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| 1. |
Find area of quadrilateral whose vertices are _4;2 _3;_5 3;_2 2;3 |
| Answer» Let A(-4, -2), B(-3, -5), C(3, -2) and D(2, 3)Area of Quadrilateral ABCD = Area of Triangle ABD + Area of Triangle BCD .................... (1)Using formula to find area of triangle:\xa0Area of {tex}\\triangle{/tex}ABD{tex} = \\frac{1}{2}\\left[ {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \\right]{/tex}{tex} = \\frac{1}{2}\\left[ { - 4( - 5 - 3) - 3\\left\\{ {3 - ( - 2)} \\right\\} + 2\\left\\{ { - 2 - ( - 5)} \\right\\}} \\right]{/tex}{tex} = \\frac{1}{2}(32 - 15 + 6) = \\frac{1}{2}(23) = 11.5sq\\;units{/tex}\xa0........... (2)Again using formula to find area of triangle:Area of {tex}\\triangle{/tex}BCD\xa0{tex} = \\frac{1}{2}\\left[ {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \\right]{/tex}{tex} = \\frac{1}{2}\\left[ { - 3( - 2 - 3) + 3\\left\\{ {3 - ( - 5)} \\right\\} + 2\\left\\{ { - 5 - ( - 2)} \\right\\}} \\right]{/tex}{tex} = \\frac{1}{2}(15 + 24 - 6) = \\frac{1}{2}{/tex}\xa0(33) = 16.5 sq units ........ (3)Putting (2) and (3) in (1), we getArea of Quadrilateral ABCD = 11.5 + 16.5 = 28 sq units | |