Answer :

Given:

A = (1, 1)


B = (7, –3)


C = (12, 2)


D = (7, 21)



The quadrilateral ABCD can be divided into ∆ABC and ∆ADC


We know that,


Area of ∆ABC = 1/2|x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|


Area of ∆ABC = 1/2|1((-3) – 2) + 7(2 – 1) + 12(1 – (-3))|


Area of ∆ABC = 1/2|1(-5) + 7(1) + 12(4)|


Area of ∆ABC = 1/2|-5 + 7 + 48|


Area of ∆ABC = 1/2|50|


Area of ∆ABC = 1/2 × 50


Area of ∆ABC = 25 sq. units


Similarly,


Area of ∆ADC = 1/2|x1 (y2 – y3) + x2 (y3 – y1) + x3 (y1 – y2)|


Area of ADC = 1/2|1((21) – 2) + 7(2 – 1) + 12(1 – (21))|


Area of ADC = 1/2|1(19) + 7(1) + 12(-20)|


Area of ADC = 1/2|19 + 7 – 240|


Area of ADC = 1/2|-214|


Area of ∆ABC = 1/2 × 214


Area of ADC = 107 sq. units


Now,


Area of ABCD = Area of ∆ABC + Area of ∆ADC


Area of ABCD = 25 sq. units + 107 sq. units


Area of ABCD = 132 sq. units


Hence, Area of quadrilateral ABCD = 132 sq. units


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