Question

Calcule a área do quadrilátero ABCD, cujas coordenadas dos vértices são A (3,5), B (-1,2), C (5,3) e D (4,8) ​

Answer 1

$\boxed{\begin{array}{l}\underline{\rm Observe~a~figura~que~anexei}\\\sf A~\acute area~pode~ser~calculada~com~o~auxilio~do~determinante~\\\sf da~matriz\\\sf M=\begin{vmatrix}3&\sf5\\\sf-1&\sf2\\\sf5&\sf3\\\sf4&\sf8\\\sf3&5\end{vmatrix}\\\underline{\rm determinante~da~direita:}\\\sf det~M_{direita}=3\cdot2-1\cdot3+5\cdot8+4\cdot5\\\sf det~M_{direita}=6-3+40+20=63\\\underline{\rm determinante~da~esquerda:}\\\sf det~M_{esquerda}=5\cdot-1+2\cdot5+3\cdot4+8\cdot3\\\sf det~M_{esquerda}=-5+10+12+24=41\end{array}}$

$\Large\boxed{\begin{array}{l}\underline{\rm C\acute alculo~do~determinante~da~matriz~M:}\\\sf det~M=det~M_{direita}-det~M_{esquerda}\\\sf det~M=63-41=22\\\underline{\rm A~\acute area~e~dada~por:}\\\sf A=\dfrac{|det~M|}{2}\\\sf A=\dfrac{22}{2}\\\sf A=11~u\cdot a\end{array}}$

Answer 2

Resposta:

Explicação passo a passo:

Calcule a área do quadrilátero ABCD, cujas coordenadas dos vértices são

A (3,5), B (-1,2), C (5,3) e D (4,8) ​

∆ ABC

3 5 1 3 5

-1 2 1 -1 2

5 3 1 5 3

D = 3.2.1 + 5.1.5 +1.(-1).3 - [1.2.5+3.1.3 +5.(-1).1]

D = 6+25-3-(10+9-5)

D= 28-(14)

D = 14

A |D|/2= 14/2= 7

∆ ABD

3 5 1 3 5

-1 2 1 -1 2

4 8 1 4 8

D = 6+20-8-(8+24-5)

D = 18-(27)

D = -9

A = |d|/2= 9/2= 4,5

A = 7+4,5

A= 11,5 u.a.