----------(ii)
Adding this i and ii we get total area = 79.36694514 Sq. cm