In trapezium ABCD- as shown- AB - DC- AD - DC - BC - 20 cm and - A - 60- Distance between AB and DC in cm-

The correct option is C 17.32 cm Given, trapezium ABCD, AD=20 , ∠ A=60^∘ Draw perpendicular from D on AB , which meets AB at ...

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Standard VIII

Mathematics

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In trapezium ...

Question

In trapezium $A B C D$ , as shown, $A B / / D C$ , $A D = D C = B C = 20 c m$ and $∠ A = 60^{\circ}$ . Distance between $A B$ and $D C$ in cm.

11.95 c m

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13.84 c m

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17.32 c m

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17.39 c m

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Solution

The correct option is C $17.32 c m$
Given, trapezium ABCD, $A D = 20$ , $∠ A = 60^{\circ}$
Draw perpendicular from $D$ on $A B$ , which meets $A B$ at $E$
Now, In $△ A D E$ ,
$sin 60^{\circ} = \frac{D E}{A D}$
$\frac{\sqrt{3}}{2} = \frac{D E}{A D}$
$D E = 10 \sqrt{3}$
$D E = 17.32 c m$

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In trapezium ABCD, as shown, AB//DC, AD=DC=BC=20 cm and ∠A=60∘. Distance between AB and DC in cm.

The correct option is C 17.32 cmGiven, trapezium ABCD, AD=20, ∠A=60∘Draw perpendicular from D on AB, which meets AB at ENow, In △ADE,sin60∘=DEAD√32=DEADDE=10√3DE=17.32 cm

In trapezium $A B C D$ , as shown, $A B / / D C$ , $A D = D C = B C = 20 c m$ and $∠ A = 60^{\circ}$ . Distance between $A B$ and $D C$ in cm.

The correct option is C $17.32 c m$
Given, trapezium ABCD, $A D = 20$ , $∠ A = 60^{\circ}$
Draw perpendicular from $D$ on $A B$ , which meets $A B$ at $E$
Now, In $△ A D E$ ,
$sin 60^{\circ} = \frac{D E}{A D}$
$\frac{\sqrt{3}}{2} = \frac{D E}{A D}$
$D E = 10 \sqrt{3}$
$D E = 17.32 c m$