Find the area of the quadrilaterals, the coordinates of whose vertices are

(i) (-3, 2), (5, 4), (7, - 6) and (-5, - 4)

(ii) (1, 2), (6, 2), (5, 3) and (3, 4)

(iii) (-4, - 2), (-3, - 5), (3, - 2), (2, 3)

(i) (-3, 2), (5, 4), (7, - 6) and (-5, - 4)

Let the vertices of the quadrilateral be A (−3, 2), B (5, 4), C (7, −6), and D (−5, −4). Join AC to form two triangles ∆ABC and ∆ACD.

Area of □ ABCD = Area of ∆ABC + Area of ∆ACD

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃)

= |x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|

Area of ∆ABC

= |-3(4 – (-6))+5(-6 - 2)+7(2 - 4)|

= |-30 – 40 -14|

= 42 sq. units

Area of ∆ACD

= |-3(-6 – 4)+7(-4 - 2) - 5(2 + 6)|

= |6 – 42 - 40|

= 38 sq. units

Area of □ ABCD = 42 + 38 = 80 sq. units

(ii) (1, 2), (6, 2), (5, 3) and (3, 4)

Let the vertices of the quadrilateral be A (1, 2), B (6, 2), C (5, 3), and D (3, 4). Join AC to form two triangles ∆ABC and ∆ACD.

Area of □ ABCD = Area of ∆ABC + Area of ∆ACD

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃)

= |x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|

Area of ∆ABC

= |1(2 – 3)+6(3 - 2)+5(2 - 2)|

= |-1 + 6|

= sq. units

Area of ∆ACD

= |1(3 – 4)+5(4 - 2) + 3(2 - 3)|

= |-1 + 10 -3|

= 3 sq. units

Area of □ ABCD = + 3 = sq. units

(iii) (-4, - 2), (-3, - 5), (3, - 2), (2, 3)

Let the vertices of the quadrilateral be A (−4, −2), B (−3, −5), C (3, −2), and D (2, 3). Join AC to form two triangles ∆ABC and ∆ACD

Area of □ ABCD = Area of ∆ABC + Area of ∆ACD

Area of the triangle having vertices (x₁,y₁), (x₂,y₂) and (x₃,y₃)

= |x₁(y₂-y₃)+x₂(y₃-y₁)+x₃(y₁-y₂)|

Area of ∆ABC

= |-4(-5 – (-2)) - 3(-2 – (-2)) + 3(-2 – (-5))|

= |12 + 0 +9|

= sq. units

Area of ∆ACD

= |-4(-2 – 3) - 3(3 – (-2)) + 2(-2 – (-2))|

= |20 + 15 +0|

= sq. units

Area of □ ABCD = + = 28 sq. units