9396 /www/nrich/html/content/id/9396/ 1 2012-09-14T15:19:11 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <br></br> The smaller of two similar rectangles has height $2$ units; the larger rectangle has length $6$ units.<br></br> <br></br> If one rectangle has twice the area of the other, find the length of the smaller rectangle.<br></br> <br></br> <br></br> <br></br> <br></br> <br></br> If you liked this problem, <a href="/7385">here is an NRICH task</a> which challenges you to use similar mathematical ideas.<br></br> <br></br> <br></br> <em>This problem is taken from Tony Gardiner&#39;s &#39;Extension Mathematics Gamma&#39; book.</em></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> Call the length of the smaller rectangle $x$.<br></br> <br></br> Charlie then said that the area of the smaller rectangle is $2x$, so the area of the larger rectangle is $4x$, and the height of the larger rectangle is $4x/6$. He then argued that the height:length ratio of both rectangles must be the same, since they are similar.<br></br> <br></br> So $\frac{2}{x} = \frac{4x}{36}$<br></br> $x^{2} = 18$<br></br> $x=\sqrt{18}$<br></br> <br></br> Emily&#39;s method was slightly different. She called the height of the larger rectangle $y$.<br></br> <br></br> Then, comparing the areas, $4x = 6y$<br></br> <br></br> By similarity, $\frac{y}{2} = \frac{6}{x}$<br></br> i.e. $y = \frac{12}{x}$<br></br> <br></br> Then by substitution, $4x = \frac{72}{x}$<br></br> $x^{2} = 18$<br></br> $x = \sqrt{18}$</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Enlargements and Scale factors - Short Problem Tony Gardiner Gamma p181 Q0</mdoxml> 2 3 0 0 1 1 0 Can you find the missing length? Admin Short problems Secondary Mapping Document DisplayCabinet Secondary Mapping Document Enlargements and Scale factors