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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.

The coordinate plane with horizontal axis pointing to the right labelled x, and vertical axis pointing up labelled y. The two axes meet at point O. There are eight vertical and eight horizontal grid lines. Point A is on the vertical axis, 4 units above O. Point B is on the horizontal axis, 3 units to the right of O. Triangle AOB is highlighted.

  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.

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

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.

    Point C is on the vertical axis, 8 units above O. Point D is on the horizontal axis, 6 units to the right of O.

  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 \(\triangle AOB\).

    One of the smaller triangles is the original highlighted triangle AOB. Two of the other smaller triangles are copies of triangle AOB but one is moved 4 units above and the other is moved 3 units to the right. The final small triangle is the mirror image of triangle AOB, reflected in its slanted side AB.

    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.

Triangle 1 has base x units and height y units. Triangle 2 is identical to triangle 1 but moved x units to the right. Triangle 3 is identical to triangle 1 but moved y units down. Triangle 4 is identical to triangle 1 but moved x units to the right then y units down.

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.