Question

the area of a equilateral triangle ABC is 50cm^2. Find AB. Please show the workout.

Answer 1

area of equilateral triangle= sqrt3/4 * (side)^2

Answer 2

\[50^2= \frac{ \sqrt3 }{ 4 }* (side AB)^2\]

Answer 3

@nafis.irteza can you do this?

Answer 4

let me see

Answer 5

is that some kind of formula you wrote?

Answer 6

yes...that is the area of equilateral triangle

Answer 7

area = 0.5 * base * height
= 0.5*x *h where x = length of each side
height h = sqrt(x^2 - (0.5x)^2)
50 = sqrt(x^2 - 0.25x^2)
50 = sqrt(0.75x^2)

solve for x
x = AB

Answer 8

@cwrw238 i guess AB is a side....so we don't need to do all these stuff....just area of equilateral triangle

Answer 9

sorry the equation should be
50 = 0.5 * x * sqrt(0.75x^2)

yes - but i think its instructive to see how the formula can be derived

Answer 10

cool :)

Answer 11

how did u get 0.75?
and is sqrt =square root?

Answer 12

sqrt = square root
x^2 - 0.25x^2 = 0.75x^2

(like 1 - 0.25 = 0.75)

Answer 13

@cwrw238 how did u figure this out (h = sqrt(x^2 - (0.5x)^2)

Answer 14

Area of the equilateral triangle = 50 square cm.
Area of the equilateral triangle = \[\frac{ \sqrt{3} }{4}*(side)^{2} = 50\]
\[(side)^{2} = 50*\frac{ 4 }{ \sqrt{3} }\]
\[side = \sqrt{50*\frac{ 4 }{ \sqrt{3} }}\]
\[side = \sqrt{200/\sqrt{3}}\]
Side = 10.75 approximately.

Answer 15

@Fall12-13 this really helped...but can u please work it out using trigonometry (sin,cos,tan)

Answer 16

@Nafis: I don't exactly remember that, sorry :(