a.
the area of triangle is equal to area of rectangle minus the area of three right triangles
A=(3,4)
B=(8,5)
C=(7,8}
sides of rectangle are:
from A to vertical side contains B, distance is 5 units (the length of rectangle)
from A to vertical side contains C, distance is 4 units (the width of rectangle)
area of rectangle is 5*4=20
right triangle with hypotenuse AC has an area of (4*4)/2=16/2=8
right triangle with hypotenuse AB has an area of (5*1))/2=5/2=2.5
right triangle with hypotenuse BC has an area of (3*1))/2=3/2=1.5
add all: 8+2.5+1.5=12
then area of triangle ABC is: 20-12=8
b.
A=(1,3)
B=(4,1)
C=(10,4)
horizontal distance from A to C is 9 units
vertical distance from B to C is 3 units
sides of rectangle are:9 and 4
area of rectangle is: 9*3=27
right triangle with hypotenuse AC has an area of (9*1)/2=9/2
right triangle with hypotenuse AB has an area of (3*2))/2=3
right triangle with hypotenuse BC has an area of (6*3))/2=9
then area of triangle ABC is: 27-(9/2+3+9)=10.5
c)
using formula (1/2)(x1*y2-x2*y1 +x2*y3-x3*y2+x3*y1-x1*y3)
A=(1,3)=(x1,y1)
B=(4,1)=(x2,y2)
C=(10,4)=(x3,y3)
area of triangle ABC is:
(1/2)(x1*y2-x2*y1 +x2*y3-x3*y2+x3*y1-x1*y3)
=(1/2)(1*1-4*3 +4*4-10*1+10*3-1*4)
=(1/2)(21)
=10.5
