The diagram shows a trapezium inscribed in a semi-circle. If O is the mid-point of WZ and |WX| = |XY| = |YZ|, calculate the value of m
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Correct Answer: Option B
Explanation:
In the diagram, < WOZ = 180\(^o\) (angle on a straight line)
< WOX = < XOY = < YOZ
(|WX| = |XY| = |YZ|)
\(\frac{180^o}{3}\) = 60\(^o\)
= 60\(^o\)
M + m =2m (base angles of isosceles \(\bigtriangleup\), |OY| and |OZ| are radii)
< YOZ + 2m (base angles of a \(\bigtriangleup\))
60\(^o\) + 2m = 180\(^o\) (sum of a \(\bigtriangleup\))
60\(^o\) + 2m = 180\(^o\)
2m = 180\(^o\) - 60\(^o\)
2m = 120\(^o\)
m = \(\frac{120^o}{2}\)
= 60\(^o\)
In the diagram, < WOZ = 180\(^o\) (angle on a straight line)
< WOX = < XOY = < YOZ
(|WX| = |XY| = |YZ|)
\(\frac{180^o}{3}\) = 60\(^o\)
= 60\(^o\)
M + m =2m (base angles of isosceles \(\bigtriangleup\), |OY| and |OZ| are radii)
< YOZ + 2m (base angles of a \(\bigtriangleup\))
60\(^o\) + 2m = 180\(^o\) (sum of a \(\bigtriangleup\))
60\(^o\) + 2m = 180\(^o\)
2m = 180\(^o\) - 60\(^o\)
2m = 120\(^o\)
m = \(\frac{120^o}{2}\)
= 60\(^o\)