find an equilateral triangle that has the same area as its perimeter?

Question

Please assist me with this problem as well...

Answer 1

Area of a triangle = 1/2bh, b=base, h=height.

Perimeter = 3b, for an equilateral triangle, since all sides have the same length, equal to the length of the base

To get the height: an equilateral triangle has all internal angles equal, so they must be 60 degrees (since sum of the angles must be 180). You can bisect the triangle into two 30-60-90 triangles by drawing a line segment from the top to the base, and its length is the height. The hypotenuse has length b, and the length of the other leg is half the base, b/2. From the Pythagorean theorem, h^2 + (b/2)^2 = b^2, so h=bsqrt(3)/2.

Therefore, the area is (sqrt(3)/4)b^2.

For the perimeter to equal the area, (sqrt(3)/4)b^2 = 3b, or (sqrt(3)/4)b = 3. Therefore, b=4*sqrt(3).

Answer 2

it isn't the comparable because of the fact the fringe. if the fringe is eighteen, then, b/c that's equilateral, the two area is 6 (18/3=6). Then use the Pythagorean Theorem to locate the top of the triangle, (the part of a triangle is comparable to a million/2 the backside expanded by the top), so which you draw a right now line down from the acceptable perspective of the triangle to the backside, to create 2 ideal triangles. for that reason, 3^2 (a million/2 base ^2) + b^2 (the triangle's top) = 6^2 (the scale of the hypotenuse of the desirable triangle you created), for that reason 9 + b^2 = 36, so b^2=27, so the top of the triangle is the sq. root of 27, or 3 * the sq. root of three. Then plug this into the section formula for a triange: a million/2 base * top, to get a million/2 * 6 * 3 * the sq. root of three, to get 9 * the sq. root of three as your answer. --------------------------------------... RANDY-- i understand you could in basic terms use the Pythagorean theorem on a ideal triangle. besides the undeniable fact that, in case you draw a line from the acceptable vertice of the equilateral triangle, to the backside of the triangle, so as that the line is perpendicular to the final diagnosis of the triangle, you have created 2 ideal triangles. the line you drew will bisect the backside of the triangle, arising 2 completely equivalent halves, for that reason you could divide 6 by 2 to get 3, so each of those halves of the backside measures 3. then you definately use between the desirable triangles created by bisecting the backside and you have the backside and the hypotenuse (one area of the triangle), so which you plug them into the Pythagorean theorem to get the unknown area of the desirable triangle, this is the top of the equilateral triangle.

Answer 3

I assume you mean the same numerical answer because area is square measure & perimeter is linear measure.

p=3s
A=1/2 bh
1/2*s*h
h sqrt(3)/2 * s
A=sqrt(3)/4 * s^2=3s
Sqrt(3)/4s=3s
s=12/sqrt(3)=4sqrt(3)=6.928

Answer 4

area for equliateral triangle is sqrt(3)/4* b^2

3b = sqrt(3)/4 * b^2, b = 4*sqrt(3)

Answer 5

...the two are measured in completely different units