# Find the area of

Given:

A = (1, 1)

B = (7, –3)

C = (12, 2)

D = (7, 21) 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 ∆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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