Recall that the perimeter is the sum of all the exterior sides of a shape. The sides that add up to the perimeter are highlighted in red.

Since the equilateral triangle shares a side with the square, each of the five sides that are outlined have the same length.

Recall that the height of an equilateral triangle splits the triangle into \(\displaystyle 2\) congruent \(\displaystyle 30-60-90\) triangles.

We can then use the height to find the length of the side of the triangle.

Recall that a \(\displaystyle 30-60-90\) triangle has sides that are in ratios of \(\displaystyle 1:\sqrt3:2\). The smallest side in the given figure is the base, the second longest side is the height, and the longest side is the side of the triangle itself.

Thus, we can use the ratio and the length of the height to set up the following equation:

\(\displaystyle \frac{\text{height}}{\sqrt3}=\frac{\text{side}}{2}\)

\(\displaystyle \text{side}=\frac{2(\text{height})}{\sqrt3}=\frac{2\sqrt3(\text{height})}{3}\)

Plug in the given height to find the length of the side.

\(\displaystyle \text{side}=\frac{2\sqrt3(18)}{3}=12\sqrt3\)

Now, since the perimeter of the shape consists of \(\displaystyle 5\) of these sides, we can use the following equation to find the perimeter.

\(\displaystyle \text{Perimeter}=5(\text{side})\)

\(\displaystyle \text{Perimeter}=5(12\sqrt3)=60\sqrt3=103.92\)

Below is regular Pentagon \(\displaystyle PENTA\) with center \(\displaystyle C\), a segment drawn from \(\displaystyle C\) to each vertex - that is, each of its radii drawn.

The measure of each angle of a regular pentagon can be calculated by setting \(\displaystyle N\) equal to 5 in the formula

\(\displaystyle \frac{(N-2)180^{\circ }}{N}\)

and evaluating:

\(\displaystyle \frac{(5-2)180^{\circ }}{5} = \frac{3 \cdot 180^{\circ }}{5} = \frac{540^{\circ }}{5} = 108^{\circ }\)

By symmetry, each radius bisects one of these angles. Specifically, 

\(\displaystyle m \angle CTN = \frac{1}{2} m \angle ATN = \frac{1}{2} \cdot 108^{\circ } = 54 ^{\circ }\).

An equilateral triangle has three angles of measure \(\displaystyle 60^{\circ }\), so \(\displaystyle \bigtriangleup CTN\) is not equilateral.

Assume \(\displaystyle m \angle A = 60^{\circ }\). Then, since \(\displaystyle AB = AC\), it follows by the Isosceles Triangle Theorem that their opposite angles are also congruent. Since the measures of the angles of a triangle total \(\displaystyle 180^{\circ }\), letting \(\displaystyle t = m \angle B = m \angle C\):

\(\displaystyle m \angle A + m \angle B + m \angle C = 180^{\circ }\)

\(\displaystyle 60^{\circ }+ t+t = 180^{\circ }\)

\(\displaystyle 60^{\circ }+2t = 180^{\circ }\)

\(\displaystyle 60^{\circ }+2t - 60^{\circ } = 180^{\circ } - 60^{\circ }\)

\(\displaystyle 2t = 120^{\circ }\)

\(\displaystyle \frac{2t }{2}= \frac{120^{\circ }}{2}\)

\(\displaystyle t = 60^{\circ }\)

All three angles have measure \(\displaystyle 60^{\circ }\), making \(\displaystyle \bigtriangleup ABC\) equiangular and, as a consequence, equilateral. Therefore, \(\displaystyle BC = AB = AC = 18\), and the perimeter, or the sum of the lengths of the sides, is

\(\displaystyle P= AB + BC + AC = 18 + 18+18 = 54\)

However, the perimeter is given to be 56. Therefore, \(\displaystyle m \angle A \ne 60^{\circ }\).

