Q22 of 52 Page 1

Find the area of the quadrilateral ABCD whose vertices are A(–3, 1), B(–2, –4), C(4, –1) and D(3, 4).


Given,


Vertices of the quadrilateral ABCD,


A(-3, 1)


B(-2, -4)


C(4, -1)


D(3, 4)


Area of triangle = 1/2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]


∴ Area of ∆ ABC = 1/2[(-3)(-4 + 1) + (-2)(-1 + 1) + 4(-1 + 4)]


= 1/2[9 + 0 + 12]


= 21/2 Sq. units …. (i)


Area of ∆ACD,


Area of triangle = 1/2[x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)]


∴ Area of ∆ ACD = 1/2[(-3)(-1-4) + 4(4 + 1) + 3(-1 + 1)]


= 1/2 [15 + 20 + 0]


= 1/2 [15 + 20] = 35/2 ……….(ii)


From (i) and (ii),


21/2 + 35/2 = 56/2


= 28 sq. units


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