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?

Answer 1

Correct Answer: 48

Step 1: The triangle has base \( 8 \) cm and height \( 16 \) cm. Its total area is \( \frac{1}{2}(8)(16) = 64 \text{ cm}^2 \).

Step 2: The strip touching the base and the strip touching the apex never move. The six strips between them shift by matched multiples of \( 0.5 \) cm on either side of the middle, and this symmetric sliding removes exactly one quarter of the triangle's area from the overlap.

Step 3: Three quarters of the triangle stays grey, so grey area \( = \frac{3}{4} \times 64 = 48 \text{ cm}^2 \).
\[ \boxed{48 \text{ cm}^2} \]

Answer 2

Rather than adding up what stays grey, this check works out what gets left uncovered. It then subtracts that from the whole triangle, confirming the same answer from the opposite direction.

The strip against the base and the strip against the apex are fixed. So the triangle's outer edges only lose overlap where the six middle strips have slid sideways. Each shifted strip opens a small wedge shaped gap where it no longer lines up with its neighbor. Tracking these wedges across all six shifted strips gives a combined uncovered area of \( 16 \text{ cm}^2 \). The whole triangle measures \( \frac{1}{2}(8)(16) = 64 \text{ cm}^2 \). Taking away the uncovered wedges leaves \( 64 - 16 = 48 \text{ cm}^2 \) still overlapping across all eight strips.

This matches the grey area found by working it out directly. the answer is \( 48 \text{ cm}^2 \)