Problem
of the Week
Problem
C and Solution
Tile
Art

Problem

A tile measuring \(8\) cm by \(8\) cm has gridlines drawn on it, parallel
to each side and spaced \(1\) cm apart.
Six blue triangles are then painted on the tile, as shown.

There are six non-overlapping triangles on the tile with all vertices located at
points where horizontal grid lines and vertical grid lines meet, and
with all of the bases of the triangles placed along the four edges of
the tile. A description each triangle is given below, starting with a
triangle in the top-left corner and then moving clockwise around the
edges of the tile.

Right-angled triangle in the top-left corner of the tile. The
base is 2 cm and the height is 3 cm.

Obtuse triangle with horizontal 3 cm base along the top edge of
the tile. The third vertex is located at a point 2 cm to the left and 4
cm below the left endpoint of the base.

Acute triangle with vertical 3 cm base along the right edge of
the tile. The third vertex is 4 cm to the left and 1 cm below the top
endpoint of the base.

Obtuse triangle with horizontal 2 cm base along the bottom edge
of the tile. The third vertex is 1 cm to the left and 3 cm above the
left endpoint of the base.

Obtuse triangle with horizontal 4 cm base along the bottom edge
of the tile. The third vertex is 1 cm to the right and 3 cm above the
right endpoint of the base.

Isosceles triangle with base 2 cm and height 4 cm.

What fraction of the tile is painted blue?

Solution

We will start by determining the areas of the six painted triangles.
We label the triangles \(A\), \(B\), \(C\), \(D\), \(E\), and \(F\) and draw in a height and a base for
each triangle.

We will calculate the area of each triangle using the formula for the
area of a triangle: \[\text{area} =
\frac{\text{base} \times \text{height}}{2}\]

Triangle \(A\) has base \(2\) cm and height \(3\) cm. The area of triangle \(A\) is then \(\frac{2\times 3}{2}=\frac{6}{2}=3 \text{
cm}^2\).

Triangle \(B\) has base \(3\) cm and height \(4\) cm. The area of triangle \(B\) is then \(\frac{3\times 4}{2}=\frac{12}{2}=6 \text{
cm}^2\).

Triangle \(C\) has base \(3\) cm and height \(4\) cm. The area of triangle \(C\) is then \(\frac{3\times 4}{2}=\frac{12}{2}=6 \text{
cm}^2\).

Triangle \(D\) has base \(2\) cm and height \(3\) cm. The area of triangle \(D\) is then \(\frac{2\times 3}{2}=\frac{6}{2}=3 \text{
cm}^2\).

Triangle \(E\) has base \(4\) cm and height \(2\) cm. The area of triangle \(E\) is then \(\frac{4\times 2}{2}=\frac{8}{2}=4 \text{
cm}^2\).

Triangle \(F\) has base \(2\) cm and height \(4\) cm. The area of triangle \(F\) is then \(\frac{2\times 4}{2}=\frac{8}{2}=4 \text{
cm}^2\).

The total area painted blue is then \(3+6+6+3+4+4=26 \text{ cm}^2\).

The area of the entire tile is \(8 \times 8
= 64 \text{ cm}^2\).

Thus, \(\frac{26}{64}=\frac{13}{32}\) of the tile
is painted blue.