If PQRS is a trapezium such that PQ || RS, (angle)P: S = 3:2, (angle)Q : R = 7:2,
then find the angles of the trapezium.
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Answer:
I guess I just answered it previously,
But here you go again...
Step-by-step explanation:
P=108°, Q=140°,R=72°, S=40°
Let the angles P and S be 3x and 2x
And angles Q and R be 7y and 2y.
Sum of all angles of a quadrilateral = 360°
3x+2x + 7y+2y = 360°
5x + 9y = 360°
Also,
PQ||RS,
Q+R = 180°= P+S ( Pair of co-interior angles)
7y+2y = 180° = 3x+2x
5x = 180°, 9y = 180°
x= 36° , y = 20°
Therefore,
angles are:
P=108°, Q=140°,R=72°, S=40°
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