Notifications
Clear all
Coordinate Geometry
1
Posts
2
Users
1
Likes
335
Views
0
10/06/2021 1:50 pm
Topic starter
Find the area of the quadrilateral whose vertices, taken in order, are
(-4, -2), (-3, -5), (3, -2) and (2, 3).
1 Answer
1
10/06/2021 1:52 pm
Let the vertices of the quadrilateral be A (- 4, – 2), B (- 3, -5), C (3, – 2), and D (2, 3).
Join AC and divide the quadrilateral into two triangles.
We have two triangles ΔABC and ΔACD.
Area of a triangle = 1/2 × [x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]
Area of ΔABC = 1/2 [(-4) {(-5) – (-2)} + (-3) {(-2) – (-2)} + 3 {(-2) – (-5)}]
= 1/2 (12 + 0 + 9)
= 21/2 square units
Area of ΔACD = 1/2 [(-4) {(-2) – (3)} + 3{(3) – (-2)} + 2 {(-2) – (-2)}]
= 1/2 (20 + 15 + 0)
= 35/2 square units
Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD
= (21/2 + 35/2) square units = 28 square units
This post was modified 4 years ago by Raavi Tiwari