Question

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Answer 1

Here is your answer in the attachment given above

Answer 2

$\huge\boxed{\texttt{\fcolorbox{blue}{aqua}{Solution:-}}}$

$\huge{\bold{\boxed{\boxed{\mathbb{\red{Explanation}}}}}}$

Let (3, 0), (4, 5), ( - 1, 4) and ( - 2, - 1) are the Vertices A, B, C, D

Of a rhombus ABCD

$Length\:of\:Diagonal\:BD=\sqrt{[3-(-1)]^{2} +\:[(0-4)^{2}] }$

$=\sqrt{16+16=4\sqrt{2} }$

$Length\:of\:Diagonal\:BD=\sqrt{[4-(-2)]^{2} +\:[(5-(-1)^{2}] }$

$=\sqrt{36+36}=6\sqrt{2}$

$Therefore\:Area\:Of\:Rhombus=\frac{1}{2}\:\times\: 4\sqrt{2}\:\times\: 6\sqrt{2}$

$=24\:square\:unit$