Question

Calculati aria triunghiului format din pnctele A(-2,2) B(1.0) C(-1,1)

Answer 1

$A(-2,2)$

$B(1,0)$

$C(-1,1)$

$AB=\sqrt{{(x_{B}-x_{A})}^{2}+{(y_{B}-y_{A})}^{2}}$

$AB=\sqrt{{(1 + 2)}^{2}+{(0-2)}^{2}}$

$AB=\sqrt{{3}^{2}+{(-2)}^{2}}$

$AB=\sqrt{9+4}$

$AB=\sqrt{13}$

$AC=\sqrt{{(x_{C}-x_{A})}^{2}+{(y_{C}-y_{A})}^{2}}$

$AC=\sqrt{{( - 1 + 2)}^{2}+{(1-2)}^{2}}$

$AC=\sqrt{{1}^{2}+{(-1)}^{2}}$

$AC=\sqrt{1 + 1}$

$AC=\sqrt{2}$

$BC=\sqrt{{(x_{C}-x_{B})}^{2}+{(y_{C}-y_{B})}^{2}}$

$BC=\sqrt{{( - 1-1)}^{2}+{(1-0)}^{2}}$

$BC=\sqrt{{( - 2)}^{2}+{1}^{2}}$

$BC=\sqrt{4+1}$

$BC=\sqrt{5}$

Pentru arie folosim formula lui Heron :

$AB=a$

$AC=b$

$BC=c$

$p = \frac{a + b + c}{2}$

$p = \frac{ \sqrt{13} + \sqrt{2} + \sqrt{5} }{2}$

$A=\sqrt{p(p-a)(p-b)(p-c)}$

$A=\sqrt{ \frac{ \sqrt{13} + \sqrt{2} + \sqrt{5} }{2} ( \frac{ \sqrt{13} + \sqrt{2} + \sqrt{5} }{2} - \sqrt{13} )( \frac{ \sqrt{13} + \sqrt{2} + \sqrt{5} }{2} - \sqrt{2} )( \frac{ \sqrt{13} + \sqrt{2} + \sqrt{5} }{2} - \sqrt{5} )}$

$A=\sqrt{ \frac{ \sqrt{13} + \sqrt{2} + \sqrt{5} }{2} \times \frac{ - \sqrt{13} + \sqrt{2} + \sqrt{5} }{2} \times \frac{ \sqrt{13} - \sqrt{2} + \sqrt{5} }{2} \times \frac{ \sqrt{13} + \sqrt{2} - \sqrt{5} }{2} }$

$A = \sqrt{ \frac{( \sqrt{13} + \sqrt{2} + \sqrt{5} )( - \sqrt{13} + \sqrt{2} + \sqrt{5} )( \sqrt{13} - \sqrt{2} + \sqrt{5} )( \sqrt{13} + \sqrt{2} - \sqrt{5} )}{16} }$

$( \sqrt{13} + \sqrt{2} + \sqrt{5} )( - \sqrt{13} + \sqrt{2} + \sqrt{5} ) = - 1 + \sqrt{26} + \sqrt{65} - \sqrt{26} + 2 + \sqrt{10} - \sqrt{65} + \sqrt{10} + 5 = 6 + 2 \sqrt{10}$

$( \sqrt{13} - \sqrt{2} + \sqrt{5} )( \sqrt{13} + \sqrt{2} - \sqrt{5} ) = 13 + \sqrt{26} - \sqrt{65} - \sqrt{26} - 2 + \sqrt{10} + \sqrt{65} + \sqrt{10} - 5 = 6 + 2 \sqrt{10}$

$A = \sqrt{ \frac{(6 + 2 \sqrt{10} )(6 + 2 \sqrt{10}) }{16} }$

$A = \sqrt{ \frac{ {(6 + 2 \sqrt{10} )}^{2} }{ {4}^{2} } }$

$A = \frac{6 + 2 \sqrt{10} }{ 4}$

$A = \frac{2(3 + \sqrt{10} )}{4}$

$A = \frac{3 + \sqrt{10} }{2}$