Find the area of the figure formed by joining the points:A(-4,-2),B(-3,-5),C(3,-2) and D(2,4)
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A(-4,-2) (x1,y1)
B(-3,-5) (x2,y2)
C(3,-2) (x3,y3)
D(2,4) (x4,y4)
area(ABCD) = ar(ABC) + ar(ADC)
ar(∆) = ½ [x1(y2-y3) + x2(y3-y1) + x3(y1-y2)
ar(ABC) = ½ [-4(-5+2) - 3(-2+2) + 3(-2+5)]
= ½ [12+0+9]
= ½ [21]
10.5 unit²
similarly ar(ADC) = 21 unit²
therefore area of ABCD = 31.5 unit²
B(-3,-5) (x2,y2)
C(3,-2) (x3,y3)
D(2,4) (x4,y4)
area(ABCD) = ar(ABC) + ar(ADC)
ar(∆) = ½ [x1(y2-y3) + x2(y3-y1) + x3(y1-y2)
ar(ABC) = ½ [-4(-5+2) - 3(-2+2) + 3(-2+5)]
= ½ [12+0+9]
= ½ [21]
10.5 unit²
similarly ar(ADC) = 21 unit²
therefore area of ABCD = 31.5 unit²
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