Question

Show that (0,-1) (6,7) (-2,3) (8,3) are the vertices of a rectangle also find its area

Answer 1

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Answer 2

The area of a rectangle is "$40\sqrt{5}$" square units.

Step-by-step explanation:

Let the four vertices of a rectangle are A(0,-1), B(6,7), C(-2,3) and D(8,3).

To find, area of a rectangle = ?

$AB=\sqrt{(6-0)^{2} +(7+1)^{2} } =\sqrt{36+64} } =10$

$BC=\sqrt{(-2-6)^{2} +(3-7)^{2} } =\sqrt{64+16} } =\sqrt{80} } =4\sqrt{5}$

$CD=\sqrt{(8+2)^{2} +(3-3)^{2} } =\sqrt{100+0} } =10$

$DA=\sqrt{(8-0)^{2} +(3+1)^{2} } =\sqrt{64+16} } =\sqrt{80} } =4\sqrt{5}$

∴ AB = CD = 10 and BC = DA = $4\sqrt{5}$, opposite sides are equal.

Also,

$AC=\sqrt{(-2-0)^{2} +(3+1)^{2} } =\sqrt{4+16} } =\sqrt{20} } =2\sqrt{5}$

$BD=\sqrt{(8-6)^{2} +(3-7)^{2} } =\sqrt{4+16} } =\sqrt{20} } =2\sqrt{5}$

∴ AC = BD =$2\sqrt{5}$, diagonal are also equal.

Hence, the given vertices are rectangle proved.

Area of a rectangle =$10\times 4\sqrt{5}=40\sqrt{5}$ square units

Hence, the area of a rectangle is $40\sqrt{5}$ square units.