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Scale – the Mathematics of Supersizing without Fries


In this picture it seems that the golf ball is enlarged too much for our budding maths student!!

There are two ways to look at scale – one is that similar objects (that is, objects which have corresponding angles the same) also have their corresponding sides in the same ratio. For example, say we have two similar triangles (below) and one side is 5 and its corresponding side (in the other triangle) measures 10, then all corresponding sides will be in the ratio of 5:10 (or 1:2). So, the length of x is 7
Perhaps a better way to think about scale is to look at how much the first object has been enlarged (or reduced). In the above case the enlargement factor (or Scale Factor) is 2, since there has been an enlargement of two times. Therefore, x has to be twice the length of its corresponding side. The corresponding side is 3.5. So x = 2(3.5) which equals 7.

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NCEA Level 2 Algebra Problem. Using the information given, the shaded area = 9, that is:
y(y-8) = 9 –> y.y – 8y – 9 =0
–> (y-9)(y+1) = 0, therefore y = 9 (can’t have a distance of – 1 for the other solution for y)
Using the top and bottom of the rectangle,
x = (y-8)(y+2) = (9-8)(9+2) = 11
but, the left side = (x-4) = 11-4 = 7, but rhs = y+? = 9+?, which is greater than the value of the opp. side??
[I think that the left had side was a mistake and should have read (x+4)?]

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