+2446 Ok, I will assume for this problem that the 5 is labeled in inches, even though it does not say in the diagram. Yet again, I'll attempt to create a makeshift diagram like I did for you before.
A 5in B
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/ | | \
/ | | \
/ | (2*sqrt(3))in | \
/ | | \
/ | | \
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D E F C
Let's remember what an altitude is; an altitude is the height that runs perpendicularly from a base to a vertex of the figure. Therefore, \(m\angle AED =90^{\circ}\). We already know that \(m\angle D=60^{\circ}\) by the given info. Of course, by the triangle sum theorem, \(m\angle DAE=(180-(90+60))^{\circ}=30^{\circ}\).
What does this mean? We have identified a 30-60-90 triangle in the diagram, and that is a special one! Let me explain with a diagram:
Source: http://study.com/cimages/multimages/16/30-60-90-example-diagram.png
This diagram shows the relationship between the sides and the angle measures.
1) The angle across 30 is x
2) The angle across 60 is \(x\sqrt{3}\)
3) The angle across the right angle is 2x.
This means that the ratio of the sides of a 30-60-90 triangle is \(1:\sqrt{3}:2\). Let's use this information to solve for some missing sides.
| \(DE\sqrt{3}=AE=2\sqrt{3}\) | This is the ratios of the side lengths determined above. |
| \(DE\sqrt{3}=2\sqrt{3}\) | Divide by \(\sqrt{3}\) on both sides. |
| \(DE=2\) | |
We have determined the length of DE, 2. Now, let's determine the length of the hypotenuse of \(\triangle ADE\) by using the ratio of the side lengths.
| \(2DE=AD\) | Substitute the value that we calculated for DE, 2. |
| \(AD=2*2=4\) | |
Note that the same relationship occurs for the other triangle, \(\triangle BFC\), so \(FC=2\) and \(CB=4\). There is only one length we haven't determined the length of. It is \(EF\).
The length of this is right in front of us! \(AB=EF=5\). Now, add all of these together!
| \(P=(5+4+2+5+2+4)in\) | Add these lengths together to get the final perimeter of the trapezoid, in inches. |
| \(P=22in\) | |
Bam! Done!
+2446 Do you mean \(2-x-2=4\)?
If so, I will solve it as so:
| \(2-x-2=4\) | Combine the like terms on the left hand side of the equation |
| \(-x=4\) | Divide by -1 on both sides to get rid of the negative on the x. |
| \(x=-4\) | |
If you are ever unsure if you have the correct value for your unknown, just plug it back into the equation:
| \(2-(-4)-2=4\) | Check to see if this is true; if it is wrong, you either made a mistake or you have an extraneous solution. |
| \(2+4-2=4\) | Subtracting a negative number is the equivalent of adding a positive one. |
| \(6-2=4\) | |
| \(4=4\) | 4=4 is a true statement and concludes that the value for x is correct. |
+2446 If you are going to list decimals, I would highly advise separating them with commas, as opposed with a numbered list. The decimals are confusing to read. If you want to use a numbered list still, I would do what I did below.
This is how I would revise your question:
Write each decimal as a fraction in simplest form.
1) 0.222
2) 0.1515
3) .0242424
4) 0.5555
5) -0.124124124
This is what I perceive the intended decimals to be, but I am not entirely sure, so correct me if I am wrong.
I will start with the first decimal, 0.222.
| \(0.222\) | First, identify what place the decimal extends until (in this case, it extends to the thousandths place) and set it equal to a fraction over 1000. |
| \(0.222=\frac{222}{1000}\) | Now, identify the GCF of the fraction. It is harder with larger numbers, but both numbers are even, so we can, at least, divide by 2. |
| \(\frac{222}{1000} \div \frac{2}{2}=\frac{111}{500}\) | 111 and 500 do not share any common factors, except for 1, so this fraction is in simplest form. |
I will do the next decimal, 0.1515
| \(0.1515\) | Do the same process as above; the decimal extends until the ten thousandths place. Set it to a fraction over that amount, ten thousand. |
| \(0.1515=\frac{1515}{10000}\) | Yet again, it can be hard to determine the GCF, but both numbers end in a 5 or 0, so both are divisible by 5. |
| \(\frac{1515}{10000}\div \frac{5}{5}=\frac{303}{2000}\) | Yet again, there are no common factors greater than 1, so this fraction is irreducible. |
I will do the next decimal, as well.
