19 ABCD is a trapezium in which \( \overline{\mathrm{AD}} / / \overline{\mathrm{BC}}, \mathrm{AD}=27 \mathrm{~cm} \). and \( \mathrm{BC}=45 \mathrm{~cm} \). If the area of \( \triangle \mathrm{ABC}=225 \mathrm{~cm}^{2} \)., find the area of the trapezium.

To find the area of trapezium ABCD, we first use the fact that the area of triangle ABC is given as \( 225 \, \text{cm}^2 \). Since AD and BC are parallel, the height of triangle ABC (which we'll denote as \( h \)) can be calculated using the formula for the area of a triangle: \[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \] For triangle ABC, considering BC as the base: \[ 225 = \frac{1}{2} \times 45 \times h \implies 225 = 22.5 \times h \implies h = \frac{225}{22.5} = 10 \, \text{cm} \] Now, we can find the area of trapezium ABCD using the formula for the area of a trapezium: \[ \text{Area}_{trapezium} = \frac{1}{2} \times (b_1 + b_2) \times h \] Where \( b_1 = AD = 27 \, \text{cm} \), \( b_2 = BC = 45 \, \text{cm} \), and \( h = 10 \, \text{cm} \): \[ \text{Area}_{trapezium} = \frac{1}{2} \times (27 + 45) \times 10 = \frac{1}{2} \times 72 \times 10 = 360 \, \text{cm}^2 \] Therefore, the area of trapezium ABCD is \( 360 \, \text{cm}^2 \).