In a parallelogram abcd, angle b is 60°. The perpendicular distance between ad and bc is 3 cm. Find the sides of AB and BC if BC is 3/2 of AB
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Given : In a parallelogram abcd, angle b is 60°. The perpendicular distance between ad and bc is 3 cm. BC is 3/2 of AB
To Find : sides AB and BC
Solution:
AB = 2x cm
=> BC = (3/2)2x = 3x cm
Area of parallelogram = AB * BC Sin B
= 2x * 3x . Sin 60°
= 6x² (√3 / 2)
= 3√3 x² cm²
Area of parallelogram = BC * perpendicular distance between ad and bc
= 3x * 3
= 9x cm²
3√3 x² = 9x
=> x = √3
AB = 2√3 cm
BC = 3√3 cm
or another method
Sin 60° = 3/ AB
=> AB = 2√3 cm
BC = (3/2) AB = 3√3 cm
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