Question

Ache a área do triângulo ABC nos casos:

a) A(1, 1), B(4,0) e C(3,5)

b) A(-5/2, 1), B(7/2, 3/2) e C(1, -9/4)

C) A(0, 4), B(10, -2) e C(-1/2, -7/2)

D) A(0,5; 1,8), B(5; 0,5) e C(3/2; 1,5)

Answer 1

(A)

$\left[\begin{array}{ccc}1&1&1\\4&0&1\\3&5&1\end{array}\right] \left[\begin{array}{ccc}1&1&\\4&0&\\3&5&\end{array}\right]$

DET = 1*0*1 + 1*1*3 + 1*4*5 - 3*0*1 - 5*1*1 - 1*4*1 = 0 + 3 + 20 - 0 - 5 - 4 = 23 - 9 = 14

A = |DET| / 2 = 14/2 = 7 u.a.

(B)

$\left[\begin{array}{ccc}-5/2&1&1\\7/2&3/2&1\\1&-9/4&1\end{array}\right] \left[\begin{array}{ccc}-5/2&1&\\7/2&3/2&\\1&-9/4&\end{array}\right]$

DET = (-5/2)*(3/2)*1 + 1*1*1 + (1)*(7/2)*(-9/4) -1*(3/2)*1 - (9/4)*1*(5/2) - 1*(7/2)*1 = -15/4 + 1 - 63/8 - 3/2 - 45/8 - 7/2 = -30/8 + 8/8 - 63/8 - 12/8 - 45/8 - 28/8 = -168/8 + 8/8 = -160/8 = -20

A = |DET| / 2 = 20 / 2 = 10 u.a.

(C)

$\left[\begin{array}{ccc}0&4&1\\10&-2&1\\-1/2&-7/2&1\end{array}\right] \left[\begin{array}{ccc}0&4&\\10&-2&\\-1/2&-7/2&\end{array}\right]$

DET = 0*-2*1 + 4*1*-1/2 + 1*10*-7/2 - (1/2)*2*1 - (-7/2)*1*0 - 1*10*4 = 0 - 2 - 35 - 1 + 0 - 40 = -38 -40 = - 78

A = |DET| / 2 = 78 / 2 = 49 u.a.

(D)

$\left[\begin{array}{ccc}0,5&1,8&1\\5&0,5&1\\3/2&1,5&1\end{array}\right] \left[\begin{array}{ccc}0,5&1,8&\\5&0,5&\\3/2&1,5&\end{array}\right]$

DET = 0,5*0,5*1 + 1,8*1*(3/2) + 1*5*1,5 - (3/2)*0,5*1 - 1,5*1*0,5 - 1*5*1,8 = 0,25 + 2,7 + 7,5 - 0,75 - 0,75 - 9,0 = 10,45 - 10,50 = - 0,05

A = |DET| / 2 = 0,05 / 2 = 0,025 u.a