Four copies of an isosceles triangle with base 8 cm and height 8 cm are joined to form a new shape. Work out the area of the new shape.

Step-by-step solution

1. 1

   Realise the arrangement does not matter. Area is additive for pieces that meet only along edges: joining shapes without overlapping or leaving gaps means the total is simply the sum of the parts. So the answer can be found without knowing how the four triangles are placed.

2. 2

   Find the area of one triangle.

   $A_{\triangle} = \frac12 \times \text{base} \times \text{height} = \frac12 \times 8 \times 8 = 32\ \text{cm}^2$

   The triangle being isosceles matters only for the shape's outline; the area formula needs just the base and the perpendicular height to it.

3. 3

   Multiply by the number of copies.

   $A_{\text{total}} = 4 \times 32 = 128\ \text{cm}^2$

4. 4

   Sanity-check with a concrete arrangement. Place two triangles base to base to form a rhombus of diagonals $8$ and $16$: its area is $\tfrac12 \times 8 \times 16 = 64\ \text{cm}^2$, exactly two triangles. Two such rhombi give $128\ \text{cm}^2$.

5. 5

   Note the units. Both given lengths are in centimetres, so the area is in square centimetres: $128\ \text{cm}^2$.