We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.

+0

# Geometry Question

+3
226
13

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

### 13+0 Answers

#1
+2

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
+3

Haha! Thanks for informing me, Melody! I will keep that in mind in the future.

- Daisy

dierdurst  Aug 24, 2018
#3
+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
+1

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

Melody  Aug 24, 2018
#5
+2

I have figured it out! Thank you so much for your help.

- Daisy

dierdurst  Aug 24, 2018
#7
+2

Here is my diagram Melody  Aug 24, 2018
#8
+4

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
#10
+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
+4

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
#6
+3

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   Aug 24, 2018
edited by CPhill  Aug 24, 2018
edited by CPhill  Aug 24, 2018
#9
+3

It's well drawn, and it helped me comprehend the solution. Thanks!

- Daisy

dierdurst  Aug 24, 2018
edited by dierdurst  Aug 24, 2018
#12
+1

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
+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