/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 9 The "DIN A" series of paper size... [FREE SOLUTION] | 91Ó°ÊÓ

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The "DIN A" series of paper sizes is determined by two conditions. The basic requirement is that all the DIN A rectangles are similar; the second condition is that when we fold a given size exactly in half, we get the next smaller size. Hence \- a sheet of paper of size \(\mathrm{A} 3\) folds in half to give a sheet of size \(\mathrm{A} 4-\) which is similar to \(\mathrm{A} 3 ;\) and \- a sheet of size \(A 4\) folds in half to give a sheet of size \(A 5 ;\) etc.. (a) Find the constant ratio \(r=\) "(longer side length) : (shorter side length)" for all DIN A paper sizes. (b)(i) To enlarge A4 size to A3 size (e.g. on a photocopier), each length is enlarged by a factor of \(r\). What is the "enlargement factor" to get from A3 size back to A4 size? (ii) To "enlarge" A4 size to A5 size (e.g. on a photocopier), each length is "enlarged" by a factor of \(\frac{1}{r} .\) What is the enlargement factor to get from A5 size back to A4 size?

Short Answer

Expert verified
The constant ratio is \( r = \sqrt{2} \). The enlargement factor from A3 to A4 is \( \frac{1}{\sqrt{2}} \), and from A5 to A4 is \( \sqrt{2} \).

Step by step solution

01

Understanding Similarity of Rectangles

The DIN A series of paper sizes are rectangles that are always similar in shape. The ratio of the longer side (length) to the shorter side (width) of any DIN A rectangle remains constant for any paper size in the DIN A series. This constant ratio is what we need to find.
02

Setting Up the Ratio Equation

Let's define the longer side of the paper as \( L \) and the shorter side as \( S \). The ratio \( r \) is defined as \( r = \frac{L}{S} \). Since folding the paper in half should give a similar rectangle, we have \( \frac{S}{\frac{L}{2}} = r \). This is because when folded, the shorter side \( S \) becomes the longer side, and the half of the longer side \( \frac{L}{2} \) becomes the new shorter side.
03

Finding the Ratio Equation

From the last step, we have two equations: \( r = \frac{L}{S} \) and \( r = \frac{S}{\frac{L}{2}} \). By replacing \( S \) in the second equation, we have \( \frac{2S}{L} = r \). Now we equate the two expressions for \( r \): \( \frac{L}{S} = \frac{2S}{L} \).
04

Solving the Ratio Equation

Cross-multiply to solve \( \frac{L}{S} = \frac{2S}{L} \). This results in \( L^2 = 2S^2 \). Solving for \( \frac{L}{S} \), we find \( L = \sqrt{2} S \). Hence, \( r = \frac{L}{S} = \sqrt{2} \). Thus, the ratio \( r \) is \( \sqrt{2} \).
05

Enlargement Factor from A4 to A3

To enlarge an A4 size to A3, each dimension of A4 is multiplied by the ratio \( r = \sqrt{2} \). Thus, the enlargement factor is \( \sqrt{2} \).
06

Finding Enlargement from A3 to A4

To go back from A3 to A4, each dimension of A3 is divided by the ratio \( r \). The factor for reduction is given by \( \frac{1}{\sqrt{2}} \).
07

Enlargement Factor from A4 to A5

To "enlarge" from A4 to A5 (making it smaller), each dimension is multiplied by the factor \( \frac{1}{r} = \frac{1}{\sqrt{2}} \).
08

Finding Enlargement from A5 to A4

To 'enlarge' from A5 to A4, each dimension of A5 needs to be increased by the ratio \( r = \sqrt{2} \), which makes the enlargement factor equal to \( \sqrt{2} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Similarity of Rectangles
In geometry, similarity refers to figures or shapes having the same shape but differing in size. Two rectangles are similar if their corresponding angles are equal and the sides are in proportion. This principle is at the heart of the DIN A paper sizes. Every rectangle in this series maintains the same shape by keeping a constant ratio of the longer side to the shorter side. Therefore, any two rectangles in the DIN A series are similar, even though one might be larger than the other.

The fundamental property of similarity in rectangles is that their side lengths remain proportional, so as you decrease or increase the size of the paper, the shape remains unchanged. This consistent ratio facilitates operations such as folding and enlarging, which are unique to the DIN A sizes. For instance, when you fold an A3 size paper in half lengthwise, you get an A4 paper, which maintains the similarity to A3 because of the constant ratio. Conversely, unfolding or enlarging it back restores the original proportions, which means you always work within a family of similar rectangles.
Ratio and Proportion
The ratio in a rectangle is expressed as the comparison between the length of the longer side and the shorter side. In the DIN A series, this ratio is universally represented as \( r = \frac{L}{S} \), where \( L \) represents the longer side and \( S \) represents the shorter side. This ratio remains constant across all sizes in the series.

To discover this constant, recall that folding a paper of one DIN A size in half along its length creates another similar rectangle. When the longer side is halved, the new shorter side becomes \( \frac{L}{2} \). Hence, for similarity to be maintained, the ratio must hold true even after folding, leading us to \( \frac{S}{\frac{L}{2}} = r \). Equating this with \( \frac{L}{S} \) gives \( L^2 = 2S^2 \), from which the ratio \( r = \sqrt{2} \) is derived.

