Question

The length of altitude of a equilateral triangle of side a unit is ?

Answer 1

Answer:

The length of altitude of a equilateral triangle of side a unit is

Step-by-step explanation:

1/2×b×h=root 3/4×a^2

Base ='a' because the side of triangle is a

1/2×a×h=root 3/4a^2

a/2×h=root 3/4a^2

h=root 3/4a^2×2/a

h=root 3/2a

Answer 2

Step-by-step explanation:

From the given question we have to find the altitude of an equilateral triangle having side measuring ‘a’.

let us assume that ABC is an equilateral triangle and AD is an altitude on side BC in triangle ABC.

We know that the altitude of the equilateral triangle bisects the base which means altitude from A to BC bisects BC

$BD=CD= \frac{a}{2}$

we have two right angled triangles ΔADB and ΔADC.

FromΔADB, by applying pythagoras theorem.

(height)2+(base)2=(hypotenuse)2

Therefore, in ΔADB, we have

${AB}^{2} = {BD}^{2} + {AD}^{2}$

Here AB=a, BD=a/2

${AD}^{2} = {AB}^{2} - {BD}^{2}$

${AD}^{2} = {a}^{2} - ({ \frac{a}{2} })^{2}$

by simplifying

${AD}^{2} = \frac{4 {a}^{2} - {a}^{2} }{4}$

${AD}^{2} = \frac{3 {a}^{2} }{4}$

${AD} = \frac{ \sqrt{3}a }{2}$

Therefore for an equilateral triangle having each side equal to a, we get a length of an altitude as

${AD} = \frac{ \sqrt{3}a }{2}$

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