Weekly Problem 20

7168 /www/nrich/html/content/id/7168/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> The diagram shows a polygon ABCDEFG, in which $FG =6$ and $GA=AB=BC=CD=DE=EF$. Also $BDFG$ is a square. The area of the whole polygon is exactly twice the area of $BDFG$. Find the length of the perimeter of the polygon. <mdo:image alt="" height="368" src="2011%2020.PNG" width="477"></mdo:image> If you liked this problem, <a href="http://nrich.maths.org/2293&part=">here is an NRICH task</a> which challenges you to use similar mathematical ideas. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> <mdo:image width="477" height="368" alt="" src="2011%2020.PNG"></mdo:image> The area of square BDFG is $6\times 6 = 36$ square units. So the total area of the three triangles ABG, BCD and DEF is also $36$ square units. These three triangles are congruent and so each has an area of $12$ square units. The area of each triangle is $\frac{1}{2}\times base \times \ height$ and the base is $6$ units and hence we have $\frac{1}{2}\times 6 \times \ height = 12$, so the height is $4$ units. Let $X$ be the midpoint of BD. Then CX is perpendicular to the base BD (since BCD is an isosceles triangle). By Pythagoras' Theorem, $BC = \sqrt{3^2+4^2} = 5$ units. Therefore the perimeter of ABCDEFG is $6\times 5+ 6 = 36$ units. <mdo:image width="435" height="315" alt="" src="2011%2020a.PNG"></mdo:image> </mdoxml> 2 3 0 1 1 0 0 Weekly Problem 20 - 2011 What is the perimeter of this unusually shaped polygon... 2D Geometry, Shape and Space Squares 2D Geometry, Shape and Space Right angled triangles 2D Geometry, Shape and Space Pythagoras' theorem Secondary Mapping Document Pythagoras’s Theorem