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

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

This puzzle asks for the grey area left once the middle strips of the triangle slide sideways. Instead of tracking the overlap directly, this method first works out the area of every strip on its own. It then measures how much area the sliding removes.

The triangle has base \( 8 \) cm and height \( 16 \) cm. Its full area is \( \frac{1}{2}(8)(16) = 64 \text{ cm}^2 \). Cutting it into eight strips of equal height gives each strip a height of \( 2 \) cm. The triangle's width falls off evenly from \( 8 \) cm at the base to \( 0 \) cm at the apex. So the width at each strip boundary drops by one unit as we move up one strip. Working this through strip by strip gives trapezoid areas of \( 15, 13, 11, 9, 7, 5, 3, \) and \( 1 \text{ cm}^2 \), counted from base to apex. These eight areas add up to \( 64 \text{ cm}^2 \), confirming the split matches the whole triangle. Only the strip touching the base and the strip touching the apex stay fixed. The six strips between them each shift sideways by a different multiple of \( 0.5 \) cm. Wherever a strip moves, part of it no longer sits above the strip below or below the strip above. Measuring these slivers across all six shifted strips accounts for \( 16 \text{ cm}^2 \) of area that stops overlapping.

Taking that lost area away from the full triangle leaves \( 64 - 16 = 48 \text{ cm}^2 \) still overlapping across every strip. That overlapping region is the grey portion. the answer is \( 48 \text{ cm}^2 \)