Geometry: finding the area of a triangle
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 AD such that AY = 2DY. Find the area of the triangle CXY if the area of triangle BCD is 100cm(squared) and the area of triangle BAD is 120cm(squared).
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Picture a trapezoid with a horizontal base and top, A is bottom left, B top left, C top right, D bottom right.
This trapezoid has height h, top length a, base length b.
The triangle ABD has area 120, triangle BCD has area 100, and together these comprise the entire quadrilateral. So:
`(1)/(2)bh=120` (thus `(bh)/(6)=40` )
`(1)/(2)ah=100` (thus `(ah)/(4)=50` )
Triangle CXY is what remains when you start with quadrilateral ABCD and cut off triangles BCX, AXY, and CDY
So we want to figure out the areas of each of these, and subtract them from 220.
We start with BCX:
This triangle has base length a, and height `(h)/(2)` , so its area is `(1)/(2)*a*(h)/(2)=50`
Next, triangle AXY:
If AY=2DY, then we divide AD into thirds: Y is `(2)/(3)` of the way to D from A. Thus triangle AXY has base `(2)/(3) b` and height `(h)/(2)`
So triangle AXY has area `(1)/(2)*(2)/(3) b * (h)/(2) = (bh)/(6) = 40`
Finally, triangle CDY
This triangle has base `(b)/(3)` and height `h`
So its area is: `(1)/(2)*(b)/(3)*h=40`
Thus: the area of triangle CXY = 220-50-40-40=90 `"cm"^2`
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