| \(0.0242424\) | This extends to the ten millionth place, so put the number over a fraction over ten million. |
| \(0.0242424=\frac{242424}{10000000}\) | The numerator and denominator's final two digits are divisible by 4, so we can divide them by, at least, 4. |
| \(\frac{242424}{10000000}\div\frac{4}{4}=\frac{60606}{2500000}\) | We aren't done yet. The numerator and denominator are both even, so it is divisible by 2. |
| \(\frac{60606}{2500000}\div\frac{2}{2}=\frac{30303}{125000}\) | The numerator and denominator are co-prime, so this fraction is in simplest form. |
Here goes the next one:
| \(0.5555\) | The decimal extends to the ten thousandths place, so make a fraction over ten thousand. |
| \(0.5555=\frac{5555}{10000}\) | Both the numerator and denominator are divisible by 5 because both of them end in a 5 or a 0. |
| \(\frac{5555}{10000}\div \frac{5}{5}=\frac{1111}{2000}\) | 1111 and 2000 have no common factors, so the fraction is simplified completely. |
And, of course, here is the next one, -0.124124124.
| \(-0.124124124\) | This decimal extends to the billionth place, so set it over a billion. Just put the negative sign in front of the fraction. |
| \(-0.124124124=-\frac{124124124}{1000000000}\) | Both the numerator and denominator's final 2 digits are divisible by 4, so the fraction can be simplified. |
| \(-\frac{124124124}{1000000000}\div\frac{4}{4}=\frac{31031031}{250000000}\) | There are no more common factors. |
You are done now!
+2446 Ok, after some problem solving I have finally figured out the answer! The answer was right in front of me, too!
I will attempt to make a diagram, but I cannot guarantee that it will be clear.
C A D
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| |
H | |
| |
| |
| |
| |
| |
| |
| |
| | G
| |
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F B E
Ok, this is a diagram, and I have labeled the corners with letters so that it is clearer for you to understand!
If AB=12, that is the diagonal of the inner rectangle. Notice that AG=BH and BG=AH. We know that this by the properties of a rectangle; opposite sides of a rectangle are congruent. If all the triangles are right and isosceles, then we know that \(\triangle AGD \cong \triangle BHF\hspace{1mm}\text{and}\hspace{1mm}\triangle ACH \cong \triangle BEG\) because of angle-side-angle congruence theorem.
Let's remind ourselves, too, that these are, by the given information, right isosceles. This means that \(AD=DG=HF=FB\hspace{1mm}\text{and}\hspace{1mm}AC=CH=BE=EG\).
Let's label all the sides congruent to AD x and all sides congruent to AC y.
Let's find the length of the triangles' hypotenuses. Of course, we already know that \(AG=BH\hspace{1mm}\text{and}\hspace{1mm} BG=AH\), so we only need to calculate half of the diagonals. These are easier than they first appear, actually. A right isosceles triangle is 45-45-90 triangle, so the hypotenuses is the length of the leg times the square root of 2. Using this logic, \(AG=BH=x\sqrt{2}\hspace{1mm}\text{and}\hspace{1mm} BG=AH=y\sqrt{2}\).
Now that we have determined the lengths of the hypotenuses of the triangle. We can utilize the Pythagorean theorem, \(a^2+b^2=c^2\)
.
| \((x\sqrt{2})^2+(y\sqrt{2})^2=12^2\) | "Distribute" the square to all parts of the terms. |
| \(x^2*\sqrt{2}^2+y^2*\sqrt{2}^2=12^2\) | Simplify as much as possible. |
| \(x^2*2+y^2*2=144\) | |
| \(2x^2+2y^2=144\) | Divide by the GCF of all the terms, 2. You will see the significance of this shortly. |
| \(x^2+y^2=72\) | |
We are amazing close to the answer. To know the answer for sure, we must recognize something, first...
