+400 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
+118724 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)
+118724 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.
+130554 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
+10 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).
+118724 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.
