'a' is length of side of an equilateral triangle ABC base BC lies on x axis & vertex B at origin find cordinat

Question

find coordinates of vertex A, B and C

Answer 1

You need to use the pythagorean theorem to find the height of an equilateral triangle. You know the hypotenuse is a, and the leg is a/2. The other leg (height) is a*sqrt(3) / 2.

You have:
(a/2)^2 + h^2 = a^2
a^2 / 4 + h^2 = a^2
h^2 = a^2 ( 1 - 1/4 )
h^2 = a^2 ( 3 / 4)
h = a*sqrt(3) / 2

From there it is straightforward to determine the coordinates:
B is at (0, 0) at the origin
C is at (a, 0) along the x-axis -- over length of base, up 0
A is at (a/2, a*sqrt(3)/2) -- over half the base, up the height of the triangle.

Answer 2

The key to this problem is the fact that your triangle is equilateral. Thus, all three sides and all three angles are equal. The problem tells you that B is at the origin so the coordinates for B are (0,0). Since BC lies on the x-axis, the coordinates for C are either (a, 0) or (-a, 0). The problem doesn't specify which direction the line stretches.

Now, for the fun part. A is going to be a units from the origin at an angle of 60 degrees. What we don't know is whether it's above or below the x-axis. Imagine drawing a line from A to the base of the triangle. Since ABC is equilateral, the line will be the perpendicular bisector of BC. Thus, the x-coordinate for A will be either a/2 or
-a/2. To solve for y, we can use the Pythagorean theorem because we have a right triangle with a hypoteneuse of a units and a leg of a/2 units.

Thus, a^2 = (a/2)^2 + y^2. Solving for y gives us
y^2 = a^2 - a^2/4 = 4a^2/4 - a^2/4 = 3a^2/4
y = [a (3^.5)/2] Note that raising to a power of .5 or 1/2 is the same as taking the square root.

So, the possible cooridinates for A are
(+/- a/2, +/- [a(3^.5)/2])

Answer 3

Express the Area of an equilateral triangle as function of the length x of a side. First, find the height of the triangle. Split the triangle in half so it becomes two right triangles with side lengths x, x/2, and ?. a^2 + b^2 = c^2 a^2 + 1/4(x^2) = x^2 3x^2/4 = a^2 (sqrt(3)x)/2 = a The formula for the area of a triangle is bh/2 The base is x and the height is (sqrt(3)x)/2 So it becomes (x)((sqrt(3)x)/2)/2 =(sqrt(3)x)/4 I think that is right.

Answer 4

Since B lies at origin therefore coordinate of B is (0,0)
Since ABC is an equil. triangle therefore,AB=a,AC=a,BC=a
Draw a perpendicular AD from A to BC then D will be the midpoint of BC,therefore BD=a/2
Now ABD is a right triangle with AB=a,BD=a/2
By Pythagoras Th.
AD^2+BD^2=AB^2
AD^2=AB^2-BD^2
=a^2-(a/2)^2
=3(a^2)/4
AD=(sqrt of 3)a/2
If ABC lies in
Ist quad then A(a/2,(sqrt of 3)a/2),B(0,0),C(a,0)
2nd quad then A(-a/2,(sqrt of 3)a/2),B(0,0),C(-a,0)
3rd quad then A(-a/2,-(sqrt of 3)a/2),B(0,0),C(-a,0)
4th quad then A(a/2,-(sqrt of 3)a/2),B(0,0),C(a,0)

Answer 5

Equilateral triangle means length of AB=BC=CA=a
B is at origin means its co-ordinate is (0,0)
BC is at x-axis and B is at origin means co-ordinate of C is (a,0)
Co-ordinate of A is ((a/2),[sqrt((a^2)-((a/2)^2)])

Answer 6

The co-ordinates will be
B(0,0)

case 1:If in 1st quad:

Now the base is of length a so half=a/2
=>A(a/2,a)
=>C(a,0)

case 2:If in 2nd quad:

B(0,0)
=>C(-a,0)
=>A(-a/2,a)

case 3: if in 3rd quad:

B(0,0)
=>C(-a,0)
=>A(-a/2,-a)

case 4: if in 4th quad:

B(0,0)
=>C(a,0)
=>A(a/2,-a)

Answer 7

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

Answer 8

vertex C is (0,a) vertex B is (0,0), and vertex A is (a/2,sqrt(a^2-a/2^2)