جواب تمرین صفحه 73 درس 3 هندسه یازدهم

الف

\(\begin{array}{l}B{C^2} = A{C^2} + A{B^2} - 2AC\,.AB.\cos A\\\\B{C^2} = 36 + 100 - 2 \times 6 \times 10 \times \frac{1}{2} = 76\\\\ \Rightarrow BC = 2\sqrt {19} \end{array}\)

ب

\(S = \frac{1}{2}.AC.AB.\sin A = \frac{1}{2} \times 6 \times 10 \times \frac{{\sqrt 3 }}{2} = 15\sqrt 3 \)

پ

\(\begin{array}{l}\frac{{\sin B}}{b} = \frac{{\sin A}}{a} \Rightarrow \frac{{\sin B}}{6} = \frac{{\frac{{\sqrt 3 }}{2}}}{{2\sqrt {19} }}\\\\ \Rightarrow \sin B = \frac{3}{2}\sqrt {\frac{3}{{19}}} \Rightarrow \sin B = \frac{{3\sqrt {57} }}{{38}}\end{array}\)

\(\begin{array}{l}{S_{ABCD}} = {S_{ABD}} + {S_{BCD}}\\\\{P_{ABD}} = \frac{{4 + 13 + 15}}{2} = 16\\\\{P_{BCD}} = \frac{{11 + 13 + 20}}{2} = 22\\\\{S_{ABD}} = \sqrt {16 \times 12 \times 3 \times 1} = 24\\\\{S_{BCD}} = \sqrt {22 \times 11 \times 9 \times 2} = 66\\\\{S_{ABCD}} = 24 + 66 = 90\end{array}\)

\(\begin{array}{l}\left. \begin{array}{l}{S_{ABD}} = \frac{1}{2}AB.BD.\sin \alpha \\{S_{BCD}} = \frac{1}{2}BC.BD.\sin \beta \end{array} \right\} \Rightarrow \\\\\left. \begin{array}{l}24 = \frac{1}{2} \times 4 \times 13 \times \sin \alpha \\66 = \frac{1}{2} \times 11 \times 13 \times \sin \beta \end{array} \right\} \Rightarrow \\\\\left. \begin{array}{l}\sin \alpha = \frac{{12}}{{13}}\\\sin \beta = \frac{{12}}{{13}}\end{array} \right\} \Rightarrow \sin \alpha = \sin \beta \end{array}\)

\(\begin{array}{l}AB = AC = BC = a \Rightarrow {P_{ABC}} = \frac{{3a}}{2}\\\\{S_{ABC}} = \sqrt {\frac{{3a}}{2}{{(\frac{{3a}}{2} - a)}^3}} \\\\ = \sqrt {\frac{{3a}}{2}{{(\frac{a}{2})}^3}} = \sqrt {\frac{{3{a^4}}}{{16}}} = \frac{{\sqrt 3 }}{4}{a^2}\end{array}\)

با توجه به اینکه مثلث ADE متساوی الساقین است، پس \(D\hat AE = {60^ \circ }\) در نتیجه:

\(\begin{array}{l}B{C^2} = A{C^2} + A{B^2} - 2AC.AB.\cos A\\\\B{C^2} = 36 + 64 - 2 \times 6 \times 8 \times \frac{1}{2} = 52\\\\ \Rightarrow BC = 2\sqrt {13} \\\\{S_{BCED}} = {S_{ABC}} - {S_{ADE}}\\\\{S_{ABC}} = \frac{1}{2}AB.AC.\sin A = \frac{1}{2} \times 8 \times 6 \times \frac{{\sqrt 3 }}{2}\\ = 12\sqrt 3 \\\\{S_{ADE}} = \frac{{\sqrt 3 }}{4}{a^2} = \frac{{\sqrt 3 }}{4} \times {1^2} = \frac{{\sqrt 3 }}{4}\\\\{S_{BCED}} = 12\sqrt 3 - \frac{{\sqrt 3 }}{4} = \frac{{47\sqrt 3 }}{4}\end{array}\)

