Pole trapezu

[bocian554]

krutsza podstawa trapezu ma długosc \(\displaystyle{ 3\sqrt{6}}\) cm. Katy przy tej podstawie mają miary \(\displaystyle{ 135^{o}}\) i \(\displaystyle{ 60^{o}}\) a dłuzsze ramie ma długosc 18 cm. Oblicz pole tego trapzu.

[jaffa84]

\(\displaystyle{ |CD|=3 \sqrt{6}}\)
\(\displaystyle{ |AD|=18}\)
\(\displaystyle{ |\sphericalangle ADC|=135}\)
\(\displaystyle{ |\sphericalangle DCB|=60}\)
\(\displaystyle{ | \sphericalangle ADE|=135-90=45}\)
\(\displaystyle{ | \sphericalangle DAE|=180-135=45}\) \(\displaystyle{ Trojkat AED jest prostokatny rownoramienny}\), więc

\(\displaystyle{ h ^{2} +h ^{2}=18 ^{2}}\)

\(\displaystyle{ 2h ^{2}=324}\)

\(\displaystyle{ h ^{2}=162}\)

\(\displaystyle{ h=9 \sqrt{2}}\)

\(\displaystyle{ |AB|=|AE|+|EB|}\), \(\displaystyle{ |AE|=h=9 \sqrt{2}}\), \(\displaystyle{ |EB|=|CD|-|CF|}\)

\(\displaystyle{ \frac{h}{|CF|}=tg60}\)

\(\displaystyle{ |CF|= \frac{9 \sqrt{2}}{ \sqrt{3} } =3 \sqrt{6}}\)

\(\displaystyle{ |EB|=|CD|-|CF|}\)
\(\displaystyle{ |EB|=3 \sqrt{6} -3 \sqrt{6} =0}\)

\(\displaystyle{ |AB|=|AE|+|EB|}\)
\(\displaystyle{ |AB|=9 \sqrt{2}+0=9 \sqrt{2}}\)

\(\displaystyle{ P _{ABCD} = \frac{(|AB|+|CD|)*h}{2}}\)

\(\displaystyle{ P _{ABCD} = \frac{(9 \sqrt{2}+3 \sqrt{6})*9 \sqrt{2}}{2}}\)

\(\displaystyle{ P _{ABCD} = \frac{81*2+27 \sqrt{12}}{2} =\frac{81*2+27*2 \sqrt{3}}{2}}\)

\(\displaystyle{ P _{ABCD} = 81+27 \sqrt{3} =27(3+\sqrt{3})}\)