Catherine, I'm curious what method you used to get your answer. Obviously your method is correct, because it leads to the right numerical result. At some point you must have taken a square root, right? What happens if you just retrace your steps, but resist the urge to turn that square root into a decimal, and leave it in its "unevaluated" form?
I would attack the problem this way:
Triangle BEC is a right triangle, with one leg twice as large as the other. In any such triangle, the hypotenuse is as long as the short leg times the square root of 5. Why? Because if the legs are x and 2x, then the Pythagorean Theorem gives you (hypotenuse)^2 = x^2 + (2x)^2. Simplifying the right-hand-side of the equation, you get 5x^2. (The 5 comes from squaring the 2 in the second term, to get 4 and then adding the 1 from the first term.) So the hypotenuse is (sqrt 5)*x.
Going back to the problem, we now know that EC is as long as BE times sqrt(5), i.e. its length is sqrt(5) / 2.
From the information in the problem, that is also the length of FE.
So FB is that quantity, minus the length of BE... Or, in other words, FB = sqrt(5)/2 - 1/2. Over a common denominator, we have FB = (sqrt(5) - 1)/2.
And, lastly, BH is just as long as FB.
Extra credit writing assignment: What is the connection between this problem and the Golden Rectangle?
My attempt, trying to put in as many little steps as possible:
*) given: AD = 1 *) given: AB and AD are sides of a square, and thus are equal *) AD = 1 --> AB = 1 *) given: EB is the midpoint of AB *) EB = ½ of AB --> EB = ½
*) EBC is a right triangle, so we can use the pythagorean theorem: EB×EB + BC×BC = EC×EC *) ½*½ + 1×1 = EC*EC *) ¼ + 1 = EC×EC *) EC = sqrt (1¼)
*) given: FE = CE *) EB = ½ from above, so *) FB = sqrt(1¼) - ½
i used a decimal length from the get-go and then didn't see that if I'd used 1/2 rather than .5 I'd have a square root of 5/4 instead of a square root of 1.75
No reason to be embarrassed. The urge to "calculate" (express your answer as a decimal) can be hard to resist. But it should almost always be saved for the end of a problem. Otherwise how else would you be able to recognize that the answer to a certain problem is (to take just one made-up example) pi / sqrt(3), when all you see is 1.8148?
It is a rare person who can see a decimal and figure out what the radical would be. You really want to get in the habit of working in whole numbers and leaving square roots once you get there, especially for SAT math where the answer is rarely the decimal.
Having said that, it is worth memorizing the square root of 2 (1.41) and ln 2 (.693) so you can see them and know what the original value was.
Students familiar with problems like this expect to see sqrt(2)/2, or sqrt(5)/2, or sqrt(3)/2 in problems involving right triangles or equilateral triangles. One reason they know to leave them in that form because they expect to use the Pythagorean theorem again to solve more steps, and they'll be squaring them up again in the future. But after a while, they know to leave them in that form because that's just how the problems are usually written.
Can I just add a recommendation to do Michael's extra credit writing assignment?
As soon as I saw 0.618, I knew what the answer was and all that was left was how to get to the quadratic equation that leads to that number. A bit like a mathematical version of Jeopardy!!
For the record,
AD=1 BC=1 AE=1/2 BE=1/2 FB=x BH=x HC=1-x (not needed, but it looks incomplete without it) FE = x +1/2 = CE
Pythagoras on BEC to give
BC^2+BE^2=CE^2 1^2+(1/2)^2=(x+1/2)^2
5/4 = x^2 + x +1/4
and from that you get
x^2 + x = 1
The rest is left as "an exercise for the reader" (how I love that phrase!!)
Remember to use your test taking skill of glancing at the answers in the process of deciding your approach. Since these answers involve fractions and square roots, no need to waste precious time in converting to decimal.
E is recognizable to those who have an enrichment/honors course or math club that included the Golden Ratio topic. It is introduced in Dolciani PreAlgebra: An Accelerated Course.
BE is 1/2 (midpoint of square side, side=1)
ReplyDeleteBH is side of square, equal to BF + BE, so BH = FE - 1/2 = CE - 1/2
by Pythagoras, square(CE) = square(BE) + square(BC)
so square(CE) = 1/4 + 4/4 = 5/4 and CE = square root(5/4) = (square root 5)/2
(square root 5)/2 - 1/2 = (square root(5) - 1)/2
(sorry for poor formatting and lazy use of =)
- Andy
Catherine, I'm curious what method you used to get your answer. Obviously your method is correct, because it leads to the right numerical result. At some point you must have taken a square root, right? What happens if you just retrace your steps, but resist the urge to turn that square root into a decimal, and leave it in its "unevaluated" form?
ReplyDeleteI would attack the problem this way:
Triangle BEC is a right triangle, with one leg twice as large as the other. In any such triangle, the hypotenuse is as long as the short leg times the square root of 5. Why? Because if the legs are x and 2x, then the Pythagorean Theorem gives you (hypotenuse)^2 = x^2 + (2x)^2. Simplifying the right-hand-side of the equation, you get 5x^2. (The 5 comes from squaring the 2 in the second term, to get 4 and then adding the 1 from the first term.) So the hypotenuse is (sqrt 5)*x.
