SOLUTION: A(5;5), B(-7;1), C(1;-7) are the vertices of triangle ABC which is an isosceles triangle. D(x;y) is a point in Quadrant 4 such that ABCD is a parallelogram. Calculate the area of A

Algebra ->  Surface-area -> SOLUTION: A(5;5), B(-7;1), C(1;-7) are the vertices of triangle ABC which is an isosceles triangle. D(x;y) is a point in Quadrant 4 such that ABCD is a parallelogram. Calculate the area of A      Log On


   

Question 863143: A(5;5), B(-7;1), C(1;-7) are the vertices of triangle ABC which is an isosceles triangle. D(x;y) is a point in Quadrant 4 such that ABCD is a parallelogram. Calculate the area of ABCD correct to 3 decimal places.
Answer by mananth(16946) About Me  (Show Source):
You can put this solution on YOUR website!
Distance between two points (AB)
x1 y1 x2 y2

5 5 -7 1
d= sqrt%28%28y2-y1%29%5E2%2B%28x2-x1%29%5E2%29
d= sqrt%28%28%09%091%09-%095%09%29%5E2%09%2B%09%28%09-7%09-%095%09%29%5E2%09%29
d= sqrt%28%28%09%09-4%09%29%5E2%09%2B%09%28%09-12%09%29%5E2%09%29
d= sqrt%28%28%09%09160%09%29++%09%29
d= 12.65
Distance between two points (BC)
x1 y1 x2 y2

1 -7 -7 1
d= sqrt%28%28y2-y1%29%5E2%2B%28x2-x1%29%5E2%29
d= sqrt%28%28%09%091%09-%09-7%09%29%5E2%09%2B%09%28%09-7%09-%091%09%29%5E2%09%29
d= sqrt%28%28%09%098%09%29%5E2%09%2B%09%28%09-8%09%29%5E2%09%29
d= sqrt%28%28%09%09128%09%29++%09%29
d= 11.31

Distance between two points (AC)
x1 y1 x2 y2

-7 1 5 5
d= sqrt%28%28y2-y1%29%5E2%2B%28x2-x1%29%5E2%29
d= sqrt%28%28%09%095%09-%091%09%29%5E2%09%2B%09%28%095%09-%09-7%09%29%5E2%09%29
d= sqrt%28%28%09%094%09%29%5E2%09%2B%09%28%0912%09%29%5E2%09%29
d= sqrt%28%28%09%09160%09%29++%09%29
d= 12.65

Apply Herons formula for area of triangle
Area of triangle = sqrt%28s%28s-a%29%28s-b%29%28s-c%29%29
where s = 1/2 perimeter
Perimeter36.61
s=18.31
Area =+sqrt%2818.31%2818.31-11.31%29%5E2%2B%2818.31-12.65%29%5E2%2B%2818.31-12.65%29%5E2%29
=31.004
area of parallelogram = 2 * area of triangle
=2*31.004
=62.008 sq. units
=