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