Question

знайти периметр трикутника ABC​

Answer 1

Ответ:

$12+4\sqrt{2}+4\sqrt{3}$

Объяснение:

$\angle C=180^{\circ}-\angle BNC-\angle NBC;$

$\angle C=180^{\circ}-90^{\circ}-60^{\circ}=30^{\circ} \Rightarrow BN=\dfrac{BC}{2} \Rightarrow BC=2 \cdot BN \Rightarrow BC=2 \cdot 4=8;$

$BN^{2}+NC^{2}=BC^{2};$

$NC^{2}=BC^{2}-BN^{2};$

$NC=\sqrt{BC^{2}-BN^{2}};$

$NC=\sqrt{8^{2}-4^{2}}=\sqrt{64-16}=\sqrt{48}=\sqrt{16 \cdot 3}=\sqrt{16} \cdot \sqrt{3}=4\sqrt{3};$

$\angle ABN=180^{\circ}-\angle BNA-\angle BAN;$

$\angle ABN=180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ};$

$\angle ABN=\angle BAN=45^{\circ} \Rightarrow AN=BN=4;$

$AN^{2}+BN^{2}=AB^{2};$

$AB=\sqrt{AN^{2}+BN^{2}};$

$AB=\sqrt{4^{2}+4^{2}}=\sqrt{16+16}=\sqrt{16 \cdot 2}=\sqrt{16} \cdot \sqrt{2}=4\sqrt{2};$

$AC=AN+NC \Rightarrow AC=4+4\sqrt{3};$

$P_{\Delta ABC}=AB+BC+AC;$

$P_{\Delta ABC}=4\sqrt{2}+8+4+4\sqrt{3}=12+4\sqrt{2}+4\sqrt{3};$