Question

Question 10
Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order. [Hint: Area of a rhombus = $\frac{1}{2}$(product of its diagonals)]

Answer 1

Let (3, 0), (4, 5), ( - 1, 4) and ( - 2, - 1) are the vertices A, B, C, D of a rhombus ABCD.
Length of diagonal $AC=\sqrt{[3-(-1){]}^2+(0-4{)}^2}$
$=\sqrt{16+16}=4\sqrt{2}$
Length of diagonal $BD=\sqrt{[4-(-2){]}^2+[5-(-1){]}^2}$
$=\sqrt{36+36}=6\sqrt{2}$
Therefore, area of rhombus $ABCD=\frac{1}{2}\times 4\sqrt{2}\times 6\sqrt{2}$
= 24 square units