What is the area of a parallelogram whose vertices are A(−4, 9) , B(11, 9) , C(5, −1) , and D(−10, −1) ?
Answers
area of a parallelogram ABCD = 150 sq units whose vertices are A(−4, 9) , B(11, 9) , C(5, −1) , and D(−10, −1)
Step-by-step explanation:
area of a parallelogram ABCD
= Area of ΔABC + Area of ΔACD
Area of ΔABC
A(−4, 9) , B(11, 9) , C(5, −1)
= (1/2) | -4 ( 9 - (-1)) + 11(-1 -9) + 5(9 - 9) |
= (1/2) | -40 - 110 + 0)
= (1/2) | - 150|
= 150/2
= 75
Area of ΔACD
A(−4, 9) , C(5, −1) , and D(−10, −1)
= (1/2) | -4 (-1 - (-1)) + 5(-1 - 9) -10(9 - (-1))|
= (1/2) | 0 -50 - 100|
= 75
area of a parallelogram ABCD = 75 + 75 = 150 sq units
Learn more with following links :
Find the area of the triangle whose vertices are :(-4,6) (20,8) (9,10)
https://brainly.in/question/12361320
Find the area of the parallelogram ABCD if three of its vertices are A(2,4) ,B(2+-√3,5) ,and C(2,6).
https://brainly.in/question/2080442
Point A(3,1), B(5,1) c(a,b) and D(4,3) are vertex of parallelogram ABCD. find the value of a b
https://brainly.in/question/2080442