گام به گام تمرین صفحه 72 درس 3 هندسه (1) (چند ضلعی ها)

1)

\(\begin{array}{l}{x^2} + {\left( {3x} \right)^2} = {\left( {2\sqrt {10} } \right)^2} \Rightarrow {x^2} + 9{x^2} = 40\\\\ \Rightarrow 10{x^2} = 40 \Rightarrow {x^2} = 4\\\\ \Rightarrow x = 2 \Rightarrow \left\{ \begin{array}{l}a = 6x = 12\\b = 2x = 4\end{array} \right.\\\\ \Rightarrow S = \frac{1}{2} \times a \times b = 24\end{array} \)

2)

نکته: در فرمول های زیر (commen = مشترک) می باشد.

\(\begin{array}{l}\left. \begin{array}{l}AB = AD\\CB = CD\\AC = common\quad \quad \quad \end{array} \right\} \Rightarrow A\mathop B\limits^\Delta  C \cong A\mathop C\limits^\Delta  D \Rightarrow \widehat {{A_1}} = \widehat {{A_2}}\\\\\left. \begin{array}{l}AB = AD\\\widehat {{A_1}} = \widehat {{A_2}}\\AH = commen\quad \quad \quad \end{array} \right\} \Rightarrow A\mathop B\limits^\Delta  H \cong A\mathop D\limits^\Delta  H \Rightarrow \widehat {{H_1}} = \widehat {{H_2}} = {90^ \circ }\\\\AC \bot BD \Rightarrow {S_{A\mathop {BC}\limits^{} D}} = \frac{1}{2}AC \times BD = 24\end{array}\)

3)

فرض می کنیم فاصله دو خط موازی d و d’ باشند . در این صورت داریم :

\({S_{A\mathop {BC}\limits^{} D}} = {S_{A\mathop {BE}\limits^{} F}} = AB \times h\)

4)

عمودهای AF و BE را بر ضلع DC وارد می کنیم. چهار ضلعی ABEF مستطیل است؛ بنابراین :

\(\begin{array}{l}AB = EF = b\\\\A\mathop D\limits^\Delta  F:\quad \widehat {{A_1}} = \widehat D = {45^ \circ } \Rightarrow AF = DF = h\\\\B\mathop C\limits^\Delta  E:\quad \widehat {{B_1}} = \widehat C = {45^ \circ } \Rightarrow BE = CE = h\\\\ \Rightarrow CD = 2h + b = a \Rightarrow h = \frac{{a - b}}{2}\\\\{S_{ABCD}} = \frac{1}{2}\left( {a + b} \right)h\\\\ \Rightarrow {S_{ABCD}} = \frac{{a + b}}{2} \times \frac{{a - b}}{2} = \frac{{{a^2} - {b^2}}}{2}\end{array}\)

5)

\(\begin{array}{l}{S_{ABCD}} = \frac{1}{2}\left( {AB + CD} \right) \times AD\\ = \frac{1}{2}\left( {b + c} \right)\left( {b + c} \right) = \frac{1}{2}{\left( {b + c} \right)^2}\quad \left( i \right)\\\\{S_{ABCD}} = {S_{A\mathop B\limits^\Delta M}} + {S_{C\mathop D\limits^\Delta M}} + {S_{M\mathop B\limits^\Delta C}}\\ = \frac{1}{2}bc + \frac{1}{2}bc + \frac{1}{2}{a^2} = bc + \frac{1}{2}{a^2}\quad \left( {ii} \right)\\\\\mathop \Rightarrow \limits^{\quad \left( i \right),\left( {ii} \right)} \quad \frac{1}{2}{\left( {b + c} \right)^2} = bc + \frac{1}{2}{a^2}\\ \Rightarrow {\left( {b + c} \right)^2} = 2bc + {a^2}\\\\ \Rightarrow {b^2} + 2bc + {c^2} = 2bc + {a^2}\end{array}\)

\(\Rightarrow {b^2} + {c^2} = {a^2}\)   رابطه فیثاغورث :

6)

\(A\mathop B\limits^\Delta  C \cong A\mathop C\limits^\Delta  D \Rightarrow {S_{A\mathop B\limits^\Delta  C}} = {S_{A\mathop C\limits^\Delta  D}} = \frac{1}{2}{S_{A\mathop {BC}\limits^{} D}}\quad \left( i \right)\)

میانه های هر مثلث، آن را به شش قسمت با مساحت های مساوی تقسیم می کند؛ بنابراین :

\(\begin{array}{l}A\mathop B\limits^\Delta  C:\quad \left\{ \begin{array}{l}BM = MC\\AO = OC\end{array} \right. \Rightarrow {S_{M\mathop N\limits^\Delta  B}} = \frac{1}{6}{S_{A\mathop B\limits^\Delta  C}}\quad \left( {ii} \right)\\\\\mathop  \Rightarrow \limits^{\left( i \right),\left( {ii} \right)} \quad {S_{M\mathop N\limits^\Delta  B}} = \frac{1}{6}\left( {\frac{1}{2}{S_{A\mathop {BC}\limits^{} D}}} \right) = \frac{1}{{12}}{S_{A\mathop {BC}\limits^{} D}}\end{array}\)

