# 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.

1. Label the coordinates of the points $$A$$, $$O$$, and $$B$$.

2. 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 $$D$$.

3. 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.

## Solution

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

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

3. 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.