This constant \( \sqrt{2} \) is fascinating because it not only preserves similarity but also ensures that each subsequent size in the DIN A series retains the same shape, whether you scale up or down.
Mathematical Problem Solving
Mathematical problem solving in geometry often involves finding hidden connections between shapes and sizes, much like the relation among the DIN A paper sizes. Here, the problem involves transitions between different paper sizes, which require understanding both mathematical ratios and geometric properties.

To solve problems related to paper size changes by copying or scaling, you need to accurately apply the constant ratio \( \sqrt{2} \). For example, enlarging an A4 sheet to an A3 requires multiplying every dimension of the A4 by \( \sqrt{2} \). This same ratio \( r = \sqrt{2} \) works in reverse to scale down from A3 to A4, performed by multiplying by \( \frac{1}{\sqrt{2}} \).

Effective problem solving also considers inverse operations, like reducing an A4 to A5. This seems counterintuitive since it involves reducing the paper size by "enlarging" with the factor \( \frac{1}{\sqrt{2}} \), maintaining the constant shape. To go from A5 back to A4, multiply the dimensions of A5 by \( \sqrt{2} \), the constant unfolding factor.

Understanding these principles simplifies the geometric transformations necessary to handle the DIN A series efficiently. Whether scaling up or down, maintaining the shape while changing the size is a fundamental technique in mathematical problem solving in geometry.

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Most popular questions from this chapter

(a) Expand and simplify in your head: (i) \((\sqrt{2}+1)^{2}\) (ii) \((\sqrt{2}-1)^{2}\) (iii) \((1+\sqrt{2})^{3}\) (b) Simplify: (i) \(\sqrt{10+4 \sqrt{6}}\) (ii) \(\sqrt{5+2 \sqrt{6}}\) (iii) \(\sqrt{\frac{3+\sqrt{5}}{2}}\) (iv) \(\sqrt{10-2 \sqrt{5}}\) The expressions which occur in exercises to develop fluency in working with surds often appear arbitrary. But they may not be. The arithmetic of surds arises naturally: for example, some of the expressions in the previous problem have already featured in Problem \(\mathbf{3}(\mathrm{c}) .\) In particular, surds will feature whenever Pythagoras' Theorem is used to calculate lengths in geometry, or when a proportion arising from similar triangles requires us to solve a quadratic equation. So surd arithmetic is important. For example: \- A regular octagon with side length 1 can be surrounded by a square of side \(\sqrt{2}+1\) (which is also the diameter of its incircle); so the area of the regular octagon equals \((\sqrt{2}+1)^{2}-1\) (the square minus the four corners). \- \(\sqrt{2}-1\) features repeatedly in the attempt to apply the Euclidean algorithm, or anthyphairesis, to express \(\sqrt{2}\) as a "continued fraction". -\(\sqrt{10-2 \sqrt{5}}\) may look like an arbitrary, uninteresting repeated surd, but is in fact very interesting, and has already featured as \(4 \sin 36^{\circ}\) in Problem \(\mathbf{3}(\mathrm{c})\) \- One of the simplest ruler and compasses constructions for a regular pentagon \(A B C D E\) (see Problem 185) starts with a circle of radius 2 , centre \(O,\) and a point \(A\) on the circle, and in three steps constructs the next point \(B\) on the circle, where \(\underline{A B}\) is an edge of the inscribed regular pentagon, and $$ \underline{A B}=\sqrt{10-2 \sqrt{5}} $$

(a) Using only mental arithmetic: (i) Determine which is bigger: $$ \frac{1}{2}+\frac{1}{5} \quad \text { or } \quad \frac{1}{3}+\frac{1}{4} ? $$ (ii) How is this question related to the observation that \(10<12 ?\) (b) [This part will require some written calculation and analysis.] (i) For positive real numbers \(x\), compare $$ \frac{1}{x+2}+\frac{1}{x+5} \quad \text { and } \quad \frac{1}{x+3}+\frac{1}{x+4} $$ (ii) What happens in part (i) if \(x\) is negative?

Let \(\Delta=\operatorname{area}(\triangle A B C)\) (a) Prove that $$ \Delta=\frac{1}{2} \cdot a b \cdot \sin C . $$ (b) Prove that \(4 R \Delta=a b c\).

(a) Joining the midpoints of the edges of an equilateral triangle \(A B C\) cuts the triangle into four identical smaller equilateral triangles. Removing one of the three outer small triangles (say \(A M N\), with \(M\) on \(A C\) ) leaves three-quarters of the original shape in the form of an isosceles trapezium \(M N B C .\) Show how to cut this isosceles trapezium into four congruent pieces. (b) Joining the midpoints of opposite sides of a square cuts the square into four congruent smaller squares. If we remove one of these squares, we are left with three-quarters of the original square in the form of an L-shape. Show how to cut this L-shape into four congruent pieces.

(a)(i) Explain why any integer that is a factor (or a divisor) of both \(m\) and \(n\) must also be a factor of their difference \(m-n,\) and of their sum \(m+n\). (ii) Prove that $$ H C F(m, n)=H C F(m-n, n) $$ (iii) Use this to calculate in your head \(H C F(1001,91)\) without factorising either number. (b)(i) Prove that: \(H C F(m, m+1)=1\). (ii) Find \(H C F(m, 2 m+1)\). (iii) Find \(H C F\left(m^{2}+1, m-1\right)\).

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