+128474
59
We have this relationship :
PS / SR = SR / SQ ....so...
4 / SR = SR / 16 cross-multiply
4 * 16 = SR * SR
64 = SR^2 take the square root of both sides
8 = SR
60
We can us the Pythagorean Theorem to find PR
PR = √[PS^2 + SR^2 ] = √[ 4^2 + 8^2 ] = √[16 + 64 ] = √ 80 = √[16 * 5 ] = 4√5 units ≈ 8.9 units
+2441 59. This is a very specific case where we can use this relationship, but it turns out that the hypotenuse of \(\triangle PRQ\) can be calculated using the geometric mean of the individual segments \(\overline{PS}\text{ and }\overline{SQ}\). The geometric mean is the nth root of the product of n numbers. In other words,
| \(RS=\sqrt{PS\cdot SQ}\) | We know the length of both of these, so we can solve for the missing length. |
| \(RS=\sqrt{4*16}\) | Now, simplify. |
| \(RS=\sqrt{64}=8\) | In the context of geometry where lengths are involved, only the positive answer is ever considered. |
We can find the other lengths using the Pythagorean's Theorem.
| \(PS^2+SR^2=RP^2\) | Plug in the values we know and solve for the unknown. |
| \(4^2+8^2=RP^2\) | |
| \(16+64=RP^2\) | |
| \(80=RP^2\) | Take the square root of both sides and reject the negative answer. |
| \(RP=\sqrt{80}=4\sqrt{5}\approx8.9 \) | |
| \(QS^2+SR^2=RQ^2\) | Do the same process as before to solve for the missing length. |
| \(16^2+8^2=RQ^2\) | This is now a matter of simplification. |
| \(256+64=RQ^2\) | |
| \(320=RQ^2\) | Take the principal root of both sides. |
| \(RQ=\sqrt{320}=8\sqrt{5}\approx17.9\) | |
Note to CPhill: Your ratio was close. It should have been \(\frac{RQ}{PR}=\frac{SQ}{RS}\). Visually, it is tricky to determine similar side lengths, which is why I generally avoid them. The similarity statement should be \(\triangle PRQ\sim \triangle PSR\sim \triangle RSQ\).
+2441 Geometers detest making assumptions; it is generally frowned upon because the idea to construct sound logic using previous theorems and postulate to formulate new ideas. However, rectangles cannot have lengths--sides can. Because of this, I have had to dissect the given information and have had to make an educated guess as to where the given information fits. I will let this diagram demonstrate this:
We can solve for the remaining side length, \(BC\), by understanding the perimeter formula for a rectange.
| \(2(AB+BC)=\text{Perimeter}_{ABCD}\) | Let's plug in the known values and solve for the missing one. |
| \(2(20\text{cm}+BC)=60\text{cm}\) | Instead of distributig the 2, we can divide it from both sides. |
| \(20\text{cm}+BC=30\text{cm}\) | Subtract 20cm from both sides to finish the algebra. |
| \(BC=10\text{cm}\) | |
We know that \(EFGH\) is similar to \(ABCD\), which means that the sides are proportional. We can use this information to our advantage.
| \(\frac{AB}{BC}=\frac{EF}{FG}\) | Yet again, substitute in the known values and solve for the unknown. |
| \(\frac{20\text{cm}}{10\text{cm}}=\frac{32\text{cm}}{FG}\) | Let's simplify the left hand side of this proportion so that the calculation is easier. |
| \(2=\frac{32\text{cm}}{FG}\) | We can use cross-multiplication here to clear all fractions. |
| \(2FG=32\text{cm}\) | Divide by 2 from both sides. |
| \(FG=16\text{cm}\) | We are not done yet! We must find the area of the rectange. That's what the question is asking for, after all! |
| \(\text{Area}_{EFGH}=EF\cdot FG\) | Plug those values in. |
| \(\text{Area}_{EFGH}=32\text{cm}*16\text{cm}=512\text{cm}^2\) | This is your final answer! |
+2448 The answer for 58 is actually 5/2 cm.^2 but I don't understand how to get that.
Also, what about 61?
+2441 To be fair, the original question is ambiguous because rectangles do not have lengths; sides do. When the question says "Rectangle ABCD has a length of 20cm," what is that supposed to mean? I noted this as a disclaimer in my previous answer. I attempted to transform obscure information to something known, but I guess I was unsuccessful. I am sorry that I was unable to assist you. RainbowPanda, if there is any explanation on how to arrive at the answer, then I would like to see it.
I have already solved 61; it is in my previous answer. I just did not label it. \(RQ\approx 17.9\)
