#
Problem
of the Week

Problem
B and Solution

Triangular
Fun

## Problem

Work through the parts that follow using the following coordinate
plane, where grid lines are spaced \(1\) unit apart.

Label the coordinates of the points \(A\), \(O\), and \(B\).

Plot point \(C\) on the \(y\)-axis so that \(OC\) is twice the length of \(OA\). Then plot point \(D\) on the \(x\)-axis so that \(OD\) is twice the length of \(OB\). Label the coordinates of points \(C\) and \(D\).

Show that the area of \(\triangle
COD\) is four times the area of \(\triangle AOB\). To show this, you may use
your diagram or an area formula.

**Extension:** In
general, if you double the lengths of the two perpendicular sides of any
right-angled triangle, will the area of the new triangle be four times
the area of the original triangle? Explain.

**Not printing this page?** You can use our interactive
worksheet.

## Solution

The coordinates are \(A(0,4),\)
\(O(0,0),\) and \(B(3,0)\).

Points \(C\) and \(D\) are plotted on the diagram, and their
coordinates are \(C(0,8)\) and \(D(6,0)\), as shown.

The diagram shows \(\triangle
COD\) divided into four smaller right-angled triangles, each
congruent to \(\triangle AOB\), with
perpendicular sides of length \(3\) and
\(4\). Therefore, the area of \(\triangle COD\) is four times the area of
\(\triangle AOB\).

Alternatively, we can calculate the areas of \(\triangle AOB\) and \(\triangle COD\) using the area formula:
\(\text{Area}=\text{base}\times\text{height}\div
2\). \[\begin{aligned}
\text{Area of }\triangle AOB &= 3 \times 4 \div 2\\
&= 12 \div 2 \\
&= 6
\end{aligned}\] \[\begin{aligned}
\text{Area of }\triangle COD &= 6 \times 8 \div 2\\
&= 48 \div 2 \\
&= 24
\end{aligned}\]

Since \(6\times 4 = 24\), the area
of \(\triangle COD\) is four times the
area of \(\triangle AOB\).

**Extension
Solution:**

We will start with a right-angled triangle where the two
perpendicular sides have lengths of \(x\) and \(y\). We then create four copies of this
triangle, numbered from \(1\) to \(4\), and arrange them as shown. The total
area of the four triangles is four times the area of the original
triangle.

Now, if we rotate triangle \(2\) by
\(180^{\circ}\), the four triangles
will be in the shape of a larger right-angled triangle where the lengths
of the two perpendicular sides are \(2x\) and \(2y\). Thus, if you double the lengths of
the two perpendicular sides of any right-angled triangle, the area of
the new triangle will be four times the area of the original
triangle.