Задача 7 клас

[Гост]

Задача 7 клас

[ammornil]

Screenshot 2023-03-16 084613.png (18.45 KiB) Прегледано 475 пъти

[tex]\triangle ABC, AC=BC, s_{BC} \cap AC = P, P \in s_{BC} \Rightarrow BP=CP[/tex]

(а) [tex]\because \angle{CBP}=46 ^\circ \rightarrow \angle{ABC}=?, \angle{BCA}=?, \angle{BAC}=?[/tex]
[tex]\triangle{BPC} \text{ -равнобедрен} \Rightarrow \angle{PCB}=\angle{PBC}=46 ^\circ[/tex]
[tex]\triangle{ABC} \text{ -равнобедрен} \Rightarrow \angle{ABC}= \angle{BAC} \rightarrow[/tex]
[tex]\angle{ABC}+ \angle{BCA}+ \angle{BAC}=180 ^\circ \Leftrightarrow 2\cdot \angle{BAC} = 180 ^\circ -46 ^\circ \Rightarrow[/tex]
[tex]\angle{BAC}=\angle{ABC}=67 ^\circ[/tex] $$ \angle{ABC}=67^\circ, \angle{BCA}=46^\circ, \angle{BAC}=67^\circ $$

(б) [tex]\because \angle{ABP}=18 ^\circ \rightarrow \angle{ABC}=?, \angle{BCA}=?, \angle{BAC}=?[/tex]
[tex]\triangle{ABC} \text{ -равнобедрен} \Rightarrow \angle{ABC}= \angle{BAC}[/tex]
[tex]\triangle{BPC} \text{ -равнобедрен} \Rightarrow \angle{PCB}=\angle{PBC} \Rightarrow \angle{ABC}=\angle{BAC}=\angle{BCA}+18 ^\circ \Rightarrow \angle{BCA}=\angle{ABC}-18 ^\circ[/tex]
[tex]\angle{ABC}+ \angle{BCA}+ \angle{BAC}=180 ^\circ \Leftrightarrow \angle{BAC} +\angle{BAC} +\angle{BAC} -18 ^\circ = 180^\circ \Rightarrow[/tex]
[tex]3\cdot \angle{BAC} = 198^\circ \Rightarrow \angle{BAC}=\angle{ABC}=66^\circ \Rightarrow \angle{BCA}=\angle{ABC}-18 ^\circ=48^\circ[/tex] $$ \angle{ABC}=66^\circ, \angle{BCA}=48^\circ, \angle{BAC}=66^\circ $$

(в) [tex]\because AB=8\>[cm], P_{\triangle{ABC}}=38\>[cm], P_{\triangle{ABP}}=?[/tex]
[tex]AB + 2\cdot BC=P_{\triangle{ABC}} \Rightarrow BC=\frac{P_{\triangle{ABC}}-AB}{2}=\frac{38-8}{2}=15\>[cm][/tex]
[tex]AC=BC=15\>[cm], BP=CP, AC=AP+PC \Rightarrow AP+BP=15\>[cm][/tex] $$P_{\triangle{ABP}}=AP+BP+AB=15+8=23\>[cm]$$

(г) [tex]\because AB=8\>[cm], P_{\triangle{ABP}}=21\>[cm], P_{\triangle{ABC}}=?[/tex]
[tex]AB+BP+AP=P_{\triangle{ABP}} \Rightarrow AP+BP=P_{\triangle{ABP}}-AB=21-8=13\>[cm][/tex]
[tex]CP=BP \Rightarrow AP+BP=AP+PC \Rightarrow AP+PC=13\>[cm]=AC, AC=BC=13\>[cm][/tex] $$P_{\triangle{ABC}}=AB+BC+AC=8+13+13=34\>[cm]$$

[Гост]

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