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Length and area

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Explainer Video

Tutor: Daniel

Summary

Length and area

​In a nutshell

The length between two points can be calculated using the distance formula. This formula allows you to solve length and area problems.



The distance formula

Pythagoras' theorem forms the basis of the distance formula. 

Maths; Straight line graphs; KS5 Year 12; Length and area

From Pythagoras' theorem, you can see that d2=(x2−x1)2+(y2−y1)2d^2 = (x_2-x_1)^2 + (y_2-y_1)^2d2=(x2​−x1​)2+(y2​−y1​)2. Taking the square root of ddd gives the distance formula.

​d=(x2−x1)2+(y2−y1)2\boxed{d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}}d=(x2​−x1​)2+(y2​−y1​)2​​​​


Procedure

1.

Label the coordinates of two points (x1,y1)(x_1,y_1)(x1​,y1​) and (x2,y2)(x_2,y_2)(x2​,y2​)​. 

2.

Substitute into the distance formula d=(x2−x1)2+(y2−y1)2d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}d=(x2​−x1​)2+(y2​−y1​)2​.

3.

Solve for distance ddd.​


Example 1

A line starts at the point (3,6)(3,6)(3,6)​ and ends at the point (6,10)(6,10)(6,10)​. What is the distance of the line?


Label the points

x1=3 ,y1=6 ,x2=6 ,y2=10x_1 = 3\space, y_1 = 6\space, x_2=6\space, y_2 =10x1​=3 ,y1​=6 ,x2​=6 ,y2​=10


Substitute into the distance formula. Solve for d.

​d=(x2−x1)2+(y2−y1)2d=(6−3)2+(10−6)2d=(3)2+(4)2d=9+16d=25d=5‾\begin{aligned}&d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\\&d=\sqrt{(6-3)^2 + (10-6)^2}\\&d=\sqrt{(3)^2 + (4)^2}\\&d=\sqrt{9 + 16}\\&d=\sqrt{25}\\&\underline{d=5}\\\end{aligned}​d=(x2​−x1​)2+(y2​−y1​)2​d=(6−3)2+(10−6)2​d=(3)2+(4)2​d=9+16​d=25​d=5​​



Solving for area

A shape formed from several lines will have an area. With the distance formula, you can find that area. You must use the area formula for the relevant shape.


Example 2

A rectangle has the points A,(1,1) B,(0,3) C,(3,5)A, (1,1)\space B,(0,3)\space C,(3,5)A,(1,1) B,(0,3) C,(3,5) and  D,(4,3).\space D,(4,3). D,(4,3). What is the area of the rectangle? Give your answer to two decimal places.


Maths; Straight line graphs; KS5 Year 12; Length and area


Find the distance of the lines using the distance formula.

A→B(1,1) (0,3)x1=1 ,y1=1 ,x2=0 ,y2=3d=(x2−x1)2+(y2−y1)2d=(0−1)2+(3−1)2d=(−1)2+(2)2d=1+4d=5\begin{aligned}&A\rightarrow B\\(&1,1)\space (0,3)\\x&_1 = 1\space, y_1 = 1\space, x_2=0\space, y_2 =3\\d&=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\\d&=\sqrt{(0-1)^2 + (3-1)^2}\\d&=\sqrt{(-1)^2 + (2)^2}\\d&=\sqrt{1+4}\\d&=\sqrt5\\\end{aligned}(xddddd​A→B1,1) (0,3)1​=1 ,y1​=1 ,x2​=0 ,y2​=3=(x2​−x1​)2+(y2​−y1​)2​=(0−1)2+(3−1)2​=(−1)2+(2)2​=1+4​=5​​


​A→Dx1=1 ,y1=1 ,x2=4 ,y2=3d=(x2−x1)2+(y2−y1)2d=(4−1)2+(3−1)2d=(3)2+(2)2d=9+4d=13\begin{aligned}&A\rightarrow D\\x&_1 = 1\space, y_1 = 1\space, x_2=4\space, y_2 =3\\d&=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}\\d&=\sqrt{(4-1)^2 + (3-1)^2}\\d&=\sqrt{(3)^2 + (2)^2}\\d&=\sqrt{9+4}\\d&=\sqrt{13}\\\end{aligned}xddddd​A→D1​=1 ,y1​=1 ,x2​=4 ,y2​=3=(x2​−x1​)2+(y2​−y1​)2​=(4−1)2+(3−1)2​=(3)2+(2)2​=9+4​=13​​​​

​

Find the area of the shape. Round to two decimal places.

Area =length×heightA=5×13A=65A=8.06 (2 d.p.)‾\begin{aligned}Are&a\space= length\times height\\&A=\sqrt5\times\sqrt{13}\\&{A=\sqrt{65}}\\&\underline{A=8.06 \space(2\space d.p.)}\end{aligned}Are​a =length×heightA=5​×13​A=65​A=8.06 (2 d.p.)​​

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Area and perimeter: Triangles and quadrilaterals

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Area and perimeter: Triangles and quadrilaterals

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Area and perimeter: Formulae

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FAQs - Frequently Asked Questions

Where does the distance formula come from?

The formula has its basis in Pythagoras' theorem.

How do you find the area of a rectangle on a plane?

First, use the distance formula, d^2=(x2-x1)^2 + (y2-y1)^2 to find the length and height of the rectangle. Then multiply those distances to give the area.

How do you find a line's length?

Use the distance formula, d^2=(x2-x1)^2 + (y2-y1)^2.

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