Question

What is the area of the parallelogram represented by the vertices A(-6, -1), B(-2, -1), C(-1,-4) and D(-5, -4)?

Answer 1

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Answer 2

Answer:

12

Step-by-step explanation:

1) Vector AB (-2-(-6); -1-(-1))= (4; 0). Vector module is module AB=sqrt (4*4+0*0)= 4

2) The module of Vector BC is sqrt((-1-(-2))*(-1 - (-2))+ (-4-(-1))* (-4-(-1))= sqrt (9+1)= sqrt 10.

3) The module of AC is sqrt ((-1-(-6))*(-1-(-6))+ (-4-(-1))*(-4-(-1))= sqrt (25+9)= sqrt 34.

4) Having the triangle with the sides AB, BC, AC use the theorem of cos:

sqrt34*sqrt34= sqrt 10*sqrt 10 + 4*4- 2*sqrt10*4*cosB

34-10-16= -8 sqrt 10*cosB

c0sB= (-1)/ sqrt10

5) Find out sinB that is equal to Sqrt (1- ((-1)/sqrt 10)* ((-1)/sqrt10))= sqrt (9/10)= 3/sqrt10.

6)S= 4*sqrt10*3/sqrt10= 4*3=12. The area is equal to 12.