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# Help asap ty

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Help asap ty Apr 23, 2018

#1
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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   Apr 23, 2018
edited by CPhill  Apr 23, 2018
edited by CPhill  Apr 25, 2018
#2
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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 $$\triangle PRQ\sim \triangle PSR\sim \triangle RSQ$$

Apr 23, 2018
edited by TheXSquaredFactor  Apr 23, 2018
edited by TheXSquaredFactor  Apr 23, 2018
#3
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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!
Apr 23, 2018
#4
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The answer for 58 is actually 5/2 cm.^2 but I don't understand how to get that.

RainbowPanda  Apr 24, 2018
#5
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The final answer is actually, 512!! LaughingFace  Apr 25, 2018
#6
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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. $$RQ\approx 17.9$$

TheXSquaredFactor  Apr 26, 2018