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At a Glance

The area of a regular pentagon looks about twice as a big as the pentangle star drawn within it. Is it?

Six Discs

Six circular discs are packed in different-shaped boxes so that the discs touch their neighbours and the sides of the box. Can you put the boxes in order according to the areas of their bases?

Equilateral Areas

ABC and DEF are equilateral triangles of side 3 and 4 respectively. Construct an equilateral triangle whose area is the sum of the area of ABC and DEF.

Dividing the Field

Age 14 to 16
Challenge Level

A trapezium has area equal to half the sum of the lengths of the parallel edges times the perpendicular distance between them. The original field has area $h(a+b)/2$ and we are looking for a field with area $h(a+b)/4$.

One solution would be to join the midpoints of the parallel edges with a straight line.

There are lots of other solutions. The farmer might want to start from a gate on the edge of length $a$ , at a distance $d$ from one corner, and $(a-d)$ from the other. The dividing line would have to be drawn to a point on the opposite side of the field at a distance $(a+b)/2$ from the corner.

Some people tried to solve a harder problem, that is to split the field by a line parallel to the parallel edges of the field. A particular solution to this was sent in by Jack , Henry, Paul and Matthew from Smithdon High School. It is shown below.

Area of original field $= A = 8 \times(10 + 20) / 2 = 120$m$^2$

Therefore area of each new field $= 60$ m$^2$

Let the `height' of the first field be $y$ metres and the bottom edge be $x$ metres.

This means that the `height' of the second field must be $(8 - y)$ and the top edge must be $x$ metres.

For the first field:

$60=y \times (10-x )/2$- (1)

and for the second field:

$60 = (8-y) \times(20+x)/2$ - (2)

Equation (1) can be rearranged to give:

y = 120 / (10+ x ) - (3)

and equation (2) gives:

120 = (8 - y ) (20 + x ) - (4)

Combining equations (3) and (4) gives:

$120 = (8-120/[10+x])(20+x)$ - (5)





which leads to


and $150+15x=x^2+15x-100)$

so $x^2 = 250$m$^2$

which gives $x=15.8$m (3 s.f.)

and in equation (3) $y=120/(10+x)$ so $y= 4.65$m (3 s.f.)


A further challenge: The solution $x^2 = (a^2 + b^2)/2$ is very special mathematically because it is the same whatever the value of the distance $h$. The language mathematicians use is to say that "the solution is independent of $h$''. Can you explain why this happens in this particular problem? If you want a hint then you could use the fact that when you enlarge a shape with a linear scale factor $s$ then the area is enlarged by a scale factor $s^2$. Look out for more about this problem in a future article on areas and scaling.

Correct solutions to the problem were received from Nicholas - South Greenhoe Middle School ;Heacham Middle School; Lucy and Sarah - Archbishop Sancroft High School