Question

in the diagram shown, abcd is an isosceles trapezoid with altitude \\(\overline{be}\\), \\(ab = 10\\), \\(ad = 15\\), and \\(be = 12\\). find the area of trapezoid \\(abcd\\).\
\
\\(\bigcirc\\) \\(108\\) units\\(^2\\)\
\\(\bigcirc\\) \\(228\\) units\\(^2\\)\
\\(\bigcirc\\) \\(256\\) units\\(^2\\)\
\\(\bigcirc\\) \\(186\\) units\\(^2\\)

Explanation:

Step1: Find length of DE

First, we use the Pythagorean theorem in right triangle ADE (where AD=15, BE=12, and BE is the height, so the vertical side is 12, hypotenuse AD=15).
$$DE = \sqrt{AD^2 - BE^2} = \sqrt{15^2 - 12^2} = \sqrt{225 - 144} = \sqrt{81} = 9$$

Step2: Calculate base DC

Since ABCD is isosceles trapezoid, DE = EC = 9. AB = 10, so:
$$DC = AB + 2 \times DE = 10 + 2 \times 9 = 28$$

Step3: Compute trapezoid area

Use trapezoid area formula $\frac{1}{2} \times (sum\ of\ parallel\ sides) \times height$:
$$Area = \frac{1}{2} \times (AB + DC) \times BE = \frac{1}{2} \times (10 + 28) \times 12 = \frac{1}{2} \times 38 \times 12 = 228$$

Answer:

228 units²