A parallelogram has vertices $A (1, 2)$ , $B (7, 5)$ , $C (10, 1)$ , and $D (4, - 2)$ . Find its area in square units.

hard 4 marks Class 9 · Mathematics · Quadrilaterals

Answer

The correct answer is 33 — Step 6: Divide by $2$ to obtain the area: $A = \frac{66}{2} = 33$ square units.

Step-by-step solution

Step 1: Write the vertices in order: $A (1, 2)$ , $B (7, 5)$ , $C (10, 1)$ , $D (4, - 2)$ .
Step 2: Apply the shoelace formula: $A = \frac{1}{2} \sum_{i = 1}^{4} x_{i} y_{i + 1} - \sum_{i = 1}^{4} y_{i} x_{i + 1}$ with $(x_{5}, y_{5}) = (x_{1}, y_{1})$ .
Step 3: Compute the products for the first sum: $1 \times 5 = 5$ , $7 \times 1 = 7$ , $10 \times (- 2) = - 20$ , $4 \times 2 = 8$ ; sum = $0$ .
Step 4: Compute the products for the second sum: $2 \times 7 = 14$ , $5 \times 10 = 50$ , $1 \times 4 = 4$ , $(- 2) \times 1 = - 2$ ; sum = $66$ .
Step 5: Subtract and take absolute value: $∣0 - 66∣ = 66$ .
Step 6: Divide by $2$ to obtain the area: $A = \frac{66}{2} = 33$ square units.

Common mistake: Using wrong order of vertices or miscomputing product terms leading to negative area

Concept tested: Shoelace formula for polygon area