Going back to the problem, we now know that EC is as long as BE times sqrt(5), i.e. its length is sqrt(5) / 2.
From the information in the problem, that is also the length of FE.
So FB is that quantity, minus the length of BE... Or, in other words, FB = sqrt(5)/2 - 1/2. Over a common denominator, we have FB = (sqrt(5) - 1)/2.
And, lastly, BH is just as long as FB.
Extra credit writing assignment: What is the connection between this problem and the Golden Rectangle?
One of the hardest things to learn is the strength to resist evaluating things numerically in the middle of a problem.
ReplyDeleteMy attempt, trying to put in as many little steps as possible:
ReplyDelete*) given: AD = 1
*) given: AB and AD are sides of a square, and thus are equal
*) AD = 1 --> AB = 1
*) given: EB is the midpoint of AB
*) EB = ½ of AB --> EB = ½
*) EBC is a right triangle, so we can use the pythagorean theorem: EB×EB + BC×BC = EC×EC
*) ½*½ + 1×1 = EC*EC
*) ¼ + 1 = EC×EC
*) EC = sqrt (1¼)
*) given: FE = CE
*) EB = ½ from above, so
*) FB = sqrt(1¼) - ½
*) sqrt(1¼) = sqrt(5/4)
*) sqrt(5/4) = sqrt(5)/sqrt(4)
*) sqrt(5)/sqrt(4) = sqrt(5)/2
*) FB = sqrt(5)/2 = 1/2
*) FB = [sqrt(5) - 1]/2
*) FB and BH are sides of a square, so BH = FB
*) BH = [sqrt(5) - 1]/2
-Mark Roulo
I've got to print these out & read!
ReplyDeleteNow I'm thinking...I wonder if the issue is that I was using decimals from the get-go ---
hang on
I'll look at my work
I was doing it late last night, so cognitive flexibility wasn't at maximum strength---
that reminds me of something else I want to post
ok, you guys
ReplyDeletedon't laugh
i used a decimal length from the get-go and then didn't see that if I'd used 1/2 rather than .5 I'd have a square root of 5/4 instead of a square root of 1.75
oy
No reason to be embarrassed. The urge to "calculate" (express your answer as a decimal) can be hard to resist. But it should almost always be saved for the end of a problem. Otherwise how else would you be able to recognize that the answer to a certain problem is (to take just one made-up example) pi / sqrt(3), when all you see is 1.8148?
ReplyDeleteThe embarassing part isn't that I used decimal calculations, it's that I didn't perceive that's what I came up with .618 as the answer.
ReplyDeleteI couldn't see how you would end up with a radicand in your answer (radicand?? that's the term, right?)
It didn't jump out at me that 1/2 x 1/2 would produce a _/5
(squre -rt 5 written out as sqr rt 5)
inflexible thinking
major, major inflexible thinking
I DON'T LIKE IT!
It is a rare person who can see a decimal and figure out what the radical would be. You really want to get in the habit of working in whole numbers and leaving square roots once you get there, especially for SAT math where the answer is rarely the decimal.
ReplyDeleteHaving said that, it is worth memorizing the square root of 2 (1.41) and ln 2 (.693) so you can see them and know what the original value was.
Students familiar with problems like this expect to see sqrt(2)/2, or sqrt(5)/2, or sqrt(3)/2 in problems involving right triangles or equilateral triangles. One reason they know to leave them in that form because they expect to use the Pythagorean theorem again to solve more steps, and they'll be squaring them up again in the future. But after a while, they know to leave them in that form because that's just how the problems are usually written.
ReplyDeleteCan I just add a recommendation to do Michael's extra credit writing assignment?
ReplyDeleteAs soon as I saw 0.618, I knew what the answer was and all that was left was how to get to the quadratic equation that leads to that number. A bit like a mathematical version of Jeopardy!!
For the record,
AD=1
BC=1
AE=1/2
BE=1/2
FB=x
BH=x
HC=1-x (not needed, but it looks incomplete without it)
FE = x +1/2 = CE
Pythagoras on BEC to give
BC^2+BE^2=CE^2
1^2+(1/2)^2=(x+1/2)^2
5/4 = x^2 + x +1/4
and from that you get
x^2 + x = 1
The rest is left as "an exercise for the reader" (how I love that phrase!!)
Richard I, you saw that .618 and knew it was Phi, didn't you? :)
ReplyDeleteRemember to use your test taking skill of glancing at the answers in the process of deciding your approach. Since these answers involve fractions and square roots, no need to waste precious time in converting to decimal.
ReplyDeleteE is recognizable to those who have an enrichment/honors course or math club that included the Golden Ratio topic. It is introduced in Dolciani PreAlgebra: An Accelerated Course.