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.

We are given a trapezium (in some countries called a trapezoid) ABCD in which the sides AD and BC are parallel. Their lengths are   AD = 27 cm  and  BC = 45 cm. Also, the area of triangle ABC is 225 cm². We wish to find the area of the trapezium. One way to approach the problem is to place the figure in the coordinate plane in a convenient way. Step 1. Choose a Coordinate System • Place the line containing BC horizontally. For convenience, let   B = (0, 0)  and  C = (45, 0) so that the length BC = 45 cm. • Since AD is parallel to BC, it is also horizontal. Let the line containing AD be y = k for some constant k. We will determine k using the given area of triangle ABC. Step 2. Locate Point A Using the Area of Triangle ABC Triangle ABC has vertices A, B, and C. Since B and C lie on the line y = 0, the altitude of the triangle from A down to line BC will just be the vertical distance |y_A|. The area of triangle ABC is given by:   Area = ½ × (base) × (height) Here, the base is BC = 45 cm. We are told that the area is 225 cm²:   ½ × 45 × |y_A| = 225 Solve for |y_A|:   45 × |y_A| = 450  ⟹  |y_A| = 10 Without loss of generality, take y_A = 10. (A can be placed above BC.) Now, A = (x_A, 10) where x_A is not yet fixed. (Its x‐coordinate does not affect the height.) Step 3. Locate Point D Since AD must lie on the same horizontal line as A (because AD ∥ BC) and its length is 27 cm, if we choose A = (x_A, 10), then we need   |x_D – x_A| = 27. For concreteness, let:   D = (x_A + 27, 10). Step 4. Compute the Area of the Trapezium The formula for the area of a trapezium with parallel sides (bases) of lengths b₁ and b₂ and height h is   Area = ½ × (b₁ + b₂) × h. Here, the two bases are AD and BC. Their lengths are:   AD = 27 cm  and  BC = 45 cm. The height is the vertical distance between the lines containing AD and BC, which is |10 – 0| = 10 cm. Thus, the area is:   Area = ½ × (27 + 45) × 10      = ½ × 72 × 10      = 36 × 10      = 360 cm². Step 5. Conclude Therefore, the area of the trapezium ABCD is 360 cm².