The sides of an equilateral triangle ABC are 12 cm each. D is the foot of the perpendicular from A to BC and E is the mid-point of AD. BE is __________________.

12 cm

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6 \sqrt{2} c m

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\sqrt{63} c m

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None of the above

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Solution

The correct option is C $\sqrt{63} c m$
AD=12sin60∘=12×3√2=63√cm
∴ED=12⋅AD=12×63√=33√cm
In △BED,
<img>
BE=BD2+ED2−−−−−−−−−−√=62+27−−−−−−√=63−−√cm

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The sides of an equilateral triangle ABC are 12 cm each. D is the foot of the perpendicular from A to BC and E is the mid-point of AD. BE is __________________.

The correct option is C √63cmAD=12sin60∘=12×3√2=63√cmAD=12sin⁡60∘=12×32=63cm∴ED=12⋅AD=12×63√=33√cm∴ED=12⋅AD=12×63=33cmIn △BED△BED,<img>BE=BD2+ED2−−−−−−−−−−√=62+27−−−−−−√=63−−√cm

The correct option is C $\sqrt{63} c m$
AD=12sin60∘=12×3√2=63√cm
∴ED=12⋅AD=12×63√=33√cm
In △BED,
<img>
BE=BD2+ED2−−−−−−−−−−√=62+27−−−−−−√=63−−√cm