4.2.3 Polygons, PT3 Practice

Question 7:
In diagram below, PQRSTU is a hexagon. UPE, PQF, QRG, RSH and UTJ are straight lines.

Find the value of x.

Solution:

\begin{array}{l} Sum of all the exterior angles of any polygon = 360^{o} \\ 25^{o} + 3 x^{o} + 2 x^{o} + 40^{o} + 75^{o} + 55^{o} = 360^{o} \\ 5 x^{o} + 195^{o} = 360^{o} \\ 5 x^{o} = 360^{o} - 195^{o} \\ = 165^{o} \\ x^{o} = 165^{o} \div 5 \\ = 33^{o} \\ x = 33 \end{array}

$\begin{array}{l} Sum of all the exterior angles of any polygon = 360^{o} \\ 25^{o} + 3 x^{o} + 2 x^{o} + 40^{o} + 75^{o} + 55^{o} = 360^{o} \\ 5 x^{o} + 195^{o} = 360^{o} \\ 5 x^{o} = 360^{o} - 195^{o} \\ = 165^{o} \\ x^{o} = 165^{o} \div 5 \\ = 33^{o} \\ x = 33 \end{array}$

Question 8:
In Diagram below, A, B, C, D and E are vertices of a 9 sided regular polygon.

Find the value of x.

Solution:

\begin{array}{l} Exterior angle = \frac{360^{o}}{9} = 40^{o} \\ Interior angle = 180^{o} - 40^{o} = 140^{o} \\ Sum of interior angles of pentagon A B C D E \\ = (5 - 2) \times 180^{o} \\ = 540^{o} \\ x^{o} + 140^{o} + 140^{o} + 140^{o} + x^{o} = 540^{o} \\ 2 x^{o} = 540^{o} - 420^{o} \\ = 120^{o} \\ x^{o} = 60^{o} \\ x = 60 \end{array}

$\begin{array}{l} Exterior angle = \frac{360^{o}}{9} = 40^{o} \\ Interior angle = 180^{o} - 40^{o} = 140^{o} \\ Sum of interior angles of pentagon A B C D E \\ = (5 - 2) \times 180^{o} \\ = 540^{o} \\ x^{o} + 140^{o} + 140^{o} + 140^{o} + x^{o} = 540^{o} \\ 2 x^{o} = 540^{o} - 420^{o} \\ = 120^{o} \\ x^{o} = 60^{o} \\ x = 60 \end{array}$

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