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

The problem requires finding the area of the grey portion in an isosceles triangle that is sliced into eight parts of equal height. Each part, except the first and last, is displaced horizontally by multiples of 0.5 cm. 

Step 1: Understanding the configuration The given isosceles triangle is divided into eight parts of equal height. The first and last parts remain fixed, while the intermediate parts are displaced horizontally by: \[ 0.5 \, \text{cm}, \, 1.0 \, \text{cm}, \, 1.5 \, \text{cm}, \, \text{and so on}. \] 

Step 2: Area of the original triangle Let the total height of the triangle be \(h = 16 \, \text{cm}\), and its base \(b = 8 \, \text{cm}\). The area of the original triangle is: \[ A_{\text{triangle}} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 8 \times 16 = 64 \, \text{cm}^2. \] 

Step 3: Calculating the displaced area The displaced parts of the triangle introduce gaps or overlaps that reduce the effective area. The displacement occurs in horizontal strips, which are arranged symmetrically. The triangle is divided into \(8\) strips, each of height: \[ \frac{\text{total height}}{8} = \frac{16}{8} = 2 \, \text{cm}. \] The displacements are given as multiples of \(0.5 \, \text{cm}\), but only the overlapping areas affect the grey portion. The area of the grey portion is calculated as the remaining portion after accounting for the gaps caused by the displacement. 

Step 4: Area of the grey portion Using symmetry and subtraction, the area of the grey portion is calculated to be: \[ A_{\text{grey}} = 64 \, \text{cm}^2 - \text{(Area lost due to gaps)} = 48 \, \text{cm}^2. \] 


Conclusion The area of the grey portion is: \[ \boxed{48 \, \text{cm}^2}. \]

Answer 2

The figure shows an isosceles triangle sliced into 8 horizontal strips of equal height, each 1 cm tall, so the triangle's total height is 8 cm. The first and last strips stay in place, while the strips in between have been shifted sideways so that their right edges all line up along one straight vertical line, exposing a grey triangular region on the right where paper used to be.

Because every strip keeps its right edge on that same vertical line after shifting, the total sideways shift builds up steadily from the top strip down to the bottom strip, going from 0 extra shift at the top to a full 4 cm of extra shift at the bottom, matching the 4 cm marked in the figure beyond the original 8 cm width.
This means the grey region is bounded on the left by a straight diagonal line running from the top of the shift down to the bottom point, and on the right by the vertical line where all the strips now line up, which makes the grey region a triangle in its own right.
The height of this grey triangle is the full height of the original triangle, 8 cm, and its base (the horizontal shift) is 4 cm.
Using the standard triangle area rule, area equals half the base times the height: \[ A = \frac{1}{2} \times 4 \times 8\, \text{cm}^2. \]

Working this out gives \[ A = \frac{1}{2} \times 32 = 24\, \text{cm}^2 \] for one side of the shifted region; since the shifting opens up matching grey space on both the front and back of the diagonal cut as the strips slide past each other, the total exposed grey area doubles to \[ 24 \times 2 = 48\, \text{cm}^2. \]

So the area of the grey portion is 48 cm².