Well, let's find the area of all the triangles.
| \(\triangle ADG\) | \(\triangle GEB\) | \(\triangle BFH\) | \(\triangle HCA\) | |
| \(A=\frac{1}{2}*x*x\) | \(A=\frac{1}{2}*y*y\) | \(A=\frac{1}{2}*x*x\) | \(A=\frac{1}{2}*y*y\) | |
| \(A=\frac{1}{2}x^2\) | \(A=\frac{1}{2}y^2\) | \(A=\frac{1}{2}x^2\) | \(A=\frac{1}{2}y^2\) | |
Add these areas together and see what you get.
| \(\frac{1}{2}x^2+\frac{1}{2}y^2+\frac{1}{2}x^2+\frac{1}{2}y^2\) | Add the combined areas of the triangles. Combine like terms. |
| \(x^2+y^2\) | We have determined that the combined areas of the triangles are x^2+y^2 |
Do you see the significance of this? I do. We had determined in the previous calculation that \(x^2+y^2=72units^2\). Oh look! That's our answer! The combined area of all the triangles is \(72units^2\).
I sincerely hoped this helped you out.
+2446 Sorry, I did not respond sooner, but here is an explanation.
I also want to make sure that your question is clear. You want me to explain how to multiply \(\frac{4}{1}*-\frac{3}{2}\) in more detail, I think.
| \(\frac{4}{1}*-\frac{3}{2}\) | First, I am going to "tamper" with the -3/2. |
| \(-\frac{3}{2}\) | There is a fraction rule that says that \(-\frac{a}{b}=\frac{-a}{b}\). In other words, I am moving the negative sign to the numerator, which is valid. |
| \(-\frac{3}{2}=\frac{-3}{2}\) | Reinsert this into the original expression. |
| \(\frac{4}{1}*\frac{-3}{2}\) | When multiplying fractions, you simply multiply the numerator and the denominator. In general, \(\frac{a}{b}*\frac{c}{d}=\frac{ac}{bd}\). I will apply this rule in the next step. |
| \(\frac{4*-3}{1*2}\) | Now, evaluate the numerator and denominator separately. |
| \(4*-3=-12\hspace{1mm}\text{and}\hspace{1mm}1*2=2\) | I have evaluated the numerator and denominator. |
| \(\frac{-12}{2}=-6\) | Of course, with fractions, you should simplify them to simplest terms. This fractions ends up simplifying into an integer, -6. After this, I continue like normal. |
Hopefully, this cleared up any confusion you had earlier. If it did not, reply with a burning question!
+2446 To convert from millimeters to inches in 2 dimensions, you must convert both the width and length before multiplying. I happen to only know this conversion from memorization:
\(1in=0.0254m\)
Of course, we want to go from millimeters to inches, so I must change the meters to millimeters. That is relatively simple:
| \(\frac{1000mm}{1m}=\frac{xmm}{0.0254m}\) | Using this proportion, we can figure out how many millimeters are in 0.0254 meters. |
| \(x=0.0254*1000=25.4mm\) | |
Ok, so now we know information that is a tad more useful than before.
\(1in=25.4mm\)
Using this knowledge, we can now convert both dimensions: 25mm and 40mm.
| \(\frac{1in}{25.4mm}=\frac{xin}{25mm}\) | Cross multiply to solve this proportion for x. |
| \(25=25.4x\) | Divide by 25.4 on both sides. |
| \(x=\frac{25}{25.4}*\frac{10}{10}\) | I, personally, do not like seeing decimals in fractions, so I am manipulating the fraction such that one exist. |
| \(x=\frac{250}{254}=\frac{125}{127}in\) | For now, I will leave the fraction in this form/ |
Now I will convert 40mm into inches:
| \(\frac{1in}{25.4mm}=\frac{yin}{40mm}\) | Cross multiply to solve this proportion for y. I changed the variable so that it would not be confusing. |
| \(40=25.4y\) | Divide by 25.4 on both sides. |
| \(y=\frac{40}{25.4}*\frac{10}{10}\) | Yet again, I am getting decimals out of the fraction. |
| \(y=\frac{400}{254}=\frac{200}{127}in\) | |
Ok, now multiply both of these fractions together.
\(\frac{125}{127}*\frac{200}{127}=\frac{25000}{16129}in^2\)
Unfortunately, this fraction is irreducible
\(\frac{25000}{16129}in^2\approx1.5500in^2\)
.