Question

Calculati aria unui trapez isoscel ABCD,AB||CD,AB,CD
a)AB=8 CM,CD=32cm BC=20cm
b)AB=8radicaldin 3 cm,AD=AB ,masura lu C= 60 grade

Answer 1

$a) Construim AA' perpendicular pe CD si BB' perpendicular pe CD.$
$AB=A'B' =\ \textgreater \ A'D=B'C= \frac{DC-A'B'}{2} = \frac{32-8}{2} = \frac{24}{2} =12$
$BB'C, T.P. =\ \textgreater \ BB'^{2} = BC^{2} - B'C^{2} =\ \textgreater \ BB'=16$
$A= \frac{(B+b)*h}{2} = \frac{(8+32)*16}{2} = \frac{40*16}{2} = 20*16= 320 cm^{2}$
La punctul b realizam aceleeasi constructii ca la punctul a.
AB=AD=BC=A'B'
In triunghiul BB'C sinus de C = $\frac{BB'}{BC} =\ \textgreater \ \frac{ \sqrt{3} }{2} = \frac{BB'}{8 \sqrt{3} } =\ \textgreater \ BB'=12$
$cos C= \frac{B'C}{BC} =\ \textgreater \ \frac{ {1} }{2} = \frac{B'C}{8 \sqrt{3} } =\ \textgreater \ B'C=4 \sqrt{3}$
$DC=DA'+A'B'+B'C= 8 \sqrt{3} +2*4 \sqrt{3} =16 \sqrt{3[tex]A= \frac{(B+b)h}{2} = \frac{(8 \sqrt{3}+16 \sqrt{3})*12 }{2} = \frac{24 \sqrt{3}*12 }{2} =144 \sqrt{3} }$