8. A rectangle is divided into three regions of equal area. One of the regions is an equilateral triangle with side-length $4 \mathrm{~cm}$; and the other two are trapezia, as shown.
What is the length of the smaller of the parallel sides of the trapezia?

A $\sqrt{2} \mathrm{~cm}$
B $\sqrt{3} \mathrm{~cm}$
C $2 \sqrt{2} \mathrm{~cm}$
D $3 \mathrm{~cm}$
E $2 \sqrt{3} \mathrm{~cm}$
SOLUTION
B

The height of the triangle is $2 \sqrt{3} \mathrm{~cm}$, which you can show using Pythagoras. This gives its area, in $\mathrm{cm}^{2} \frac{1}{2} \times 4 \times 2 \sqrt{3}$, which is $4 \sqrt{3} \mathrm{~cm}^{2}$. The area of the rectangle is then $3 \times 4 \sqrt{3} \mathrm{~cm}^{2}$ or $12 \sqrt{3} \mathrm{~cm}^{2}$. Then its length is $\frac{12 \sqrt{3}}{4} \mathrm{~cm}=3 \sqrt{3} \mathrm{~cm}$. So the length of the smallest parallel side of the trapezium is $3 \sqrt{3}-2 \sqrt{3} \mathrm{~cm}$, which is $\sqrt{3} \mathrm{~cm}$.