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 \(ACD\).
Note: this is not an answer.
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)
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)
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
Area of ACD = 2/3 * 42 = 28 unites squared.
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
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
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).
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.