Saturday, March 28, 2009

Take Home Test #5, Question 3 Discussion Post

You can use this post to discuss Question 3 on Test #5 (Take-Home).

30 comments:

victoria said...

For part a, any ideas on how to find DE? I have tried creating different triangles, but I can't seem to find the right one...or maybe I am going in the wrong direction.

Mr. Kerai said...

Have you considered triangle ADE? What kind of triangle is it? What do you know about it? There may be various ways to approach the question. Sometimes what helps is to write out all that you can find in a question, even if some of the lengths and relationships you find do not directly answer the question asked. Then see which of the things you found could be helpful.

victoria said...

I looked at ADE and knew it was isosceles...I tried thinking of AX as the altitude, but then I wasn't sure if I could actually prove that it was. Is there something that we have learned that tells us that is so in a rectangular solid?

virginia said...

this is sort of a different way if you still need help- i'm not sure if this is right, but I looked at a lot of different right triangles and then used the pythagorean theorem to find DE if that helps you at all

victoria said...

virginia, that is why I was wondering about AX being the altitude. From there I could find DE, but I didn't know if I was assuming too much if I said that.

virginia said...

i actually used some other triangles like AJE (J= 4th corner of bottom rectangle) and then DJE, i think those work but i'm not sure

virginia said...

I am not sure how to find the length of DX though- does anyone have any ideas?

Jack said...

i dont know if this will help but look at angle DAE and see what it is formed by. After you see something about that angle our good old friend pythag. can help. What does it mean by show that AC intersects DE.

virginia said...

can anyone help me with how to prove that ac and de intersect? i thought i could do it by proving they are in the same plane and intersect the same pair of parallel lines- but that isn't enough is it?

Michael said...

Is it possible to prove that AC and DE intersect by looking at the planes in which they are?

jack said...

thats what i did

jack said...

actually nevermind i dont know how to do it

Michael said...

any ideas on finding the length of DX?

jack said...

can you push me towardsn the way to do b

Michael said...

plane DCBA.

Mr. Kerai said...

To prove the intersection of the two lines, prove that they are coplanar, i.e. not skew lines. Think of the lines that they are connected to on the rectangular solid's edges.

The length of DE is an easy thing, just find the right traingle of choice and then find all the necessary lengths using some other right triangles for which the dimensions are provided.

W. Stott said...

Virginia had said that AX was the altitude of triangle ADE, which was isosceles. If that is so, then isn't the altitude of an isosceles triangle also the perpendicular bisector? That would make X the midpoint of DE...

jack m said...

actually virginia, that might work beacause by definition we know that two parallel lines form a plane. Since both DE and CA intersect two lines each on said plane and they are not parallel then they must meet

jackm said...

scott,
how would ax be the perpindicular bisector. if it is then your right but how is it

jack m said...

oops i said scott instead of Stott, Whitney

Michael said...

Is there a way one can go about proving that AX is the altitude of ADE?

Vranian said...

Whitney,
I see where you're coming from. However, how would you go about proving that AX is the altitude on DE?

jack middleton said...

thats what i was wondering

JAck said...

does anyone have another way of going about part C

amanda said...

for part c don't forget what you proved for the length of DE and see if you can prove DXC is similar to AXE

virginia said...

for c i used similar triangles by making an auxiliary line from x parallel to dj, but I don't know if that works

jackm said...

alright i got those triangles similar now what. maybe finding the proportions of the sides using AE and EB then using those proportions to find the ratio of DX XE???

Michael said...

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Michael said...

Yes Jack

jackm said...

do i not get a little credit for stealing her idea