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 abed.\
\\(\bigcirc\\) 228 units\\(^2\\)\
\\(\bigcirc\\) 54 units\\(^2\\)\
\\(\bigcirc\\) 256 units\\(^2\\)\
\\(\bigcirc\\) 174 units\\(^2\\)

Explanation:

+FE=9+10=19. Yes, that's correct. So area of ABED is $\frac{1}{2}*(AB + DE)*AF = \frac{1}{2}*(10+19)*12=174$. But wait, 228 is area of ABCD. Wait, let's check the options: 228, 54,256,174. So 174 is an option. But why did I first get 114? Because I took DE as 9, which is FD, not DE. DE is FD + FE=9+10=19. Yes, that's correct.

Wait, final correction:

Step1: Calculate EC length

Right triangle BEC: $EC = \sqrt{15^2 - 12^2} = 9$

Step2: Calculate DE length

$DE = AB + EC = 10 + 9 = 19$

Step3: Compute trapezoid area

Use trapezoid area formula:
$\text{Area} = \frac{1}{2} \times (AB + DE) \times BE$
$\text{Area} = \frac{1}{2} \times (10 + 19) \times 12 = 174$

Wait no, DE is DC - EC, DC=AB+2*EC=10+18=28, so DE=28-9=19, correct. So area of ABED is 174 units².

Wait, but another way: area of ABED is area of rectangle ABFE (F is foot from A to DC) plus area of triangle ADF. Area of ABFE=10*12=120, area of ADF=$\frac{1}{2}*9*12=54$, total 120+54=174. Yes, that's correct.

Final correct answer is 174 units².

Step1: Find horizontal segment length

Right triangle: $\sqrt{15^2 -12^2}=9$

Step2: Determine DE length

$DE = 10 + 9 = 19$

Step3: Calculate trapezoid area

$\text{Area} = \frac{1}{2}(10+19) \times12 =174$

Answer:

174 units²