Question

1. in the figure above, square cdef shares the common side cf with parallelogram abcf. the area of the square is 225, the length of ag is 9, and the length of fg is 21. what is the area of the parallelogram? a) 240 b) 300 c) 360 d) 480

Explanation:

Step1: Find the side length of the square

Since the area of square \(CDEF\) is \(225\), and the area of a square is \(s^{2}\) (where \(s\) is the side - length). Let \(s = CF\), then \(s^{2}=225\), so \(s = CF=15\).

Step2: Find the base length of the parallelogram

The base length of parallelogram \(ABCF\) is \(AF\). Given \(AG = 9\) and \(FG=21\), then \(AF=AG + FG=9 + 21=30\).

Step3: Find the height of the parallelogram

The height \(BG\) of parallelogram \(ABCF\) is equal to the side - length of the square \(CDEF\) (because \(BG\) is perpendicular to \(AF\) and \(CF\) is a side of the square). So \(BG = 15\).

Step4: Calculate the area of the parallelogram

The area formula of a parallelogram is \(A = base\times height\). Here, the base \(AF = 30\) and the height \(BG = 15\). So \(A=30\times15 = 450\) (This is wrong, let's re - check. Wait, no, the height of the parallelogram \(ABCF\): Since \(CF\) is a side of the square \(CDEF\) with area \(225\), \(CF = 15\). And in parallelogram \(ABCF\), \(BG\) (the height) is equal to \(CF\) (because \(ABCF\) is a parallelogram and \(BG\perp AF\), \(BC\parallel AF\), \(CF = BG\)). The base \(AF=AG + FG=9 + 21=30\). The area of parallelogram \(A = base\times height=30\times12\) (Wait, no, another wrong. Wait, the side of the square \(s=\sqrt{225}=15\). The base of the parallelogram \(AF = AG+FG=9 + 21=30\). The height of the parallelogram (the perpendicular distance between \(AB\) and \(CF\)): Since \(ABCF\) is a parallelogram, and \(BG\) is the height. In right - triangle \(ABG\), \(AB = CF = 15\) (opposite sides of a parallelogram are equal). Using the Pythagorean theorem in \(\triangle ABG\), if \(AB = 15\), \(AG = 9\), then \(BG=\sqrt{AB^{2}-AG^{2}}=\sqrt{15^{2}-9^{2}}=\sqrt{(15 + 9)(15 - 9)}=\sqrt{24\times6}=\sqrt{144}=12\). The base \(AF=AG + FG=9+21 = 30\). The area of the parallelogram \(A=AF\times BG\).

So \(A=(9 + 21)\times12=30\times12 = 360\)

Answer:

C. 360