7)

\(\begin{array}{l}\frac{{PC}}{{PB}} = \frac{1}{3} \Rightarrow \frac{{PC}}{{BC}} = \frac{1}{4}\\\\ \Rightarrow \frac{{{S_{A\mathop P\limits^\Delta  C}}}}{{{S_{A\mathop B\limits^\Delta  C}}}} = \frac{1}{4} \Rightarrow {S_{A\mathop B\limits^\Delta  C}} = 4{S_{A\mathop P\limits^\Delta  C}}\quad \left( i \right)\\\\A\mathop B\limits^\Delta  C:\quad MN\parallel BC \Rightarrow \left\{ \begin{array}{l}\frac{{AN}}{{AC}} = \frac{{AM}}{{AB}} = \frac{1}{3}\\A\mathop Q\limits^\Delta  N \sim A\mathop P\limits^\Delta  C\end{array} \right.\\\\ \Rightarrow \frac{{{S_{A\mathop Q\limits^\Delta  N}}}}{{{S_{A\mathop P\limits^\Delta  C}}}} = {\left( {\frac{1}{3}} \right)^2} = \frac{1}{9} \Rightarrow {S_{A\mathop P\limits^\Delta  C}} = 9{S_{A\mathop Q\limits^\Delta  N}}\quad \left( {ii} \right)\\\\\mathop  \Rightarrow \limits^{\left( i \right),\left( {ii} \right)} \quad {S_{A\mathop B\limits^\Delta  C}} = 4\left( {4{S_{A\mathop Q\limits^\Delta  N}}} \right) = 36{S_{A\mathop Q\limits^\Delta  N}}\\\\ \Rightarrow \frac{{{S_{A\mathop Q\limits^\Delta  N}}}}{{{S_{A\mathop B\limits^\Delta  C}}}} = \frac{1}{{36}}\end{array}\)

\(\begin{array}{l}\frac{{PB}}{{BC}} = \frac{3}{4} \Rightarrow \frac{{{S_{A\mathop P\limits^\Delta  B}}}}{{{S_{A\mathop B\limits^\Delta  C}}}} = \frac{3}{4} \Rightarrow {S_{A\mathop P\limits^\Delta  B}} = \frac{3}{4}{S_{A\mathop B\limits^\Delta  C}}\quad \left( i \right)\\\\A\mathop B\limits^\Delta  C:\quad MQ\parallel BP \Rightarrow A\mathop Q\limits^\Delta  M \sim A\mathop P\limits^\Delta  B\\\\ \Rightarrow \frac{{{S_{A\mathop Q\limits^\Delta  M}}}}{{{S_{A\mathop B\limits^\Delta  P}}}} = {\left( {\frac{1}{3}} \right)^2} = \frac{1}{9} \Rightarrow {S_{A\mathop B\limits^\Delta  P}} = 9{S_{A\mathop Q\limits^\Delta  M}}\\\\ \Rightarrow {S_{BPQM}} = \frac{8}{9}{S_{A\mathop B\limits^\Delta  P}}\quad \left( {ii} \right)\\\\\mathop  \Rightarrow \limits^{\left( i \right),\left( {ii} \right)} \quad {S_{BPQM}} = \frac{8}{9}\left( {\frac{3}{4}{S_{A\mathop B\limits^\Delta  C}}} \right) = \frac{2}{3}{S_{A\mathop B\limits^\Delta  C}}\\\\ \Rightarrow \frac{{{S_{BPQM}}}}{{{S_{A\mathop B\limits^\Delta  C}}}} = \frac{2}{3}\end{array}\)

8)

\(\begin{array}{l}\left. \begin{array}{l}\left. \begin{array}{l}{b_1} = 9\\{i_1} = 13\end{array} \right\} \Rightarrow {S_1} = \frac{{{b_1}}}{2} - 1 + {i_1} = \frac{{33}}{2}\\\left. \begin{array}{l}{b_2} = 5\\{i_2} = 3\end{array} \right\} \Rightarrow {S_2} = \frac{{{b_2}}}{2} - 1 + {i_2} = \frac{9}{2}\end{array} \right\}\\\\ \Rightarrow S = {S_1} - {S_2} = 12\end{array}\)

9)

مساحت به روش معمول : \(S = m \times n\)

مساحت به روش قضیه پیک :

\(\begin{array}{l}b = 2m + 2n\\\\i = \left( {m + 1} \right) \times \left( {n + 1} \right) - \left( {2m + 2n} \right)\\ = mn - m - n + 1\\\\S = \frac{b}{2} - 1 + i = \frac{{2m + 2n}}{2}\\\\ - 1 + \left( {mn - m - n + 1} \right)\\\\ = m + n - 1 + mn - m - n + 1\\\\ = mn = m \times n\end{array}\)

10)