Rectangle ABCD has diagonals AC and BD. E is the point of intersection of the two diagonals. If the area of triangle AEB is 8, what is the area of triangle ACD?
ACD has half the area if the rectangle.
AEB is less than half of the rectangle, and its fraction of the whole depends upon the aspect ratio (AB/BC) of the triangle.
More information is needed to answer the question.
In a rectangle, the diagonals bisect each other giving you four triangles of equal area.
so triangle ACD = 16
Whoops! I forgot that equal-area theorem. Thanks again, Reiny.
To find the area of triangle ACD, we need to determine the relationship between triangles AEB and ACD since they share a common base AC.
We know that triangle AEB and triangle ACD share the same height since they are within the same rectangle ABCD. Therefore, to find the area of triangle ACD, we need to find the length of its base.
Since the diagonals of a rectangle bisect each other, we can conclude that AE = EC and BE = ED. This implies that triangles AEB and CED are congruent by the Side-Side-Side (SSS) congruence property.
Therefore, triangle ACD consists of two congruent triangles CED, each with half the area of triangle AEB. So the area of triangle ACD is twice that of triangle AEB, which gives us:
Area of triangle ACD = 2 * Area of triangle AEB
= 2 * 8
= 16
Therefore, the area of triangle ACD is 16.