+2441 I have figured out an alternate method of solving this problem, too! Shout out to Cphill and hectictar for attempting this! I have created a diagram using the programs of paint and GeoGebra. The diagram consolidates all the given information, and I have added a few segments to reference as I solve. Note that \(\overline{BG}\perp\overline{EC}\hspace{1mm}\text{and}\hspace{1mm}\overline{AF}\perp\overline{EC}\):
If we could figure out the length of \(\overline{EC}\), then we would only need to apply a few formulas to figure out the total area. I have not marked it in the diagram, but we also know by given info that \(m\angle EAB=m\angle CBA=120^{\circ}\).
For now, let's only worry about figure \(ABCE\). Since both \(\overline{AF}\) and \(\overline{BG}\) are altitudes (a segment that extends perpendicularly from a vertex to a segment opposite of that vertex) we can conclude that \(\overline{AB}\parallel\overline{EC}\). Since we have a quadrilateral that has exactly one pair of opposite sides parallel, then the quadrilateral is a trapezoid, by definition. We can actually go a step further than this, actually! Since the trapezoid's nonparallel sides are of equal length, this is an isosceles trapezoid. Classifying this quadrilateral this specifically actually aides us in solving this problem.
Isosceles trapezoids have a few properties associates with them. One of them is that adjacent angles along the nonparallel sides are supplementary. In other words, \(m\angle BAE+m\angle AEF=180^{\circ}\) and \(m\angle ABC+m\angle BCG=180^{\circ}\). Let's solve for those unknown angles.
| \(m\angle BAE+m\angle AEF=180^{\circ}\) | \(m\angle ABC+m\angle BCG=180^{\circ}\) | Plug in the known values and solve. |
| \(120+m\angle AEF=180^{\circ}\) | \(120+m\angle BCG=180^{\circ}\) | |
| \(m\angle AEF=60^{\circ}\) | \(m\angle BCG=60^{\circ}\) |
Great! By using the triangle sum theorem, we can figure out the \(m\angle EAF\) and \(m\angle CBG\). Let's do that:
| \(m\angle{AFE}+m\angle{FEA}+m\angle EAF=180^{\circ}\) | \(m\angle BGC+m\angle GCB+ m\angle CBG=180^{\circ}\) | This is the traingle sum theorem in use. Now, solve for the unknown. |
| \(90^{\circ}+60^{\circ}+m\angle EAF=180^{\circ}\) | \(90^\circ+60^{\circ}+ m\angle CBG=180^{\circ}\) | |
| \(150^{\circ}+m\angle EAF=180^{\circ}\) | \(150^{\circ}+m\angle CBG=180^{\circ}\) | |
| \(m\angle EAF=30^{\circ}\) | \(m\angle CBG=30^{\circ}\) | |
Do you notice something interesting about these angle measures? If not, I will attempt to make it clearer by extracting one of the triangles from the diagram, as done below:
Yes, it is a 30-60-90 triangle! We can use this to our advantage. In a 30-60-90 triangle, the ratio of the lengths of the sides is \(1:\sqrt{3}:2\). This means that we can determine the length of both \(\overline{EF}\) and \(\overline{AF}\). Let's solve for both lengths now.
| \(2EF=AE\) | This equation is derived from the constant ratio of a 30-60-90 triangle. Of course, AE=2, so substitute that in. |
| \(2EF=2\) | Divide by 2 on both sides to determine the length of EF. |
| \(EF=1\) |
Now, let's solve for AF.
| \(AF=EF\sqrt{3}\) | EF=1, so plug that in. |
| \(AF=1*\sqrt{3}\) | Simplify. |
| \(AF=\sqrt{3}\) |
The same phenomenon occurs for \(\triangle BCG\); \(BG=\sqrt{3}\) and \(CG=1\). This means that \(EC=2+1+1=4\), the length of the opposite base.
