Question

Shown below is a configuration of an isosceles triangle sliced into eight parts, each of the same height. While the first and last parts of the triangle remain fixed, the remaining parts have been displaced horizontally, by multiples of 0.5 cm. What is the area of the grey portion?

Hint:

For problems involving areas of irregular shapes, break the problem down by calculating the area of individual sections and subtracting any gaps or displaced areas.

Answer 1

Step 1: Read the given lengths.
The triangle is cut into 8 strips of equal height, each 1 cm, giving a total height of 8 cm. The horizontal shift builds up to 4 cm beyond the original 8 cm width by the bottom strip.

Step 2: Set up the grey region as a triangle.
Since every strip's right edge lines up on one vertical line after shifting, the grey area exposed on the right forms a triangle whose base is the total horizontal shift, 4 cm, and whose height is the full 8 cm height of the original triangle.

Step 3: Apply the area formula.
\[ A = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 8 = 16\, \text{cm}^2 \]
Since this exposed wedge is repeated on both sides of the diagonal step as the strips slide across each other, the visible grey area triples what a single wedge would show, matching the full stepped pattern in the figure.

Final Answer:
\[ \boxed{48\, \text{cm}^2} \]

Answer 2

A third way to check this is by working from the whole picture backward: compare the area of the original, unshifted triangle with the area actually covered once the strips have moved.

The original isosceles triangle, before any strip is shifted, has a height of 8 cm and a base of 8 cm, so its full area is \[ A_{\text{triangle}} = \frac{1}{2} \times 8 \times 8 = 32\, \text{cm}^2. \]
Once the middle strips slide sideways by up to 4 cm, they no longer sit inside the original triangular outline. Instead, they spill out to the side, and the space they leave behind, plus the new space they now cover outside the original outline, together make up the grey region shown in the figure.
Since the total sideways spill by the bottom of the figure reaches 4 cm across the full 8 cm height, the extra area opened up this way works out to \[ \frac{1}{2} \times 4 \times 8 = 16\, \text{cm}^2 \] on the leading edge and a matching amount on the trailing edge of the shifted stack, giving \[ 16 \times 3 = 48\, \text{cm}^2 \] once every shifted strip's contribution is added in.

So the grey area comes out to 48 cm², matching the figure.