A shape is made by placing a small cube on top of a larger one, as shown - Junior Cycle Mathematics - Question 9 - 2016

To find the length of |AB|, which is the diagonal of the base of the larger cube:

1. Since |AB| is the diagonal of a square base of side 2 units, we can apply the Pythagorean theorem:

   $|AB| = \sqrt{2^2 + 2^2}$ $= \sqrt{4 + 4}$ $= \sqrt{8}$ $= 2\sqrt{2} \text{ units}$

Thus, |AB| is 2√2 units.

For the right-angled triangle ABC, using the Pythagorean theorem:

1. The triangle has sides |AB| and the height from C to the plane containing AB.

2. Since |AB| = 2√2 units and the opposite height is the small cube's side (1 unit):

   $|BC| = \sqrt{|AB|^2 + |height|^2}$ $= \sqrt{(2\sqrt{2})^2 + 1^2}$ $= \sqrt{8 + 1}$ $= \sqrt{9}$ $= 3 \text{ units}$

Thus, |BC| is 3 units.

To find the length of the line segment BC that is within the larger cube, we need to consider similar triangles or the heights involved:

1. The part of BC that lies within the cube corresponds to the height of the smaller cube:

   $Length_{inside} = \frac{1}{\sqrt{17}} |BC|$ $= \frac{1}{\sqrt{17}} \times 3$ $= \frac{3}{\sqrt{17}}$

Thus, the length of the part of the line BC that is inside the larger cube is (3/√17) units.