Find the area of a rhombus whose vertices taken in order are the points (3,0), (4,5), (-1,4) and (-2,-1).

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Solution

Area of rhombus ABCD = Area of $△$ ABC + Area of $△$ ACD
Now, Area of $△$ ABC = $\frac{1}{2} | 3 (1) + 4 (4) - 1 (- 5) | = \frac{24}{2} = 12 s q . u n i t s$

Area of $△$ ACD = $\frac{1}{2} | 3 (5) + (- 1) (- 1) + (- 2) (- 4) | = \frac{24}{2} = 12 s q . u n i t s$

Thus, area of rhombus ABCD = $12 + 12 = 24 s q . u n i t s$

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Find the area of a rhombus whose vertices taken in order are the points (3,0), (4,5), (-1,4) and (-2,-1).

Area of rhombus ABCD = Area of △ABC + Area of △ACDNow, Area of △ABC = 12|3(1)+4(4)−1(−5)|=242=12 sq.unitsArea of △ACD = 12|3(5)+(−1)(−1)+(−2)(−4)|=242=12 sq.unitsThus, area of rhombus ABCD = 12+12=24 sq.units