\(\begin{array}{l}{S_{ABC}} = {S_{ABD}} + {S_{ACD}}\; \Rightarrow \\\\\frac{1}{2}AB\;.\;AC\;.\;\sin A = \frac{1}{2}AB\; \times AD\; \times \sin \frac{A}{2} + \frac{1}{2}AC \times AD\; \times \sin \frac{A}{2}\\\\ \Rightarrow \;\;AB\;.\;AC\;.\;\sin A = AD\;.\;\sin \frac{A}{2}\left( {AB + AC} \right)\\\\ \Rightarrow \;AD = \frac{{AB\;.\;AC\;.\;\sin A}}{{\left( {AB + AC} \right)\sin \frac{A}{2}}} = \frac{{2AB\;.\;AC\;.\;\sin \frac{A}{2}\cos \frac{A}{2}}}{{\left( {AB + AC} \right)\sin \frac{A}{2}}}\end{array}\)

 \({d_a} = \frac{{2bc\;.\;\cos \frac{A}{2}}}{{b + c}}\) (نیمساز رأس A)  \( \Rightarrow \;AD = {d_a}\;\; \Rightarrow \)

\(\begin{array}{l}{S_{ABC}} = {S_{OAB}} + {S_{OAC}} + {S_{OBC}}\\\\{P_{ABC}} = \frac{{5 + 6 + 7}}{2} = 9\\\\{S_{ABC}} = \sqrt {9 \times 2 \times 3 \times 4} = 6\sqrt 6 \\\\{S_{OAB}} = \frac{1}{2} \times 6 \times 3 = 9\\\\{S_{OAC}} = \frac{1}{2} \times 5 \times 2 = 5\\\\{S_{OBC}} = \frac{1}{2} \times 7 \times x = \frac{7}{2}x\\\\6\sqrt 6 = 9 + 5 + \frac{7}{2}x\\\\ \Rightarrow x = \frac{2}{7}(6\sqrt 6 - 14) \simeq 0/2\end{array}\)

مثلث BCD متساوی الساقین است و با توجه به اندازه زاویه C ، دو زاویه دیگر هر کدام 30 درجه خواهد بود. در مثلث قائم الزاویه CHD \(\widehat {CDH} = {30^ \circ }\)،  در نتیجه: \(CH = \frac{7}{2}\)

\(\begin{array}{l}\left. \begin{array}{l}{S_{BCD}} = \frac{1}{2} \times \frac{7}{2} \times BD = \frac{7}{4}BD\\{S_{BCD}} = \frac{1}{2} \times 7 \times 7 \times \sin {120^ \circ } = \frac{{49\sqrt 3 }}{4}\end{array} \right\}\\\\ \Rightarrow \frac{7}{4}BD = \frac{{49\sqrt 3 }}{4} \Rightarrow BD = 7\sqrt 3 \\\\{P_{ABD}} = \frac{{11 + 13 + 7\sqrt 3 }}{2} = 12 + \frac{{7\sqrt 3 }}{2}\\\\{S_{ABD}} = \sqrt {(12 + \frac{{7\sqrt 3 }}{2})(12 + \frac{{7\sqrt 3 }}{2} - 7\sqrt 3 )(12 + \frac{{7\sqrt 3 }}{2} - 11)(12 + \frac{{7\sqrt 3 }}{2} - 13)} \\\\{S_{ABD}} = \sqrt {(12 + \frac{{7\sqrt 3 }}{2})(12 - \frac{{7\sqrt 3 }}{2})(\frac{{7\sqrt 3 }}{2} + 1)(\frac{{7\sqrt 3 }}{2} - 1)} \\\\{S_{ABD}} = \sqrt {(144 - \frac{{174}}{4})(\frac{{147}}{4} - 1)} = \frac{{143}}{4}\sqrt 3 \quad \left( 1 \right)\\\\{S_{ABD}} = \frac{1}{2} \times 11 \times 13 \times \sin A = \frac{{143}}{2}\sin A\quad \left( 2 \right)\\\\\mathop \Rightarrow \limits^{\left( 1 \right)\;,\;\left( 2 \right)} \;\;\frac{{143}}{2}\sin A = \frac{{143}}{4}\sqrt 3 \Rightarrow \sin A = \frac{{\sqrt 3 }}{2} \Rightarrow \widehat A = {60^ \circ }\\\\{S_{ABCD}} = {S_{ABD}} - {S_{BCD}} = \frac{{143}}{4}\sqrt 3 - \frac{{49}}{4}\sqrt 3 = \frac{{47}}{2}\sqrt 3 \\\end{array}\)

