The parallel sides of an isosceles trapezium are 12 cm and 8 cm respectively and the non-parallel side measures 10 cm and makes an angle of 120- with the parallel side- Then the distance between the parallel sides is -sin 60-0-86-A- 7-5 cmB- 8-6 cmC- 9-5 cmD- 10-6 cm

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Question

The parallel sides of an isosceles trapezium are 12 cm and 8 cm respectively and the non-parallel side measures 10 cm and makes an angle of $120^{\circ}$ with the parallel side. Then the distance between the parallel sides is $[s i n 60^{\circ} = 0.86]$

7.5 cm

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8.6 cm

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9.5 cm

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10.6 cm

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Solution

The correct option is B
8.6 cm

Let ABCD be the given trapezium having AB = 12 cm, CD = 8 cm ,AD = 10 cm and $∠ D = 120^{\circ}$

Draw DE perpendicular to AB

$∠ A = 180 - 120^{\circ} = 60^{\circ}$ (As AB||CD )

In right angle AED

$s i n 60^{\circ} = \frac{D E}{A D}$

$D E = A D . s i n 60^{\circ}$

$D E = 10 \times 0.86 = 8.6 c m$

Hence, the distance between the parallel sides is DE = 8.6cm

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The parallel sides of an isosceles trapezium are 12 cm and 8 cm respectively and the non-parallel side measures 10 cm and makes an angle of 120∘ with the parallel side. Then the distance between the parallel sides is [sin 60∘=0.86]

The parallel sides of an isosceles trapezium are 12 cm and 8 cm respectively and the non-parallel side measures 10 cm and makes an angle of $120^{\circ}$ with the parallel side. Then the distance between the parallel sides is $[s i n 60^{\circ} = 0.86]$