geometry hw help plz :)?

Question

verify that the points (0,0), (a,0) and (a/2,square root of 3a/2) are the vertices of an equilateral triangle. then show that the midpoints of the three sides are the vrtices of a seconmd equilateral triangle. I was wondering if any one can walk me through this plz :)

Answer 1

use the distance formula

d² = (x1-x2)² + (y1-y2)²

A (0,0)
B (a,0)
C (a/2, sqrt(3) a /2)

lets take care of the easy one first AB = a
AC² = (a/2)² + (sqrt(3) a/2)² = a² thus AC = a
BC² = (a/2 - a)² + (sqrt(3) a/2)² = a² thus BC = a

three equal sides - equilateral triangle

now find the midpoints of those sides and do the same operation to solve it... goodluck

Answer 2

Let A(0,0), B(a,0), C(a/2, sq rt 3a/2) are three vertices's of a triangle ABC. Therefore
AB^2=(a-0)^2 + (0-0)^2=a^2
BC^2 = (a/2-a)^2 + (sq rt 3a/2-0)^2=a^2/4+3a^/4=a^2
CA^2 = (0-a/2)^2+ (0-sq rt3/2)^2=a^2/4+3a^/4=a^2
Since AB^2=BC^2=CA^2 therefore triangle ABC is a equilateral
Let D, E, F are mid points of AB,BC,CA then D(0,a/2), E(3a/4,sq rt 3/4) F(a/4.sq rt 3/4)
Now you find the distances DE, EF, FD if they are equal then triangle DEF is equilateral.

Answer 3

Distance formula

(0,0) and (a,0)
D = sqrt((a - 0)^2 + (0 - 0)^2)
D = sqrt(a^2)
D = a

(a,0) and ((a/2),(sqrt((3a)/2))
D = sqrt(((a/2) - a)^2 + ((sqrt(3)a)/2) - 0)^2)
D = sqrt(((a/2) - (2a/2))^2 + ((sqrt(3)a)/2)^2)
D = sqrt((-a/2)^2 + ((3a^2)/4))
D = sqrt(((a^2)/4) + ((3a^2)/4))
D = sqrt((a^2 + 3a^2)/4)
D = sqrt((4a^2)/4)
D = sqrt(a^2)
D = a

(0,0) and ((a/2), (sqrt(3)a/2))
D = sqrt(((a/2) - 0)^2 + ((sqrt(3)a/2) - 0)^2)
D = sqrt((a/2)^2 + ((sqrt(3)a)/2)^2)
D = sqrt(((a^2)/4) + ((3a^2)/4))
D = sqrt((a^2 + 3a^2)/4)
D = sqrt(4a^2/4)
D = sqrt(a^2)
D = a

So since they are all equal, this is an equilateral triangle.

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Midpoint formula

M = (x2 + x1)/2, (y2 + y1)/2

M = (a + 0)/2 , (0 + 0)/2)
M = (a/2),0

M = ((a/2) + a)/2 , ((sqrt(3)a/2) + 0)/2
M = ((a + 2a)/2)/2 , ((sqrt(3)a/2)/2
M = ((3a/2)/2) , ((sqrt(3)a/2)/(2/1)
M = ((3a/2)/(2/1)), ((sqrt(3)a/2)*(1/2))
M = ((3a/2)*(1/2)) , ((sqrt(3)a)/4)
M = ((3a)/4) , ((sqrt(3)a)/4)

M = ((a/2) + 0)/2 , ((sqrt(3)a)/2) + 0)/2
M = ((a/2)/2) , (sqrt(3)a)/4
M = (a/4) , (sqrt(3)a)/4

((a/2),0), ((3a)/4) , ((sqrt(3)a)/4)), ((a/4) , (sqrt(3)a)/4)


((a/2),0), ((3a)/4) , ((sqrt(3)a)/4))

D = sqrt(((3a)/4) - (a/2))^2 + (((sqrt(3)a/4) - 0)^2)
D = sqrt(((3a)/4) - (2a/4))^2 + ((sqrt(3)a/4)^2
D = sqrt(((3a - 2a)/4)^2 + ((3a^2)/16)
D = sqrt((a/4)^2 + ((3a^2)/16))
D = sqrt(((a^2)/16) + ((3a^2)/16))
D = sqrt((a^2 + 3a^2)/16)
D = sqrt((4a^2)/16)
D = sqrt((a^2)/4)
D = (a/2)

((3a)/4) , ((sqrt(3)a)/4)), ((a/4) , (sqrt(3)a)/4)

D = sqrt(((a/4) - ((3a)/4))^2 + (((sqrt(3)a)/4) - ((sqrt(3)a)/4))^2)
D = sqrt(((a - 3a)/4)^2 + 0^2)
D = sqrt((-2a/4)^2)
D = sqrt((-a/2)^2)
D = sqrt((a^2)/4)
D = (a/2)

((a/4) , (sqrt(3)a)/4) and ((a/2),0)

D = sqrt(((a/2) - (a/4))^2 + (0 - ((sqrt(3)a)/4))^2)
D = sqrt(((2a)/4) - (a/4))^2 + (-(sqrt(3)a)/4)^2
D = sqrt(((2a - a)/4)^2 + ((3a^2)/16))
D = sqrt((a/4)^2 + ((3a^2)/16))
D = sqrt(((a^2)/16) + ((3a^2)/16))
D = sqrt((a^2 + 3a^2)/16)
D = sqrt((4a^2)/16)
D = sqrt((a^2)/4)
D = (a/2)

As you can see, the midpoints all have the same value, so they too form an equilateral triangle.

Answer 4

distance formula

D= sqtr( (X2-X1) + (Y2-Y1)

A (0,0)
B (a,0)
C (a/2, sqrt(3) a /2)

lets take care of the easy one first AB = a
AC = sqrt. ( (a/2)² + (sqrt(3) a/2)² )= a
BC² = sqrt. ( (a/2 - a)² + (sqrt(3) a/2)² )= a


Mid point formula is,

(X1+x2)/2, (y1+y2)/2

A (0,0)
B (a,0)
C (a/2, sqrt(3) a /2)

M.P of AB = A/2, 0
M.P of AC= A/4, sqrt(3)a/4
M.P of BC = 2a/4 or A/2, sqrt (3)a/4

you shouldn't need to find the distance to prove they are equal. there is a proof for that. i can't remember the name of it. but im pretty sure there is a proof.