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The question is: What is the area of JKLM? Explain my work

The question is: What is the area of JKLM? Explain my work-example-1
User Waggles
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1 Answer

15 votes
15 votes

Based in our figure in the problem we can see that the shape of JKLM is a rectangle. And the formula for the Area of a rectangle is guven by the LENGTH x WIDTH. Therefore to find the Area of the Rectangle JKLM we only have to follow these steps:

1. Establish and label the corresponding coordinates (x, y) of the points J, K, L, and M.

2. Find the measurement of its Length using the Distance Formula between 2 Points.

3. Find the measurement of its WIDTH using the Distance Formula between 2 Points.

4. Use the formula AREA = LENGTH x WIDTH to find the Area of the Rectangle JKLM.

Let's start.

1. Label the points J, K, L, and M.

Using the given picture we can locate the coordinates (x, y) of each point, we can see that;

J is located at (-5, -1)

K is located at (-4, 3)

L is located at (4, 1)

M is located at (3, -3)

2. Find the LENGTH using the DISTANCE FORMULA

The Distance formula between two points (x₁, y₁) and (x₂, y₂) is given by;


d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}

By the definition of a rectangle, its should have an equal pair of length and a pair of width. And by looking at the picture, we can see that its pair of length are KL and JM. We can just use 1 of the 2 lenghts to find its measurement since they equal anyway, making the measurement of KL=JM. In our case let us use the length KL as the one in which we will be using the distance formula.

Length KL has the coordinates (-4, 3) and (4, 1) respectively, using the distance formula we have;


\begin{gathered} d_(length)=\sqrt[]{(x_2-x_1)^{2^{}}+(y_2-y_1)^{2^{}}} \\ d_(length)=\sqrt[]{(4_{}--4)^{2^{}}+(1-3)^{2^{}}} \\ d_(length)=\sqrt[]{(8_{})^{2^{}}+(-2_{})^{2^{}}} \\ d_(length)=\sqrt[]{64+4^{}^{}}=\sqrt[]{68} \\ d_(length)=\sqrt[]{68}=2\sqrt[]{17}=8.25 \end{gathered}

Therefore our Lenght = 8.25.

3. Same with the length we also have an equal pair of WIDTH which is in our case it is named Width KJ and LM. Since KJ = LM, we only have to find one of them in order for us to know the width. In our case let us use Width KJ.

Width KJ has the coordinates (-4, 3) and (-5, -1) respectively, using the same distance formula we have;


\begin{gathered} d_{\text{width}}=\sqrt[]{(x_2-x_1)^{2^{}}+(y_2-y_1)^{2^{}}} \\ d_{\text{width}}=\sqrt[]{(-5_{}--4_{})^{2^{}}+(-1_{}-3_{})^{2^{}}} \\ d_{\text{width}}=\sqrt[]{(-1)^{2^{}}+(-4_{})^{2^{}}} \\ d_{\text{width}}=\sqrt[]{1_{}^{}^{}+16}=\sqrt[]{17} \\ d_{\text{width}}=\sqrt[]{17}=4.12 \end{gathered}

Therefore our WIDTH = 4.12.

4. Now that we know that the LENGHT and WIDTH of our rectangle are 8.25 and 4.12 respectively, we can now find its Area using the formula LENGTH X WIDTH.


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User YuvShap
by
2.9k points
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