Question

find the area of rhombus if the vertices are (3,0) (4,5) (–1,4) 7and (–2,–1) taken in order?

Answer 1

Given that ( 3 , 0 ) , ( 4 , 5 ) , ( - 1 , 4 ) and ( - 2 , - 1 ). So co ordinates of points which make the diagonal of this rhombus will be ( 3 , 0 ) , ( - 1 , 4 ) and ( 4 , 5 ) , ( - 2 , - 1 )





By distance formula,
$\text{Length of 1st diagonal }= \sqrt{( x_{2}-x_{1})^{2} + ( y_{2}-y_{1})^{2}} = \sqrt{ ( - 1 - 3 )^{2} + ( 4 - 0 )^{2}}$



$\text{Length of 2nd diagonal} = \sqrt{( x_{2}-x_{1})^{2} + ( y_{2}-y_{1})^{2}) } = \sqrt{ ( - 2 - 4 ) ^{2} + ( -1 - 5 )^{2})}$




Area of rhombus = 1 / 2 × product of diagonals


$\text{Area of rhombus} = \dfrac{1}{2} \times\sqrt{ ( - 1 - 3 )^{2} + ( 4 - 0 )^{2}} \times \sqrt{ ( - 2 - 4 ) ^{2} + ( -1 - 5 )^{2})} \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{2} \times \sqrt{ {4}^{2} + {4}^{2} } \times \sqrt{ {6}^{2} + {6}^{2} }$
$\\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \dfrac{1}{2} \times \sqrt{2 \times {6}^{2} } \times \sqrt{2 \times {4}^{2} } \\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{2} \times {6} \times {4}\times \sqrt{2} \times \sqrt{2}$
$\\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{1}{2} \times 6 \times 4 \times 2\\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 6 \times 4\\ \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: =24$



Thus
Area of the rhombus is 24 unit²

Answer 2

Answer:

The diagonal AC is 4 root 2 and diagonal BD is 6root 2. And the answer which is the area is 1/2 * 4 root2 * 6root2 = 24 units

Step-by-step explanation: