Answer :

Given vertices of quadrilateral ABCD,


A (-3, 5), B (-2, -7), C (1, -8) and D (6, 3)


We know that area of quadrilateral ABCD = area of Δ ABC + area of Δ ACD and area of Δ ABC = 1/2 |x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2) |


Consider Δ ABC


Area of Δ ABC = 1/2 | (-3) (-7 – (-8)) + (-2) (-8 – 5) + (1) (5 – (-7)) |


= 1/2 | (-3) (1) + (-2) (-13) + 1 (12) |


= 1/2 | -3 – 26 + 12|


= 1/2 |-17|


= 1/2 (17)


Area of Δ ABC = 17/2 square units


Now in Δ ACD,


Area of Δ ACD = 1/2 | (-3) (-8 – 3) + (1) (3 – 5) + (6) (5 – (-8)) |


= 1/2 | (-3) (-11) + (1) (-2) + 6 (13) |


= 1/2 | -33 – 2 + 78|


= 1/2 |43|


Area of Δ ACD = 43/2


Area of quadrilateral = 17/2 + 43/2 = 60/2 = 30 square units


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Glimpses of IndiaGlimpses of IndiaGlimpses of India39 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
caricature
view all courses