ABCD is a quadrilateral with AB = 9 cm, BC = 40 cm, CD = 28 cm, DA = 15 cm, and angle ABC is a right angle. What is the area of triangle ADC (in square centimetres)?

Introduction / Context:
This question involves a quadrilateral with one right angle and known side lengths. We are asked to find the area of one of the triangles formed by its vertices, namely triangle ADC. By placing the figure on a coordinate plane in a convenient way and using distance relations, we can determine the coordinates of point D and then compute the area of triangle ADC using base and height. This approach highlights how coordinate geometry can simplify area calculations for composite figures.

Given Data / Assumptions:
• Quadrilateral ABCD has side lengths AB = 9 cm, BC = 40 cm, CD = 28 cm, and DA = 15 cm.
• Angle ABC is a right angle (90 degrees).
• We are required to find the area of triangle ADC.
• No other angles are explicitly given, but the distances and right angle condition determine the geometry uniquely enough for this area.

Concept / Approach:
Because angle ABC is 90 degrees, we can choose a coordinate system with B at the origin, BC on the x axis, and BA on the y axis. Then we can locate points A and C easily. Point D must satisfy two distance conditions: its distance from C is 28 cm and from A is 15 cm. Solving these conditions shows that D lies on segment BC. Once we know the coordinates of A, D, and C, triangle ADC has a horizontal base DC and a vertical height equal to the y coordinate of A, making area computation straightforward using the formula (1/2) * base * height.

Step-by-Step Solution:
Step 1: Place point B at (0, 0). Let BC lie along the x axis with C at (40, 0) so that BC = 40 cm. Step 2: Since angle ABC is 90 degrees and AB = 9 cm, let A be at (0, 9) on the y axis. Step 3: Point D must satisfy CD = 28 cm and DA = 15 cm. Let D = (x, 0) on the x axis. Then distance CD = |40 - x| must equal 28, and distance AD must be 15. Step 4: From CD = 28, we get |40 - x| = 28, so x = 12 or x = 68. Point D must lie between B and C, so x = 12 is valid and D is at (12, 0). Step 5: Verify AD: distance from A(0, 9) to D(12, 0) is √[(12 - 0)² + (0 - 9)²] = √(144 + 81) = √225 = 15 cm, which matches the given DA. Step 6: Now triangle ADC has vertices A(0, 9), D(12, 0), and C(40, 0). The base DC lies along the x axis with length DC = 40 - 12 = 28 cm. Step 7: The height of triangle ADC from A to base DC is the vertical distance from A to the x axis, which is 9 cm. Step 8: Area of triangle ADC = (1/2) * base * height = (1/2) * 28 * 9 = 14 * 9 = 126 square centimetres.

Verification / Alternative check:
We can verify the area using the coordinate formula for area of a triangle with vertices (x₁, y₁), (x₂, y₂), and (x₃, y₃). Applying that formula to A(0, 9), D(12, 0), and C(40, 0) gives area = (1/2) * |x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂)|. Substituting the coordinates and simplifying yields (1/2) * |0(0 - 0) + 12(0 - 9) + 40(9 - 0)| = (1/2) * |0 - 108 + 360| = (1/2) * 252 = 126. This matches the earlier base height method and confirms the answer.

Why Other Options Are Wrong:
124, 122, 120, and 130 square centimetres are all close but do not match the exact computed area of 126 square centimetres. They might arise from arithmetic mistakes in computing distances or in applying the area formula, but they are not consistent with either the base height approach or the coordinate geometry formula.

Common Pitfalls:
Typical errors include misplacing the points in the coordinate plane, miscomputing the length of DC, or incorrectly calculating the area using wrong base or height. Some learners might also attempt to split the quadrilateral in a different way and end up with more complicated calculations. Using the right angle at B to set up simple coordinates and then applying basic triangle area formulas is the cleanest method and avoids unnecessary complexity.

Final Answer:
The area of triangle ADC is 126 SQ.CM.