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ABCD is a trapezium in which AD || BC and AB = DC = 10 m. then the distance of AD from BC is :
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- 10√2 m
- 4√2 m
- 5√2 m
- 6√2 m
- 10√2 m
Correct Option: C
According to question , we draw a figure of trapezium ABCD in which AD || BC and AB = DC = 10 m
AE ⊥ BC; DF ⊥ BC
∴ ∠DCB = 45°
In ∆ CDF,
| sin 45° = | |
| DC |
| ⇒ | = | ||
| √2 | 10 |
| ⇒ DF = | = 5√2 metre. | |
| √2 |
