The sides BC and AD of a quadrilateral ABCD are parallel. X is the midpoint of AB and Y is a point on the side CD such that CY = 3DY. Find the area of the triangle XYA if the area of the triangle ABC is 120cm squared and the area of the triangle ABD is 160cm squared.
I worked out that:
angle ABD = angle CBD
angle DAC = angle ACB
That's all I could come up with.
Thanks a lot!
August 4th 2005, 12:08 AM
Too bad you have no idea on Geometry. I could have asked you some more information---like, is AB perpendicular to both BC and AD---and then I could have shown a solution, but you wouldn't understand it, so no use.
Anyway, if AB is perpendicular to BC and AD, then the area of triangle XYA is 65 sq.cm.
If AB is not perpendicular to BC and AD, then the solution is more complicated. I dared not venture into any solution here.
If a picture/diagram/figure can be shown here, we might be able to solve question. Since you have no idea on Geometry, then you have no idea also on the usefulness/importance of a picture/diagram/figure in solving questions like your question here.
August 4th 2005, 01:36 AM
There was no diagram with the question. Sorry. Thanks for your help though. We have no idea what quadrilateral it is.
August 4th 2005, 03:31 AM
Let me just correct my mistake, in case someone want to play with your question.
The area of triangle XYA is 75 sq.cm.---not 65---if AB is perpendicular to both BC and AD.
August 5th 2005, 12:32 AM
Okay thanks a lot. Any working with that?
August 5th 2005, 04:26 AM
You mean we assume that AB is perpendicular to both BC and AD?
I don't know. You said you have no idea on Geometry, so I think you won't understand the workings. But let me show one. See if you can follow.
ABCD is a trapezium/trapezoid where AD is longer than BC---because right triangle ABD is larger in area than right triangle ABC.
AX is (1/2) of AB
CY = (3/4) of CD
Let u = AB
And BC = v
And AD = v+w ---if we project BC on AD, then AD minus BC equals w.
Area of triangle = (1/2)(base)(altitude)
For right triangle ABC,
area = (1/2)(u)(v) = 120
So, uv = 2(120) = 240 ---------***
For right triangle ABD,
area = (1/2)(u)(v+w) = 160
So, u(v+w) = 2(160) = 320
uv +uw = 320
uw = 320 -uv
Since uv = 240, then,
uw = 320 -240 = 80 -------***
In triangle XYA:
base = (1/2)u
altitude = v +(3/4)w -------this is by proportion, because CY = (3/4) of CD.
area = (1/2)(u/2)(v +3w/4)
area = (u/4)(v +3w/4)
area = uv/4 +3uw/16
area = (1/4)uv +(3/16)uw
Since uv=240 and uw=80, then,
area = (1/4)(240) +(3/16)(80)
area = 60 +15
area = 75 sq.cm. -------answer.
August 6th 2005, 12:13 AM
I just got some more information. The quadrilateral is a trapezium.