572 Найдите: а) һ, а и ь, если b = 25, а = 16; б) һ, а и ь, если b = 36, а = 64; в) а, с и а, если b = 12, b = 6; г) б, с и ь, если а = 8, а = 4; д) һ, b, а и в, если а = 6, с = 9.

а) За дано: b = 25, ac = 16

$$h = \frac{a \cdot b}{c}$$, $$a_c + b_c = c$$, $$h^2 = a_c \cdot b_c$$, $$a^2 = a_c \cdot c$$, $$b^2 = b_c \cdot c$$

$$a_c = 16$$

$$a^2 = a_c \cdot c$$ => $$c = \frac{a^2}{a_c}$$,

$$c = \frac{256}{16} = 16$$

$$b_c = c - a_c= 16-16 = 0$$

$$b^2 = b_c \cdot c = 0 \cdot 16 = 0 => b=0 $$ - невозможно

б) За дано: b = 36, ac = 64

$$a_c = 64$$

$$a^2 = a_c \cdot c$$ => $$c = \frac{a^2}{a_c}$$,

$$c = \frac{36^2}{64} = \frac{1296}{64} = 20,25$$

$$b_c = c - a_c= 20,25 - 64 = -43,75$$ - невозможно, т.к. меньше 0

в) За дано: b = 12, bc = 6

$$b^2 = b_c \cdot c$$ => $$c = \frac{b^2}{b_c}$$ => $$c = \frac{144}{6} = 24$$

$$a_c = c- b_c = 24 - 6 = 18$$

$$a^2 = a_c \cdot c= 18 \cdot 24= 432$$ => $$a=\sqrt{432} \approx 20,78$$

г) За дано: a_c = 4, a = 8

$$a^2 = a_c \cdot c => c= \frac{a^2}{a_c}= \frac{64}{4} = 16$$

$$b_c=c-a_c = 16-4=12$$

$$b^2 = b_c \cdot c= 12 \cdot 16= 192=> b = \sqrt{192} \approx 13,86$$

д) За дано: a=6, c=9

$$a^2 + b^2= c^2 => b = \sqrt{c^2-a^2}= \sqrt{81-36} = \sqrt{45} \approx 6,71$$

$$a_c = \frac{a^2}{c} = \frac{36}{9} = 4$$

$$b_c = c - a_c= 9-4=5$$

$$h = \frac{a \cdot b}{c} = \frac{6 \cdot \sqrt{45}}{9} = \frac{2 \cdot \sqrt{45}}{3} \approx 4,47$$