Review question

# Can we show that the area of this hexagon is $2(x^2 + xy + y^2)$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R8987

## Solution

The lengths of the two smaller sides of a right-angled triangle are $x$ in. and $y$ in. A square is constructed on each side of the triangle and the exterior points of the squares are joined to form a hexagon as in the diagram above. Show that the area of the hexagon is $2(x^2 + xy + y^2)$ sq. in.

The area, $A$, of the hexagon is the sum of the seven constituent parts that are labelled in the diagram below. We have also labelled some angles.

We will write $A_n$ for the area of the $n$th region, so we have that $\begin{equation*} A_1 = \frac{xy}{2}, \quad A_2 = z^2, \quad A_4 = x^2, \quad A_5 = \frac{xy}{2}, \quad A_6 = y^2 \end{equation*}$

We need to express $A_{2}$ in terms of $x$ and $y$. By Pythagoras’ theorem, $x^2 + y^2 = z^2$, so $A_{2}=x^2+y^2$.

We still need to find $A_3$ and $A_{7}$ in terms of $x$ and $y$.

If a triangle has an angle of size $\alpha$ with adjacent sides of lengths $a$ and $b$, its area is $\dfrac{1}{2} ab \sin \alpha$.

We also have that $\sin(\pi - \phi) = \sin \phi$ and we can write $\sin \phi$ as $\dfrac{y}{z}.$ Similarly, $\sin(\pi-\theta)=\dfrac{x}{z}$.

Consequently, $\begin{equation*} A_3 = \frac{1}{2} x z \sin(\pi - \phi) = \frac{1}{2} x z \sin \phi = \frac{1}{2} x z \dfrac{y}{z} = \frac{xy}{2}. \end{equation*}$ Similarly, $\begin{equation*} A_7 = \frac{1}{2} y z \sin(\pi - \theta) = \frac{1}{2} y z\sin \theta = \frac{1}{2} y z\dfrac{x}{z} =\frac{xy}{2}. \end{equation*}$ Thus, the area of the hexagon is $\begin{equation*} A = A_{1}+A_{2}+\cdots+A_{7} = \frac{xy}{2} + (x^2 + y^2) + \frac{xy}{2} + x^2 + \frac{xy}{2} + y^2 + \frac{xy}{2} = 2(x^2 + xy + y^2) \end{equation*}$

as required.

Given that $x + y = 10$, determine the minimum area of the hexagon.

As $y = 10 - x$, we can substitute this into the above expression for $A$: \begin{align*} A &= 2(x^2 + x(10-x) + (10-x)^2) \\ &= 2(x^2 + 10x - x^2 + 100 - 20x + x^2) \\ &= 2(x^2 - 10x + 100). \end{align*}

Hence, $\begin{equation*} \frac{dA}{dx} = 4x - 20 \end{equation*}$ and $\begin{equation*} \frac{d^2A}{dx^2} = 4. \end{equation*}$

As $\dfrac{d^2A}{dx^2}>0$, any stationary point is, indeed, a minimum.

We have shown that $\begin{equation*} \frac{dA}{dx} = 0 \iff x = 5, \end{equation*}$ so it follows that the minimum area is attained when $x = 5$ and $y = 10 - x = 5$. Hence, the minimum area is $\begin{equation*} 2(5^2 + 5^2 + 5^2) = 2 \times 75 = 150 \,\,\,\text{sq. in.} \end{equation*}$