23 uždavinys
Taškas C priklauso pusapskritimiui su centru O. AB ⊥ CD, AD = 4, DB = 9. Apskaičiuokite atkarpos CD ilgį.
Sprendimas.
AB = AD + DB = 4 + 9 = 13.
AO = AB / 2 = 6.5
OC = AO = 6.5.
DO = AO - AD = 6.5 - 4 = 2.5.
Pagal pitagoro teoremą $$OC^{2} = DO^{2}+CD^{2}$$.
$$CD = \sqrt {OC^{2}-DO^{2}}$$
saknis(
6.5^2- 2.5^2) =
6.5^2- 2.5^2) = |
|
saknis( 6.5^2- 2.5^2) = $$\sqrt {6.5^{2}-2.5^{2}}$$ =
6.5^2- 2.5^2) = $$\sqrt {6.5^{2}-2.5^{2}}$$ = $${\normalsize 6.5^{2}}$$ = $${\normalsize 42.25}$$
saknis(42.25- 2.5^2) = $$\sqrt {42.25-2.5^{2}}$$ =
42.25- 2.5^2) = $$\sqrt {42.25-2.5^{2}}$$ = $${\normalsize 2.5^{2}}$$ = $${\normalsize 6.25}$$
saknis(42.25-6.25) = $$\sqrt {42.25-6.25}$$ =
42.25-6.25) = $$\sqrt {42.25-6.25}$$ = $${\normalsize 42.25-6.25}$$ = $${\normalsize 36}$$
saknis(36) = $$\sqrt {36}$$ =
36) = $$\sqrt {36}$$ = $${\normalsize \sqrt {36}}$$ = $${\normalsize 6}$$
6$$6$$
$$\sqrt {6.5^{2}-2.5^{2}}$$ = $$$$
$$\sqrt {42.25-6.25}$$ = $$$$
$$\sqrt {36}$$ = $$$$
$$6$$ $$$$
Atsakymas: 6