In the figure above, the area of rectangle ABCD is 72 and AE = 1/3 of

Too many side and segment names are hard for me to follow.

Let the base (AE) of ∆ ADE = \(x\)

\(x\) is \(\frac{1}{3}\) of \(EB\)

\(\frac{1}{3} = \frac{x}{EB}\)

\(EB = 3x\)

Let the height of ∆ ABE and short side of rectangle ABCD, (AD) = \(y\)

Area of rectangle ABCD is

\(4x * y =\)
\(4(xy) = 72\)
\(xy = 18\)

Area of ∆ ADE =

\(\frac{(x*y)}{2}\)
\(xy = 18\)
\(\frac{18}{2} = 9\)

Answer B

Since the area of a rectangle is base times height, we see that the area of rectangle ABCD is AD x AB, which is given to be 72.

Since the area of a triangle is one-half times base times height, we see that the area of triangle ADE is ½(AD x AE). Since AE = ⅓(EB), 3(AE) = EB. Since AB = AE + EB, AB = AE + 3(AE) = 4(AE) or ¼(AB) = AE.

Thus, the area of triangle ADE is ½(AD x AE) = ½(AD x ¼(AB)) = (½) x (AD) x (¼) x AB = ⅛(AD x AB). Since we know AD x AB = 72, the area of triangle ADE is ⅛(72) = 9.

Alternate Solution:

Let’s denote the length of the segment AE by x. Since AE = 1/3 of EB, the length of EB is 3x. Thus, the length of AB is x + 3x = 4x. Let’s express the length of AD in terms of x. Since the area of the rectangle is 72, |AD|(4x) = 72; thus |AD| = 72/(4x) = 18/x. Now, the area of the triangle ADE is (1/2)(x)(18/x) = 18/2 = 9.

Answer: B