Two pieces of land, the first is in the shape of a trapezium in which the lengths of its two parallel bases are 76 metres and 64 metres and the perpendicular distance between them is 45 metres and the second is in the shape of a rhombus whose diagonals lengths are 74 metres and 90 metres. The two pieces are exchanged by a rectangular piece of land whose area equals the sum of areas of the trapezium and rhombus pieces. The ratio between its length and its width is \( 5: 4 \) Find its dimensions.

To solve for the dimensions of the rectangular piece of land, we first need to calculate the areas of the trapezium and the rhombus. The area \( A \) of a trapezium can be calculated using the formula: \[ A = \frac{1}{2} \times (b_1 + b_2) \times h \] where \( b_1 \) and \( b_2 \) are the lengths of the parallel bases and \( h \) is the height. For our trapezium: \[ A = \frac{1}{2} \times (76 + 64) \times 45 = \frac{1}{2} \times 140 \times 45 = 3150 \text{ square metres} \] For the rhombus, the area can be calculated using the formula: \[ A = \frac{1}{2} \times d_1 \times d_2 \] where \( d_1 \) and \( d_2 \) are the lengths of the diagonals. Thus, for our rhombus: \[ A = \frac{1}{2} \times 74 \times 90 = \frac{1}{2} \times 6660 = 3330 \text{ square metres} \] Next, we sum the areas of the trapezium and rhombus: \[ A_{\text{total}} = 3150 + 3330 = 6480 \text{ square metres} \] Now, let the dimensions of the rectangular piece of land be \( 5x \) and \( 4x \) according to the ratio provided: \[ \text{Area of the rectangle} = 5x \times 4x = 20x^2 \] Setting this equal to the total area gives us the equation: \[ 20x^2 = 6480 \] This leads to: \[ x^2 = \frac{6480}{20} = 324 \implies x = 18 \] Now substituting \( x \) back to find the dimensions: \[ \text{Length} = 5x = 5 \times 18 = 90 \text{ metres} \] \[ \text{Width} = 4x = 4 \times 18 = 72 \text{ metres} \] Therefore, the dimensions of the rectangular piece of land are 90 metres by 72 metres.