Question

Find the area of the quadrilateral whose vertices are at (9,-2), (-8,-4), (2, 2) and (1,-3) (ii) ( -9, 0),
(-8, 6), (-1,-2) and (-6,-3)​

Answer 1

Answer:

Step-by-step explanation:

area of the quadrilateral whose vertices are at (9,-2), (-8,-4), (2, 2) and (1,-3)

= Sum of Area of two triangles whose vertices are

(9,-2), (-8,-4), (2, 2)   & (9,-2), (2, 2) ,  (1,-3)

= (1/2)| 9(-4 - 2) + (-8)(2 -(-2)) + 2(-2 - (-4))|  + (1/2)| 9(2 - (-3)) + 2(-3 -(-2)) + 1(-2 - 2)|  

= (1/2)|-54 -32 + 4| + (1/2)|45 - 2 -4|

= 41 + 39/2

= 121/2

Similarlry ( -9, 0), (-8, 6), (-1,-2) and (-6,-3)​

( -9, 0), (-8, 6), (-1,-2)    & ( -9, 0), (-1,-2) , (-6,-3)​

= (1/2)| -9(6 -(-2)) -8(-2 - 0) -1(0 - 6) |  + (1/2)| -9(-2 -(-3)) -1 (-3 -0) -6(0-(-2))|

= (1/2)| -72 + 16 + 6|  + (1/2)|-9 + 3 - 12|

= 25 + 9

= 34