Exercise solution 2, page 479 - GCSE Mathematics for OCR Higher Student Book

⭐ What you need to know:

A) Pixels per inch (ppi) tells you how many pixels fit along one side of a length of 1 inch. So if a screen has 15 ppi, that means there are 15 pixels across 1 inch and 15 pixels down 1 inch. In a square of side 1 inch, the total number of pixels is found by multiplying the number along the length by the number along the width:

\[ \text{number of pixels in }1\text{ in}^2 = 15 \times 15 = 225 \]

This is because the pixels form rows and columns.

B) You need to know the conversion between inches and centimetres. The key fact is:

\[ 1 \text{ inch} = 2.54 \text{ cm} \]

So a square measuring \(2.54 \text{ cm} \times 2.54 \text{ cm}\) is exactly the same size as a square measuring \(1 \text{ inch} \times 1 \text{ inch}\). This is very useful because ppi is based on inches, not centimetres.

C) To find the area of one pixel, first find the side length of one pixel. If the density is 25 ppi, then 25 pixels fit along 1 inch, so the width of one pixel is

\[ \frac{1}{25}\text{ inch} \]

Since a pixel is square in this situation, its area is:

\[ \left(\frac{1}{25}\right)^2 = \frac{1}{625}\text{ in}^2 \]

If the question wants the area in \(\text{cm}^2\), convert \(1\text{ in}^2\) into \(\text{cm}^2\):

\[ 1\text{ in}^2 = 2.54^2\text{ cm}^2 \]

So the area of one pixel can be written as

\[ \frac{2.54^2}{625}\text{ cm}^2 \]

D) For a rectangular screen, find the number of pixels along each side, then multiply. For example, on a screen that is 6 inches by 2 inches with a density of 12 ppi:

\[ \text{pixels along the length} = 6 \times 12 = 72 \] \[ \text{pixels along the width} = 2 \times 12 = 24 \]

Then the total number of pixels is

\[ 72 \times 24 \]

This uses the same idea as finding the number of small squares in a rectangle: rows \(\times\) columns.

E) Standard form is used for very large or very small numbers. A number in standard form is written as

\[ a \times 10^n \]

where \(1 \le a < 10\) and \(n\) is an integer.

For example:

\[ 0.00048 = 4.8 \times 10^{-4} \]

If you are asked for 3 significant figures, keep the first 3 important digits. For instance,

\[ 0.007216 = 7.22 \times 10^{-3} \]

because the first three significant figures are 7, 2 and 2.

F) When multiplying numbers in standard form, multiply the ordinary numbers together and add the powers of 10:

\[\begin{aligned} (3 \times 10^4)(5 \times 10^{-2}) &= (3 \times 5)\times 10^{4+(-2)} \\ &= 15 \times 10^2 \\ &= 1.5 \times 10^3 \end{aligned}\]

This is especially useful when finding the area of a pixel from two side lengths written in standard form, or when finding the total area of many pixels.

G) To find the area of a rectangle, multiply length by width:

\[ \text{Area} = \text{length} \times \text{width} \]

If one tiny rectangle has width \(2 \times 10^{-3}\) cm and length \(6 \times 10^{-5}\) cm, then its area is found by multiplying those two measurements. Always remember that the unit for area is squared, such as \(\text{cm}^2\).

H) If you know the area of one pixel and want the total area of many pixels, multiply:

\[ \text{total area} = \text{number of pixels} \times \text{area of one pixel} \]

For example, if one pixel has area \(8 \times 10^{-6}\text{ cm}^2\), then the area of \(4 \times 10^2\) pixels is:

\[\begin{aligned} (4 \times 10^2)(8 \times 10^{-6}) &= 32 \times 10^{-4} \\ &= 3.2 \times 10^{-3}\text{ cm}^2 \end{aligned}\]

This combines area, multiplication, and standard form.