1171 Вправа 1171Геометрія 8 клас Бурда 2025 · § 21 · стор. 234

1171. Знайдіть висоти трикутника зі сторонами a, b і c та площею S за таблицею 39.

S = ${\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}$aha = ${\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}$bhb = ${\displaystyle \frac{\mathbf{1}}{\mathbf{2}}}$chc, звідси

ha = ${\displaystyle \frac{\mathbf{2}\mathbf{S}}{\mathbf{a}}}$; hb = ${\displaystyle \frac{\mathbf{2}\mathbf{S}}{\mathbf{b}}}$; hc = ${\displaystyle \frac{\mathbf{2}\mathbf{S}}{\mathbf{c}}}$.

1) ha = ${\displaystyle \frac{\mathbf{2}\mathbf{\cdot }\mathbf{84}}{\mathbf{13}}}$ = ${\displaystyle \frac{\mathbf{168}}{\mathbf{13}}}$ (см);

hb = ${\displaystyle \frac{\mathbf{2}\mathbf{\cdot }\mathbf{84}}{\mathbf{14}}}$ = 12 (см);

hc = ${\displaystyle \frac{\mathbf{2}\mathbf{\cdot }\mathbf{84}}{\mathbf{15}}}$ = ${\displaystyle \frac{\mathbf{168}}{\mathbf{15}}}$ = 11,2 (см).

2) ha = ${\displaystyle \frac{\mathbf{2}\mathbf{\cdot }\mathbf{126}}{\mathbf{13}}}$ = ${\displaystyle \frac{\mathbf{252}}{\mathbf{13}}}$ (см);

hb = ${\displaystyle \frac{\mathbf{2}\mathbf{\cdot }\mathbf{126}}{\mathbf{20}}}$ = 12,6 (см);

hc = ${\displaystyle \frac{\mathbf{2}\mathbf{\cdot }\mathbf{126}}{\mathbf{21}}}$ = 12 (см).

3) ha = ${\displaystyle \frac{\mathbf{2}\mathbf{\cdot }\mathbf{42}}{\mathbf{7}}}$ = 12 (см);

hb = ${\displaystyle \frac{\mathbf{2}\mathbf{\cdot }\mathbf{42}}{\mathbf{15}}}$ = ${\displaystyle \frac{\mathbf{84}}{\mathbf{15}}}$ = 5,6 (см);

hc = ${\displaystyle \frac{\mathbf{2}\mathbf{\cdot }\mathbf{42}}{\mathbf{20}}}$ = 4,2 (см).

4) ha = ${\displaystyle \frac{\mathbf{2}\mathbf{\cdot }\mathbf{36}}{\mathbf{9}}}$ = 8 (см);

hb = ${\displaystyle \frac{\mathbf{2}\mathbf{\cdot }\mathbf{36}}{\mathbf{10}}}$ = 7,2 (см);

hc = ${\displaystyle \frac{\mathbf{2}\mathbf{\cdot }\mathbf{36}}{\mathbf{17}}}$ = ${\displaystyle \frac{\mathbf{72}}{\mathbf{17}}}$ (см).