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Find the area of a rhombups if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order. |
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Answer» Solution :Let A(3, 0) , B(4, 5), C(-1, 4) and D(-2, -1) be the vertices of the rhombus ABCD. `therefore"""DIAGONAL, "AC=SQRT((-1-3)^(2)+(4-0)^(2))` `""[because "distance "=sqrt((x_(2)-x_(1))^(2)+(y_(2)-y_(1))^(2))]` `""=sqrt((-4)^(2)+4^(2))=sqrt(16+16)=sqrt(32)=4sqrt(2)` `"""Diagonal, "BD=sqrt((-2-4)^(2)+(-1-5)^(2))=sqrt((-6)^(2)+(-6)^(2))` `""=sqrt(36+36)=sqrt(72)=6sqrt(2)` `therefore` Area of the rhombus ABCD `=(1)/(2)xxACxxBD=(1)/(2)xx4sqrt(2)xx6sqrt(2)` `""=2xx6xxsqrt(2)xxsqrt(2)=12xx2=24` SQUARE units |
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