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

dierdurst Aug 23, 2018

#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)

Melody Aug 23, 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.

Melody 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

CPhill 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).

modus Aug 24, 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