Question

Fie trapezul ABCD, cu AB paralel CD cu AB = 30 cm, BC = 20 cm, CD = 5 cm, AD = 15cm. Aria trapezul ABCD este:​

Answer 1

$h {}^{2} = d {}^{2} - [a - (b + x)] {}^{2} \\ c {}^{2} - x {}^{2} = d {}^{2} - [a - (b + x)] {}^{2} \\ x = \frac{(a - b) {}^{2} + c {}^{2} - d {}^{2} }{2(a - b)} \\ \\ x = \frac{(30 - 5) {}^{2} + 15 {}^{2} - 20 {}^{2} }{2(30 - 5)} \\ \\ x = \frac{625 + (15 - 20)(15 + 20)}{50} \\ \\ x = \frac{625 - 175}{50} \\ \\ x = 9 \\ \\ \\ \implies \: h = \sqrt{c {}^{2} - x {}^{2} } \\ \implies \: h = \sqrt{15 {}^{2} - 9 {}^{2} } \\ \implies \: h = \sqrt{144} \\ \implies \: h = 12 \\ \\ \iff\tt{aria} = \frac{(B + b) \times h}{2} \\ \\ \iff\tt{aria} = (30 + 5) \times 6 \\ \iff\boxed{\tt{aria} = 210 \: cm {}^{2} }$