# Trapezoid ABCD v2

Trapezoid ABCD has length of bases in ratio 3:10. The area of riangle ACD is 825 dm2. What is the area of trapezoid ABCD?

Result

S =  1072.5 dm2

#### Solution:

$a/c = 3/10 \ \\ S_{ACD}= 825 = ch / 2 \ \\ h = 2S_{ACD}/c \ \\ \ \\ S = \dfrac{a+c}{2} h = \dfrac{a+c}{2} \dfrac{2 S_{ACD}}{c} \ \\ S = (a+c)S_{ACD}/c \ \\ S = S_{ACD}(a/c+1) = 825\cdot (3/10+1) \ \\ S = 1072.5 \ \text{dm}^2$

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