The side of an equilateral triangle is 4-3 cm- The length of a perpendicular drawn from any vertex to the opposite side will be

The correct option is B 6 cm Given, AB=4√( 3 ) and nbsp;BD=2√( 3 ) By Pythagoras theorem,we get, ∴ AD=√([ 4√(3) ]^2 [ 2√(3) ]^2) =√(48 1 ...

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Standard VII

Mathematics

Pythagorean Theorem

The side of a...

Question

The side of an equilateral triangle is $4 \sqrt{3} c m$ . The length of a perpendicular drawn from any vertex to the opposite side will be

4 c m

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6 c m

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8 c m

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9 c m

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Solution

The correct option is B $6 c m$
Given,
$A B = 4 \sqrt{3}$ and
$B D = 2 \sqrt{3}$
By Pythagoras theorem,we get,
$∴ A D = \sqrt{{(\begin{matrix} 4 \sqrt{3} \end{matrix})}^{2} - {(\begin{matrix} 2 \sqrt{3} \end{matrix})}^{2}}$
$= \sqrt{48 - 12}$
$= \sqrt{36}$

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The side of an equilateral triangle is 4√3cm. The length of a perpendicular drawn from any vertex to the opposite side will be

The correct option is B 6cmGiven,AB=4√3 andBD=2√3By Pythagoras theorem,we get,∴AD=√(4√3)2−(2√3)2=√48−12=√36

The side of an equilateral triangle is $4 \sqrt{3} c m$ . The length of a perpendicular drawn from any vertex to the opposite side will be

The correct option is B $6 c m$
Given,
$A B = 4 \sqrt{3}$ and
$B D = 2 \sqrt{3}$
By Pythagoras theorem,we get,
$∴ A D = \sqrt{{(\begin{matrix} 4 \sqrt{3} \end{matrix})}^{2} - {(\begin{matrix} 2 \sqrt{3} \end{matrix})}^{2}}$
$= \sqrt{48 - 12}$
$= \sqrt{36}$