Set 1.4 Question 34 See attachment. Work out (a) through (c). [ATTACH=full]117[/ATTACH]

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

I will post questions like this one from time to time. Due to lack of time, I don't have time to completely deep into a problem like this one but at least I have a legitimate reason: working full-time.