Question

In the diagram,ABCD is a rectangular wall that casts a shadow CDEF, on the ground. The wall has a height of 4 feet. The shadow is a parallelogram that has a height, FG, that is twice the height of the wall. If the area of the wall is 29.2 square feet, find the area of the shadow.

Answer 1

We know that the height of the rectangular shape is BC=AD= 4 feet and the the area is A=29.2 ft^2. Since the area of our rectangle is given by

$A=(DC)\times(BC)$

we get

$29.2=DC\times4$

By moving the number 4 to the left hand side, we have

$\begin{gathered} \frac{29.2}{4}=DC \\ \text{then} \\ DC=7.3 \end{gathered}$

which also is the lenght of one side of our parallelogram.

Now, the area of our parallelogram is given by

$A_P=\text{base}\times height$

where the base is given by segment DC=7.3 ft and the height FG=2 BC. Then, we get

$\begin{gathered} A_P=DC\times FG \\ A_P=DC\times2BC \end{gathered}$

by substituting our previous result and BC=4 ft, we obtain

$\begin{gathered} A_P=7.3\times2(4) \\ A_P=7.3\times8 \\ A_P=58.4ft^2 \end{gathered}$

Then, the answer is 58.4 ft^2