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Моля помогнете със задача 113

[ammornil]

Screenshot 2023-04-05 125408.png (17.32 KiB) Прегледано 1107 пъти

[tex]AM=BM, BN=ND[/tex]
[tex]MN \text{ средна отсечка в } \triangle{ABD} \Rightarrow MN=\frac{1}{2}\cdot{AD}[/tex]

[tex]\text{Понеже } \frac{BM}{BA}=\frac{BN}{BD}=\frac{MN}{AD}=\frac{1}{2} \Rightarrow \triangle{MBN} \sim \triangle{ABD} \Rightarrow \frac{S_{\triangle{MBN}}}{S_{\triangle{ABD}}}=\left(\frac{1}{2}\right)^{2} \Rightarrow S_{\triangle{ABD}}=4\cdot{S_{\triangle{MBN}}}[/tex]

[tex]S_{\triangle{ABD}}=S_{\triangle{MBN}}+S_{AMND} \Leftrightarrow 4\cdot{S_{\triangle{MBN}}}=S_{\triangle{MBN}}+18 \Leftrightarrow 3\cdot{S_{\triangle{MBN}}}=18[/tex] $$\text{(a) } S_{\triangle{MBN}}= 6\>[cm^{2}]$$

[tex]S_{\triangle{ABD}}=S_{\triangle{MBN}}+S_{AMND}=6+18=24\>[cm^{2}]; \hspace{2em} S_{ABCD}=2\cdot{S_{\triangle{ABD}}} \Rightarrow[/tex] $$\text{(б) } S_{ABCD}= 48\>[cm^{2}]$$

[tex]\text{Построяваме } DD_{1} \bot AB, \hspace{0.5em} (D_{1} \in AB)[/tex]
[tex]\begin{cases} S_{\triangle{AMD}}=\frac{\normalsize{1}}{\normalsize{2}}\cdot{AM}\cdot{DD_{1}} \\ \\ S_{\triangle{BMD}}=\frac{\normalsize{1}}{\normalsize{2}}\cdot{BM}\cdot{DD_{1}} \\ \\ AM=BM \end{cases} \Rightarrow S_{\triangle{AMD}}= S_{\triangle{BMD}}, \hspace{2em} S_{\triangle{ABD}}= S_{\triangle{AMD}}+ S_{\triangle{BMD}} \Rightarrow S_{\triangle{BMD}} =\frac{1}{2}\cdot{S_{\triangle{ABD}}}=12\>[cm^{2}][/tex]

[tex]S_{\triangle{CBD}}=S_{\triangle{ABD}}, \hspace{2em} S_{MBCD}=S_{\triangle{CBD}}+ S_{\triangle{BMD}}=24+12 \Rightarrow[/tex] $$(в) S_{MBCD}=36\>[cm^{2}] $$

[tex]\text{Аналогично на лицата на триъгълници } S_{\triangle{AMD}}= S_{\triangle{BMD}}, S_{\triangle{BNC}}= S_{\triangle{CND}} (\text{ -равни основи и една и съща височина }) \Rightarrow S_{\triangle{CNB}}=\frac{1}{2}\cdot{S_{\triangle{CBD}}}=12\>[cm^{2}][/tex]

[tex]S_{MBCN}=S_{\triangle{CNB}}+S_{\triangle{NMB}}=12+6 \Rightarrow[/tex] $$ (г) S_{MBCN}=18\>[cm^{2}] $$

[S.B.]

Без заглавие - 2023-04-05T221203.424.png (353.46 KiB) Прегледано 1091 пъти

Още един поглед върху задачата :
Нека $AB = CD = a , BC = DA = b $ ,
[tex]DH \bot AB , DH = h_{a } , DK \bot BC, DK = h_{b }[/tex]
$NP$ е средна отсечка за [tex]\triangle HBD \Rightarrow NP = \frac{ h_{a } }{2}[/tex] аналогично [tex]NQ[/tex] е средна отсечка за [tex]\triangle BDK \Rightarrow NQ = \frac{ h_{b } }{2}[/tex]
[tex]S_{AMND } = 18 \Leftrightarrow S_{ABD } - S_{MBN } = 18 \Leftrightarrow \frac{1}{2} a. h_{a } - \frac{1}{2}. \frac{a}{2}. \frac{ h_{a } }{2} = 18 \Leftrightarrow a h_{a } ( \frac{1}{2} - \frac{1}{8}) = 18 \Leftrightarrow \frac{3}{8}a h_{a } = 18 \Rightarrow[/tex]
$$ \Rightarrow S_{ABCD } = a h_{a } = b h_{b } = 48 $$

а)
[tex]S_{MBN } = S_{ABD } - S_{AMND } \Leftrightarrow S_{MBN } = \frac{1}{2} S_{ABCD } - S_{AMND } \Leftrightarrow S_{MBN } = 24 - 18[/tex]
$$\Rightarrow S_{MBN } = 6 $$

б) Вече получихме,че
$$S_{ABCD } = 48$$

в)
[tex]S_{MBCD } = S_{ABCD } - S_{AMD } \Leftrightarrow S_{MBCD } = 48 - \frac{1}{2} \frac{a}{2}. h_{a } = 48 - \frac{1}{4} a h_{a } = 48 - \frac{1}{4}.48[/tex]
$$\Rightarrow S_{MBCD } = 36 $$

г)
[tex]S_{MBCN } = S_{MBN } + S_{BCN } \Leftrightarrow S_{MBCN } = 6 + \frac{1}{2}.b. \frac{ h_{b } }{2} \Leftrightarrow S_{MBCN } = 6 + \frac{1}{4}b. h_{b } \Leftrightarrow S_{MBCN } = 6 + \frac{48}{4}[/tex]
$$\Rightarrow S_{MBCN } = 18 $$