Question

Помогите пожалуйста, поставлю 40!

Answer 1

Відповідь:

Пояснення:

[[Number 1]]

4)

$KL^2 = KM^2 - LM^2 = 13^2-5^2 = (13-5)(13+5) = 8 \cdot 18 = 16 \cdot 9 = 4^2 \cdot 3^2 \ ;$

$KL^2 = 4^2 \cdot 3^2 \ ;$

$KL = 4 \cdot 3 = 12 \ ;$

$S_4 = \frac{1}{2}KL \cdot LM = \frac{1}{2} \cdot 12 \cdot 5 = 30 \ ;$

1)

$S_1 = \frac{1}{2}AB \cdot AC \sin{30^o} = \frac{1}{2} \cdot 6 \cdot 9 \cdot \frac{1}{2} = 13.5 \ ;$

25)

$S_{25} = \sqrt{p(p-a)(p-b)(p-c)} = \sqrt{10 \cdot (10-5)(10-6)(10-9)} = \sqrt{10 \cdot 5 \cdot 4 \cdot 1 } = \sqrt{5 \cdot 2 \cdot 5 \cdot 4 \cdot 1 } = 5 \cdot 2 \sqrt{2} = 10 \sqrt{2} \ ;$

19)

$S_{19} = KL \cdot KN \sin{30^o} = 15 \cdot 20 \cdot \frac{1}{2} = 150 \ ;$

21)

$S_{21} = PR \cdot RQ = 25 \cdot 8 = 200 \ ;$

18)

$S_{18} = AB \cdot AD = AB^2 = (4 \sqrt{2})^2 = 32 \ ;$

1*)

$S_{1*} = \frac{1}{2} AC \cdot BD = \frac{1}{2} \cdot 6 \cdot 8 = 24 \ ;$



[[Number 3]]

$\frac{MH}{HK} = \frac{NH}{HM} \ ;$

$\frac{MH}{25} = \frac{144}{HM} \ ;$

$MH^2 = 12^2 \cdot 5^2 \ ;$

$MH = 12 \cdot 5 = 60 \ ;$

$MN^2 = 60^2 + 144^2 = 12^2 (5^2 + 12^2) = 12^2 (25 + 144) = 12^2 \cdot 169 = 12^2 \cdot 13^2 \ ;$

$MN^2 = 12^2 \cdot 13^2 \ ;$

$MN = 12 \cdot 13 = 156 \ ;$

$KM^2 = 60^2 + 25^2 = 5^2 (12^2 + 5^2) = 5^2 (144 + 25) = 5^2 \cdot 169 = 5^2 \cdot 13^2 \ ;$

$KM^2 = 5^2 \cdot 13^2 \ ;$

$KM = 5 \cdot 13 = 65 \ ;$


[[Number 2 img]]