Question

Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order.

Answer 1

Given: Given that, there are four vertices of a rhombus:

• A(3,0)

• B(4,5)

• C(-1,4)

• D(-2,-1)

To find: According to the question, we need to find the area of rhombus.

Solution: To find area of the Rhombus, we use the formula:

$\begin{gathered}\\\;\bf{\mapsto\;\;\red{Area_{(Rhombus)} = \dfrac{1}{2} \times AC \times BD}}\end{gathered}$

Here, AC and BD represents the diagonals of the Rhombus. But, how do we know the values of AC and BD? It's by using distance formula.

Coordinates of A and C:

• A = (3,0)

• C = (-1,4)

Here:

• $\sf{x_1 = 3}$

• $\sf{x_2 = -1}$

• $\sf{y_1 = 0}$

• $\sf{y_2 = 4}$

On using distance formula for AC:

$\begin{gathered}\mapsto\sf{AC = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}\\\\\\\mapsto\sf{AC = \sqrt{(3-(-1))^2+(0-4)^2}}\\\\\\\mapsto\sf{AC = \sqrt{(4)^2+(-4)^2}}\\\\\\\mapsto\sf{AC = \sqrt{16+16}}\\\\\\\mapsto\sf{AC = \sqrt{32}}\\\\\\\mapsto\orange{\frak{AC = 4 \sqrt{2}}}\end{gathered}$

Coordinates of B and D:

• B = (4,5)

• D = (-2,-1)

Here:

• $\sf{x_1 = 4}$

• $\sf{x_2 = -2}$

• $\sf{y_1 = 5}$

• $\sf{y_2 = -1}$

On using distance formula for BD:

$\begin{gathered}\mapsto\sf{BD = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}\\\\\\\mapsto\sf{BD= \sqrt{(4-(-2))^2+(5-(-1))^2}}\\\\\\\mapsto\sf{BD= \sqrt{(6)^2+(6)^2}}\\\\\\\mapsto\sf{BD = \sqrt{36+36}}\\\\\\\mapsto\sf{BD = \sqrt{72}}\\\\\\\mapsto\green{\frak{BD = 6 \sqrt{2}}}\end{gathered}$

Now, let's find the area of rhombus:

$\begin{gathered}\\\;\bf{\mapsto\;\;\red{Area_{(Rhombus)} = \dfrac{1}{2} \times AC \times BD}}\end{gathered}$

$\begin{gathered}\\\;\bf{\mapsto\;\;\red{Area_{(Rhombus)} = \dfrac{1}{2} \times 4 \sqrt{2} \times 6 \sqrt{2}}}\end{gathered}$

$\begin{gathered}\\\;\bf{\mapsto\;\;\red{Area_{(Rhombus)} = 12 \times 2}}\end{gathered}$

$\begin{gathered}\\\;\bf{\mapsto\;\;\red{Area_{(Rhombus)} = 24 \:square\:units}}\end{gathered}$

Hope it helps you!

Answer 2

To find:

• Area of rhombus

Solution:

$\tt \underline {On \: using \: distance \: formula \: for \: AC} \\ \tt\begin{gathered}\begin{gathered}\mapsto\sf{AC = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}\\\\\\\mapsto\sf{AC=\sqrt{(3-(-1))^2+(0-4)^2}}\\\\\\\mapsto\sf{AC = \sqrt{(4)^2+(-4)^2}}\\\\\\\mapsto\sf{AC = \sqrt{16+16}}\\\\\\\mapsto\sf{AC = \sqrt{32}}\\\\\\\mapsto{\tt{AC = 4 \sqrt{2}}}\end{gathered}\end{gathered} \\ \\ \tt \underline{On \: using \: distance \: formula \: for \: BD} \\ \tt \: \begin{gathered}\begin{gathered}\mapsto\sf{BD = \sqrt{(x_1-x_2)^2+(y_1-y_2)^2}}\\\\\\\mapsto\sf{BD= \sqrt{(4-(-2))^2+(5-(-1))^2}}\\\\\\\mapsto\sf{BD= \sqrt{(6)^2+(6)^2}}\\\\\\\mapsto\sf{BD = \sqrt{36+36}}\\\\\\\mapsto\sf{BD = \sqrt{72}}\\\\\\\mapsto{\tt{BD = 6 \sqrt{2}}}\end{gathered}\end{gathered} \\ \\ \begin{gathered}\begin{gathered}\\\;\bf{\mapsto\;\;{Area{(Rhombus)} = \dfrac{1}{2} \times AC \times BD}}\end{gathered}\end{gathered}\\\begin{gathered}\begin{gathered}\\\;\bf{\mapsto\;\;{Area{(Rhombus)} = \dfrac{1}{2} \times 4 \sqrt{2} \times 6 \sqrt{2}}}\end{gathered}\end{gathered} \\\begin{gathered} \begin{gathered}\\\;\bf{\mapsto\;\;{Area{(Rhombus)} = 12 \times 2}}\end{gathered} \\ \end{gathered} \\ \begin{gathered}\begin{gathered}\\\;\bf{\mapsto\;\; \purple{Area{(Rhombus)} = 24 \:square\:units}}\end{gathered}\end{gathered}$

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$\sf \: @WildCat7083$