با توجه به خواص متوازی الاضلاع داریم :

\(\begin{array}{l}{S_{ABCD}} = 2\,{S_{ABC}} = 2 \times \frac{1}{2} \times a \times b \times \sin \alpha \\\\ \Rightarrow {S_{ABCD}} = a.b\sin \alpha \end{array}\)

الف

\(\begin{array}{l}\widehat A > {90^ \circ } \Leftrightarrow \cos A < 0\;\;\mathop \Leftrightarrow \limits_{ \div bc}^{ \times bc} \;bc.\cos A < 0\\\\ \Leftrightarrow - bc.\cos A > 0\;\;\mathop \Leftrightarrow \limits_{ - \left( {{b^2} + {c^2}} \right)}^{ + \left( {{b^2} + {c^2}} \right)} \\\\{b^2} + {c^2} - bc.\cos A > {b^2} + {c^2}\\\\ \Leftrightarrow {a^2} > {b^2} + {c^2}\end{array}\)

ب

\(\begin{array}{l}\widehat A < {90^ \circ } \Leftrightarrow \cos A > 0\;\mathop \Leftrightarrow \limits_{ \div bc}^{ \times bc} bc.\cos A > 0\\\\ \Leftrightarrow - bc.\cos A < 0\;\;\mathop \Leftrightarrow \limits_{ - \left( {{b^2} + {c^2}} \right)}^{ + \left( {{b^2} + {c^2}} \right)} \\\\{b^2} + {c^2} - bc.\cos A < {b^2} + {c^2}\\\\ \Leftrightarrow {a^2} < {b^2} + {c^2}\end{array}\)

پ

\(\begin{array}{l}\widehat A = {90^ \circ } \Leftrightarrow \cos A = 0\;\;\mathop \Leftrightarrow \limits_{ \div bc}^{ \times bc} \;bc.\cos A = 0\\\\ \Leftrightarrow - bc.\cos A = 0\;\;\mathop \Leftrightarrow \limits_{ - \left( {{b^2} + {c^2}} \right)}^{ + \left( {{b^2} + {c^2}} \right)} \\\\{b^2} + {c^2} - bc.\cos A = {b^2} + {c^2}\\\\ \Leftrightarrow {a^2} = {b^2} + {c^2}\end{array}\)

الف

\(\begin{array}{l}a = 9\quad ,\quad b = 6\quad ,\quad c = 10\\\\{a^2} = 81\quad ,\quad {b^2} + {c^2} = 136\\\\ \Rightarrow {a^2} < {b^2} + {c^2} \Rightarrow \widehat A < {90^ \circ }\end{array}\)

ب

\(\begin{array}{l}a = 9\quad ,\quad b = 4\quad ,\quad c = 8\\\\{a^2} = 81\quad ,\quad {b^2} + {c^2} = 80\\\\ \Rightarrow {a^2} > {b^2} + {c^2} \Rightarrow \widehat A > {90^ \circ }\end{array}\)

پ

\(\begin{array}{l}a = 17\quad ,\quad b = 15\quad ,\quad c = 8\\\\{a^2} = 289\quad ,\quad {b^2} + {c^2} = 289\\\\ \Rightarrow {a^2} = {b^2} + {c^2} \Rightarrow \widehat A = {90^ \circ }\end{array}\)