The figure referenced is below:

\(\displaystyle \bigtriangleup ABC\) is equilateral, so \(\displaystyle \overline{AB} \cong \overline{AC} \cong \overline{BC}\). \(\displaystyle \bigtriangleup BCD\) is equilateral, so \(\displaystyle \overline{BC} \cong \overline{BD} \cong \overline{CD}\). By the Transitive Property of Congruence, \(\displaystyle \overline{AB} \cong \overline{AC} \cong \overline{BD} \cong \overline{CD}\). A quadrilateral with four congruent sides is a parallelogram and a rhombus. However, it is not a rectangle, and, consequently, not a square, since its angles are not right - \(\displaystyle \angle A\), an angle of an equilateral triangle, measures \(\displaystyle 60 ^{\circ }\). Also, a parallelogram is not a trapezoid. Therefore, the quadrilateral is a parallelogram and a rhombus only.

A midsegment of a triangle - a segment whose endpoints are the midpoints of two sides - is, by the Triangle Midsegment Theorem, parallel to the third side, and is half the length of that side. An equilateral triangle with perimeter 30 has three sides one third this, or 

\(\displaystyle \frac{1}{3} \cdot 30 = 10\).

Consequently, the length of each midsegment is half this, or

\(\displaystyle \frac{1}{2} \cdot 10 = 5\).

Equilateral triangles have sides of all equal length and angles of 60°. To find the area, we can first find the height. To find the height, we can draw an altitude to one of the sides in order to split the triangle into two equal 30-60-90 triangles.

Now, the side of the original equilateral triangle (lets call it "a") is the hypotenuse of the 30-60-90 triangle. Because the 30-60-90 triange is a special triangle, we know that the sides are x, x\(\displaystyle \sqrt{3}\), and 2x, respectively.

Thus, a = 2x and x = a/2.
Height of the equilateral triangle = \(\displaystyle \frac{a\sqrt{3}}{2}\)

Given the height, we can now find the area of the triangle using the equation:
\(\displaystyle Area = \frac{1}{2}bh=\frac{a^2\sqrt{3}}{4}\)

Equilateral triangles have sides of equal length, with angles of 60°. To find the area, we can first find the height. To find the height, we can draw an altitude to one of the sides in order to split the triangle into two equal 30-60-90 triangles.

Now, the side of the original equilateral triangle (lets call it "a") is the hypotenuse of the 30-60-90 triangle. Because the 30-60-90 triange is a special triangle, we know that the sides are x, x\(\displaystyle \sqrt{3}\), and 2x, respectively.

Thus, a = 2x and x = a/2.
Height of the equilateral triangle = \(\displaystyle \frac{a\sqrt{3}}{2}\)

Given the height, we can now find the area of the triangle using the equation:
\(\displaystyle Area = \frac{1}{2}bh=\frac{a^2\sqrt{3}}{4}\)

The answer is \(\displaystyle 81\sqrt{3}\) .  

To find the area you would first need to find what the length of each side is: 54 divided by 3 is 18 for each side.  

Then you would need to draw in the altitude of the triangle in order to get its height.  Drawing this altitude will create two 30-60-90 degree triangles as shown in the picture.  The longer leg is \(\displaystyle \sqrt{3}\) times the short leg.  Thus the height is \(\displaystyle 9\sqrt{3}\). 

Next we plug in the base and the height into the formula to get 

\(\displaystyle \frac{1}{2}\cdot \left 18\right\cdot 9\sqrt{3}=81\sqrt{3}\ in^{2}\)

The formula for the area of an equilateral triangle with side length \(\displaystyle a\) is

\(\displaystyle Area=\frac{\sqrt{3}}{4}a^2\)

So, since \(\displaystyle a=2\:cm\),

\(\displaystyle Area=\frac{\sqrt{3}}{4}(2)^2\)

\(\displaystyle Area=\frac{\sqrt{3}}{4}*4=\sqrt{3}\:cm^2\)

The formula of the area of an equilateral triangle is \(\displaystyle \frac{\sqrt{3}}{4}x^2\) if \(\displaystyle x\) is a side.

Since the sides of our triangle have doubled, they have changed from \(\displaystyle a\) to \(\displaystyle 2a\). We can substitute \(\displaystyle 2a\) into the equation and solve for the triangle's new area in terms of \(\displaystyle a\):

\(\displaystyle area=\frac{\sqrt{3}}{4}x^2=\frac{\sqrt{3}}{4}(2a)^2=\frac{\sqrt{3}}{4}4a^2=\sqrt{3}a^2\)