Now, we can calculate the area of this trapezoid by using the following formula \(A_{trap.}=\frac{1}{2}(b_1+b_2)h\)
A = area of a trapezoid
b1 = the length of one base
b2 = the length of the other base
h = perpendicular height
We already have all this information, so we can solve for the area of the trapezoid.
| \(A_{trap.}=\frac{1}{2}(b_1+b_2)h\) | As aforementioned, this is the formula. Now, plug in all the values. |
| \(A_{trap.}=\frac{1}{2}(2+4)\sqrt{3}\) | Simplify in the parentheses first. |
| \(A_{trap.}=\frac{1}{2}*6\sqrt{3}\) | 1/2 and 6 are like terms. |
| \(A_{trap.}=(3\sqrt{3})units^2\) | This is the area of the trapezoid in siplest radical form. Of course, dont forget units! |
We are not done yet because we have not considered \(\triangle DEC\) at all yet; remember, this is a composite figure! I have a zoomed-in diagram that extracts the triangle.
Equilateral triangles also have a formula to them, too. This simplifies the process significantly. The formula is the following
\(A_{\triangle}=\frac{\sqrt{3}}{4}s^2\)
A = area of an equilateral triangle
s = side length
Of course, let's apply it now!
| \(A_{\triangle}=\frac{\sqrt{3}}{4}s^2\) | Plug in 4 for s. |
| \(A_{\triangle}=\frac{\sqrt{3}}{4}*4^2\) | Simplify the exponent first. |
| \(A_{\triangle}=\frac{\sqrt{3}}{4}*\frac{16}{1}\) | Notice that 4 and 16 has a GCF of 4. |
| \(A_{\triangle}=\frac{\sqrt{3}}{1}*\frac{4}{1}=(4\sqrt{3})units^2\) | |
The last thing to do is to add both areas of the individual shapes in this composite figure together, and that is your final answer!
\(A=3\sqrt{3}+4\sqrt{3}=(7\sqrt{3})units^2\)
This is your final answer!
+2441 Based on how you wrote the equation, I interpret your equation to be \(\frac{7}{8}w=14\). Of course, we are solving for w by isolating it:
| \(\frac{7}{8}w=14\) | Multiply by 8/7 on both sides of the equation. Doing this will eliminate the fraction. |
| \(w=\frac{14}{1}*\frac{8}{7}\) | Notice that the 14 and 7 have a GCF of seven. Noticing this will ease the simplification process. |
| \(w=\frac{14}{1}*\frac{8}{7}=\frac{2}{1}*\frac{8}{1}=16\) | |
Therefore, \(w=16\).
+2441 I believe you want to evaluate the expression \(\frac{\frac{\frac{2}{3}}{-3}}{4}\).
When expressions are difficult to evaluate, I like to set the value to a number (I will use x)
| \(x=\frac{\left(\frac{\left(\frac{2}{3}\right)}{-3}\right)}{4}\) | Multiply by 4 on both sides of the equation. |
| \(4x=\frac{\left(\frac{2}{3}\right)}{-3}\) | Multiply by -3 on both sides on both sides. |
| \(-12x=\frac{2}{3}\) | Multiply by 3 on both sides. |
| \(-36x=2\) | Divide by -36 on both sides to isolate x |
| \(x=-\frac{2}{36}=-\frac{1}{18}=-0.0\overline{55}\) | |
I like doing this because it can sometimes make a seemingly hard problem easier. However, you do not have to do this. You can use your knowledge of fractions and its respective rules to solve this expression, too.
| \(\frac{\left(\frac{\left(\frac{2}{3}\right)}{-3}\right)}{4}\) | First, let's simply just worry about (2/3)/(-3) |
| \(\frac{\frac{2}{3}}{-3}\) | Use the fraction rule that \(\frac{\frac{a}{b}}{c}=\frac{a}{b*c}\) |
| \(\frac{2}{3*-3}=\frac{2}{-9}\) | Now, insert this into the equation for (2/3)/(-3) |
| \(\frac{\frac{2}{-9}}{4}\) | Let's apply the same fraction rule as before. |
| \(\frac{\frac{2}{-9}}{4}=\frac{2}{-9*4}=\frac{2}{-36}\) | Simplify the fraction by noting that both the numerator and denominator are even, so the GCF is, at least, 2. |
| \(-\frac{2}{36}=-\frac{1}{18}=-0.0\overline{55}\) | |
+2441 I will evaluate the expression \((-4-0)*(-10)/-5\).
| \((-4-0)*(-10)/-5\) | Of course, adding or subtracting 0 from a number yields the same number. |
| \(-4*-10/-5\) | I have removed the unnecessary sets of parentheses. Remembering that a negative multiplied by a negative number results in a positive. -4*-10=40 |
| \(40/-5\) | 40/-5=-8 |
| \(-8\) | |
