Question

find the area of rhombus if its vertices 3,0 4,5 -1,4 -2,-1 taken in order ​

Answer 1

Please refer to the above image for solution!!

Thanks!

Answer 2

Answer:

$\boxed {\mathsf{\blue {24 \: sq. \: units}}}$

Step-by-step explanation:

Let A(3,0), B(4,5), C(-1,4) and D(-2,-1) be the vertices of the rhombus ABCD.

$\mathsf {∴ \: Diagonal,\: AC = \sqrt {(-1,-3)^{2}+(4-0)^{2}}}$

$\mathsf {\sqrt {(-4)^{2}+4^{2}}}$

$\mathsf {\sqrt {32}}$

$\mathsf {4 \sqrt {2} units}$

$\mathsf {Diagonal,\: BD = \sqrt {(-2,-4)^{2}+(-1-5)^{2}}}$

$\mathsf {\sqrt {(-6)^{2}+(-6)^{2}}}$

$\mathsf {\sqrt {36+36}}$

$\mathsf {\sqrt {72}}$

$\mathsf {6 \sqrt {2} units}$

$\mathsf {∴ \: Area \: of \: the \: rhombus \: ABCD}$

$\mathsf {\dfrac {1}{2} \times Product \: of \: its \: diagonals }$

$\mathsf {\dfrac {1}{2} \times \: AC \: \times \: BD}$

$\mathsf {\dfrac {1}{2} \times 4 \sqrt {2} \times 6 \sqrt{2}}$

$\mathsf {2 \times 6 \times \sqrt {2} \times \sqrt{2}}$

$\mathsf {12 \times 2}$

$\mathsf {24 \: sq. \: units}$