Question

Show that the points (_4,_7) , (_1 ,2) , (8,5) , (5,4) taken in order are vertices of a rhombus and find its area

Answer 1

★Question:-

Show that the points (-4,-7) , (-1 ,2) , (8,5) , (5,4) taken in order are vertices of a rhombus and find its area.

★Answer:-

For a rhombus,

▶All the sides are equal.

▶Diagonals are not equal.

We know:

Distance between two points,

$\leadsto \underline{\boxed{\sf D=\sqrt{(x_2 -x_1)^2+(y_2 -y_1)^2} }}$

Let ABCD be a rhombus,and the points are:

A(-4,-7) , B(-1 ,2) , C(8,5) , D(5,4)

Distance between the points:-

$:\implies \sf AB=\sqrt{(-1-(-4))^2+(2-(-7))^2} \\\\:\implies \sf AB = \sqrt{9^2 +3^2} =\sqrt{90} \\\\:\implies \underline{\bold{AB=3\sqrt{10} }}$

$:\implies \sf BC=\sqrt{(8-(-1))^2+(5-2)^2} \\\\:\implies \sf BC=\sqrt{9^2+3^2} =\sqrt{90} \\\\:\implies \underline{\bold{BC=3\sqrt{10} }}$

$:\implies \sf CD = \sqrt{(5-8)^2+(-4-5)^2} \\\\:\implies \sf CD=\sqrt{3^2 + 9^2} =\sqrt{90} \\\\:\implies \underline{\bold{CD=3\sqrt{10} }}$

And,

$:\implies \sf DA =\sqrt{(-4-5)^2+(-7-(-4))^2} \\\\:\implies \sf DA = \sqrt{9^2 +3^2} =\sqrt{90} \\\\:\implies \underline{\bold{DA = 3\sqrt{10} }}$

Distance between the diagonals:-

$:\implies \sf AC=\sqrt{(8-(-4))^2+(5-(-7))^2} \\\\:\implies \sf AC = \sqrt{12^2 +12^2} =\sqrt{144 +144} \\\\:\implies \underline{\bold{AC=12\sqrt{2} }}$

$:\implies \sf BD = \sqrt{(5-(-1))^2+(-4-2)^2} \\\\:\implies \sf BD =\sqrt{6^2 +6^2} =\sqrt{72}\\\\:\implies \underline{\bold{BD = 6\sqrt{2} }}$

As we can see,

AB = BC=CD=DA & AC≠BD

i.e., The sides are equal & the diagonals are not equal.

Hence proved !

To find Area :-

We know,

$\leadsto \underline{\boxed{\sf Area\ of\ rhombus=\frac{1}{2} \times (Product\ of\ diagonals) }}$

Now,

$:\implies \sf Area = \frac{1}{2} \times AC \times BD \\\\:\implies \sf \frac{1}{2} \times 12\sqrt{2} \times 6\sqrt{2} \\\\:\implies \sf \frac{1}{\cancel{2}} \times 72 \times \cancel{2}\\\\:\implies \boxed{\boxed{\sf Area=72sq.units.}}$

Hence, the area of rhombus is 72sq.units.

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Answer 2

★Question:-

Show that the points (-4,-7) , (-1 ,2) , (8,5) , (5,4) taken in order are vertices of a rhombus and find its area.

★Answer:-

For a rhombus,

▶All the sides are equal.

▶Diagonals are not equal.

We know:

Distance between two points,

$\leadsto \underline{\boxed{\sf D=\sqrt{(x_2 -x_1)^2+(y_2 -y_1)^2} }}$

Let ABCD be a rhombus,and the points are:

A(-4,-7) , B(-1 ,2) , C(8,5) , D(5,4)

Distance between the points:-

$:\implies \sf AB=\sqrt{(-1-(-4))^2+(2-(-7))^2} \\\\:\implies \sf AB = \sqrt{9^2 +3^2} =\sqrt{90} \\\\:\implies \underline{\bold{AB=3\sqrt{10} }}$

$:\implies \sf BC=\sqrt{(8-(-1))^2+(5-2)^2} \\\\:\implies \sf BC=\sqrt{9^2+3^2} =\sqrt{90} \\\\:\implies \underline{\bold{BC=3\sqrt{10} }}$

$:\implies \sf CD = \sqrt{(5-8)^2+(-4-5)^2} \\\\:\implies \sf CD=\sqrt{3^2 + 9^2} =\sqrt{90} \\\\:\implies \underline{\bold{CD=3\sqrt{10} }}$

And,

$:\implies \sf DA =\sqrt{(-4-5)^2+(-7-(-4))^2} \\\\:\implies \sf DA = \sqrt{9^2 +3^2} =\sqrt{90} \\\\:\implies \underline{\bold{DA = 3\sqrt{10} }}$

Distance between the diagonals:-

$:\implies \sf AC=\sqrt{(8-(-4))^2+(5-(-7))^2} \\\\:\implies \sf AC = \sqrt{12^2 +12^2} =\sqrt{144 +144} \\\\:\implies \underline{\bold{AC=12\sqrt{2} }}$

$:\implies \sf BD = \sqrt{(5-(-1))^2+(-4-2)^2} \\\\:\implies \sf BD =\sqrt{6^2 +6^2} =\sqrt{72}\\\\:\implies \underline{\bold{BD = 6\sqrt{2} }}$

As we can see,

AB = BC=CD=DA & AC≠BD

i.e., The sides are equal & the diagonals are not equal.

Hence proved !

To find Area :-

We know,

$\leadsto \underline{\boxed{\sf Area\ of\ rhombus=\frac{1}{2} \times (Product\ of\ diagonals) }}$

Now,

$:\implies \sf Area = \frac{1}{2} \times AC \times BD \\\\:\implies \sf \frac{1}{2} \times 12\sqrt{2} \times 6\sqrt{2} \\\\:\implies \sf \frac{1}{\cancel{2}} \times 72 \times \cancel{2}\\\\:\implies \boxed{\boxed{\sf Area=72sq.units.}}$

Hence, the area of rhombus is 72sq.units.

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