1.

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

Answer» ONG>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!


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