Question 50

What is the sum of the interior angles at the vertices of a 5-pointed star as shown below? The star need not have sides of the same length.

Solution

In triangle AFJ, $$x_{1}$$+180-a+180-b=180

=> $$x_{1}$$+180=a+b

Similarly, $$x_{2}$$+180=a+e

$$x_{3}$$+180=d+e

$$x_{4}$$+180=c+d

$$x_{5}$$+180=b+c

Adding all the equations,

$$x_{1}$$+$$x_{2}$$+$$x_{3}$$+$$x_{4}$$+$$x_{5}$$+180*5= 2(a+b+c+d+e)

=> Sum of interior angles = 2*(a+b+c+d+e)-900 = 2*540-900=180 (Sum of interior angles of a pentagon = 540)

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