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Suppose that \(ABCD\) is a trapezoid in which \(AD\) is parallel to \(BC\). Given \(AC ⊥ CD\), \(AC\) bisects ∠\(BAD\), and the area of \(ABCD = 42\), then compute the area of \(ACD\).

 

- Daisy

 Aug 23, 2018
edited by dierdurst  Aug 23, 2018
edited by dierdurst  Aug 23, 2018
edited by dierdurst  Aug 24, 2018
 #1
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Note: this is not an answer.

 

Hi Daisy,

This shape is ony called a Trapezoid in USA and Canada. The rest of the English speaking world knows it as a Trapezium.

In England a Trapezoid has no parallel sides. 

To me, in Australia, a trapezoid has no meaning at all.   At least I have never known of one.

 

Anyway, you are referring to a quadrilateral with one pair of parallel sides.    (What I would call a Trapezium)

 Aug 23, 2018
 #2
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Haha! Thanks for informing me, Melody! I will keep that in mind in the future.

 

- Daisy

dierdurst  Aug 24, 2018
 #3
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+2

I believe the area if 28. 

I am sorry, right now I don't have time to write it up (and hence check my own logic) 

but

I used similar triangles. 

If you let M be the midpoint of AC then   triangle AMB= triangle CMB

and they are both similar to triangle ACD

 

ACD is bigger by a factor of 2.

hence its area is bigger by a factor of 4

So the big one has twice the area of the 2 little ones combined

so

Area of ACD = 2/3 * 42 = 28 unites squared.

 Aug 24, 2018
 #4
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+1

See if you can work it out from that. Let us know either way :)

Melody  Aug 24, 2018
 #5
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I have figured it out! Thank you so much for your help.

 

- Daisy

dierdurst  Aug 24, 2018
 #7
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Here is my diagram

 

Melody  Aug 24, 2018
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Thanks! That answered my question about the diagram! Sorry for taking up your time!

 

- Daisy

dierdurst  Aug 24, 2018
edited by dierdurst  Aug 24, 2018
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+2

Don't apologize! I give my time freely. If I help someone learn and get a thank you then it makes me very happy.  :))

 

If you still are having trouble filling in the blanks I am quite happy to elaborate :)  You only have to ask :)

Melody  Aug 24, 2018
edited by Melody  Aug 24, 2018
 #11
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Haha I'm fine, thank you very much, though! I'll keep that in mind the next time I have trouble on a question.

 

- Daisy

dierdurst  Aug 24, 2018
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I get the same answer as Melody....but....I  don't know if the included pic is accurate or not

 

 

I have assumed that  angle ABC  =  90 =  angle BAD

Since AC  bisects  angle BAD, then angle BAC =  angle CAD  = 45

 

AB = BC

 

So... AC  = √2AB  = CD

 

Area   of ABCD   =  Area  of triangle ABC   + Area of triangle ACD  = 42

 

So

Area  of triangle ABC  + area of triangle ACD  = 42

(1/2)AB*BC   + (1/2)AC * CD  = 42

(1/2)AB^2  + (1/2)√2AB * √2AB   = 42

AB^2  + 2AB^2  = 84

3AB^2  = 84

AB^2 = 28  = area of triangle ACD

 

cool cool cool

 Aug 24, 2018
edited by CPhill  Aug 24, 2018
edited by CPhill  Aug 24, 2018
 #9
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It's well drawn, and it helped me comprehend the solution. Thanks!

 

- Daisy

dierdurst  Aug 24, 2018
edited by dierdurst  Aug 24, 2018
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Despite Melody saying she didn't have time to write up and check her answer, what she wrote was nevertheless almost complete. The only omission was in the "Let M be the midpoint of AC" bit. For the sake of those who want a complete proof this bit should say:–

 

"Since AD is parallel to BC we must have CAD = ACB (as indicated in the diagram, but without justification). Drop a perpendicular BM from B to AC. At this stage we dont know that M is the midpoint of AC. But the two triangles AMB, CMB are congruent (angle, angle, side). Therefore AM=MB."

 

The second "proof" isn't really a proof at all, because it assumes that \(\alpha=45^\circ\), which isn't necessary at all, as Melody has shown.

 

By the way, we've never heard of a trapezoid in Mathematics in the UK (including England) either. It is the universal name of a bone in your hand, so if you were up in Anatomy you would have heard of it :-) The only name in the UK for a 4-sided figure with no parallel sides is a quadrilateral (admittedly an irregular one).

 Aug 24, 2018
edited by modus  Aug 25, 2018
 #13
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+1

I never suggested mine was the full answer. It was an rough outline of the answer that I had in front of me. 

Since it was never intended as a full answer, and I did make that quite clear, I certainly do not need to be excused.

I was well away that M was in place before it was discovered to be the midpoint and I would have explained that if I had been asked.

 

 

 

4 sided quadrilaterals with parallel sides can also be irregular, and they usually are.

The only 4 sided quadrilateral that is regular is a square.  Since it is the only one with all sides and all internal angles equal.

Melody  Aug 25, 2018

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