трапец

[Гост]

20220607_202014.jpg (103.28 KiB) Прегледано 374 пъти

[ammornil]

220609_002.png (9.67 KiB) Прегледано 361 пъти

[tex]ABCD, AB\|CD, AB=a, CD=b, AD=BC=v, DD_1 \bot AB, CC_1 \bot AB, CC_1=DD_1=h[/tex]
[tex]AB=6\ cm, AD=BC=2\sqrt{2}\ cm, CC_1=DD_1=2\ cm[/tex]
[tex]\angle BAD=?, b=?, S_{_{ABCD}}=?[/tex]

[tex]DD_{1}C_{1}C-[/tex] правоъгълник по построение. [tex]\Rightarrow D_{1}C_{1}=CD=b[/tex]
[tex]\triangle AD_{1}D \cong \triangle BC_{1}C \begin{cases} правоъгълни \\ DD_{1}=CC_{1} \\ AD=BC \end{cases} \Rightarrow AD_{1} = BC_{1}=x[/tex]

[tex]\triangle AD_{1}D \rightarrow AD^{2} = DD_{1}^{2} + AD_{1}^{2} \Rightarrow AD_{1}=\sqrt{AD^{2} - DD_{1}^{2}}=\sqrt{8-4}=\sqrt{4}[/tex]
[tex]\sin{\angle D_{1}AD}=\frac{2\sqrt{2}}{6}=\frac{\sqrt{3}}{2} \Rightarrow \angle D_{1}AD = 60 ^\circ[/tex]

[tex]AD_{1}+D_{1}C_{1}+C_{1}B = AB \Rightarrow 2+b+2=6 \Rightarrow b=2\ cm[/tex]

[tex]S_{_{ABCD}}=\frac{AB+CD}{2}.DD_{1}=\frac{6+2}{2}2=8\ cm^{2}[/tex]