1. **Stating the problem:** We need to calculate the area of two shapes: a right triangle with legs 3\frac{1}{2} inches and 4 inches, and a trapezoid-like quadrilateral with sides 6 cm, 5 cm, and 13 cm, and a right angle at the lower left. 2. **Right triangle area formula:** The area $A$ of a right triangle is given by $$A = \frac{1}{2} \times \text{base} \times \text{height}$$ where the base and height are the two legs forming the right angle. 3. **Calculate the right triangle area:** Convert $3\frac{1}{2}$ inches to an improper fraction or decimal: $$3\frac{1}{2} = 3 + \frac{1}{2} = \frac{7}{2} = 3.5$$ Then, $$A = \frac{1}{2} \times 3.5 \times 4$$ 4. **Simplify the multiplication:** $$A = \frac{1}{2} \times 14 = 7$$ So, the area of the right triangle is 7 square inches. 5. **Quadrilateral area:** The quadrilateral has a right angle and sides 6 cm, 5 cm, and 13 cm. Since it looks like a trapezoid with a right angle, we can split it into a right triangle and a rectangle or use the Pythagorean theorem to find the height. 6. **Check if the quadrilateral is a right trapezoid:** The side lengths 5 cm and 6 cm can be legs of a right triangle, and 13 cm is the hypotenuse. Check: $$5^2 + 12^2 = 25 + 144 = 169 = 13^2$$ But 6 cm is given, so let's check if 5, 6, and 13 satisfy Pythagoras: $$5^2 + 6^2 = 25 + 36 = 61 \neq 169$$ So 13 cm is not the hypotenuse of a right triangle with legs 5 and 6. 7. **Assuming the quadrilateral is a right trapezoid with height 5 cm and bases 6 cm and 13 cm:** Area formula for trapezoid: $$A = \frac{1}{2} (b_1 + b_2) h$$ where $b_1$ and $b_2$ are the lengths of the parallel sides, and $h$ is the height. 8. **Calculate the trapezoid area:** $$A = \frac{1}{2} (6 + 13) \times 5 = \frac{1}{2} \times 19 \times 5 = \frac{95}{2} = 47.5$$ So, the area of the trapezoid is 47.5 square centimeters. **Final answers:** - Right triangle area: 7 square inches - Quadrilateral (trapezoid) area: 47.5 square centimeters