گام به گام تمرین صفحه 42 درس 2 هندسه (1) (قضیۀ تالس، تشابه و کاربردهای آن)

1)

2)

\(1)\;BH = 9\;\;\;,\;\;\;CH = 4\;\;\;,\;\;\;AH = ?\;\;\;,\;\;\;AB = ?\;\;\;,\;\;\;AC = ?\)

\(\frac{{BH}}{{AH}} = \frac{{AH}}{{CH}} \Rightarrow \frac{9}{{AH}} = \frac{{AH}}{4}\)

\( \Rightarrow A{H^2} = 9 \times 4 = 36 \Rightarrow AH = 6\)

\(AB = \sqrt {B{H^2} + A{H^2}}  = \sqrt {{9^2} + {6^2}}  = 3\sqrt {13}\)

\(AC = \sqrt {B{H^2} + A{H^2}}  = \sqrt {{4^2} + {6^2}}  = 2\sqrt {13}\)

\(2)\;AB = 8\;\;\;,\;\;\;AC = 6\;\;\;,\;\;\;BH = ?\;\;\;,\;\;\;CH = ?\)

\(BC = \sqrt {A{B^2} + A{C^2}}  = \sqrt {{8^2} + {6^2}}  = 10\)

\(\Rightarrow \frac{1}{2}AH \times 10 = \frac{1}{2} \times 8 \times 6 \Rightarrow AH = \frac{{24}}{5}\)

\(\frac{{BH}}{{AH}} = \frac{{AB}}{{AC}} \Rightarrow \frac{{BH}}{{\frac{{24}}{5}}} = \frac{8}{6} \Rightarrow BH = \frac{{32}}{5} = 6/4\)

\(CH = BC - BH = 10 - 6/4 = 3/2\)

3)

\(\widehat B = \widehat {{A_1}}\)  ,   \(\widehat C =\) مشترک

 \(\Rightarrow A\mathop B\limits^\Delta  C \sim A\mathop D\limits^\Delta  C \Rightarrow \frac{{DC}}{{AC}} = \frac{{AC}}{{BC}} = \frac{{AD}}{{AB}} \)

\(\Rightarrow A{C^2} = DC \cdot BC\)

\(\Rightarrow {4^2} = \left( {BC - BD} \right)BC \)

\(\Rightarrow 16 = \left( {BC - 6} \right)BC \Rightarrow BC = 8\)

4)

\(\left. \begin{array}{l}{15^2} = {\left( {14 - x} \right)^2} + {y^2} = 225\\{13^2} = {x^2} + {y^2} = 169\end{array} \right\}\)

\( \Rightarrow 225 - {\left( {14 - x} \right)^2} = 169 - {x^2}\)

\(\Rightarrow 225 - 196 - {x^2} + 28x = 169 - {x^2}\)

\(\Rightarrow 28x = 140 \Rightarrow x = 5\)

\(\Rightarrow {y^2} = 169 - {x^2} = 169 - 25 = 144 \Rightarrow y = 12\)

5)

الف)

\(\frac{h}{{2/5}} = \frac{{12}}{{1/5}} \Rightarrow h = 20\)

ب)

\(\frac{h}{{1/6}} = \frac{{20}}{{2/5}} \Rightarrow h = 12/8\)

6)

الف)

اگر در مثلث ABC رابطه \({a^2} = {b^2} + {c^2}\) برقرار باشد، آن گاه زاویه A قائمه خواهد بود.

 ب)

طبق فرض: \({a^2} = {b^2} + {c^2}\)

\(\widehat {A'} = {90^ \circ } \Rightarrow \left\{ \begin{array}{l}{{a'}^2} = {{b'}^2} + {{c'}^2} = {b^2} + {c^2} = {a^2} \Rightarrow a' = a\\b' = b\\c' = c\end{array} \right.\)

\(\Rightarrow\) بنا به حالت (ض ض ض)

\(\Rightarrow A\mathop B\limits^\Delta  C \cong A'\mathop {B'}\limits^\Delta  C' \Rightarrow \widehat A = \widehat {A'} = {90^ \circ }\)

ج) اگر در مثلث ABC ،زاویه A قائمه باشد، آنگاه رابطه \({a^2} = {b^2} + {c^2}\) برقرار است و برعکس .