1. Problem a: Find the area and perimeter of a rectangle 8 cm by 5 cm with a quarter circle of radius 5 cm removed from the top-left corner. 2. Area of rectangle: $$A_{rect} = 8 \times 5 = 40\text{ cm}^2$$ 3. Area of quarter circle removed: $$A_{qc} = \frac{1}{4} \pi r^2 = \frac{1}{4} \pi (5)^2 = \frac{25\pi}{4}$$ 4. Area of colored region: $$A = A_{rect} - A_{qc} = 40 - \frac{25\pi}{4}$$ 5. Perimeter of rectangle: $$P_{rect} = 2(8 + 5) = 26\text{ cm}$$ 6. Perimeter of colored region: Replace the quarter circle arc (length $$\frac{1}{4} 2\pi r = \frac{\pi r}{2} = \frac{5\pi}{2}$$) with the two straight edges of the quarter circle (each 5 cm). So, $$P = P_{rect} - 2 \times 5 + \frac{5\pi}{2} = 26 - 10 + \frac{5\pi}{2} = 16 + \frac{5\pi}{2}$$ --- 1. Problem b: Rectangle 9 cm by 10 cm with quarter circle radius 6 cm removed from top-left corner. 2. Area rectangle: $$90\text{ cm}^2$$ 3. Area quarter circle removed: $$\frac{1}{4} \pi 6^2 = 9\pi$$ 4. Area colored: $$90 - 9\pi$$ 5. Perimeter rectangle: $$2(9+10) = 38\text{ cm}$$ 6. Perimeter colored: Replace two edges of length 6 cm with quarter circle arc length $$\frac{\pi \times 6}{2} = 3\pi$$ $$P = 38 - 12 + 3\pi = 26 + 3\pi$$ --- 1. Problem c: Rectangle 5 m by 12 m with two triangular segments removed at top-left and bottom-left corners, each with 30° angle. 2. Area rectangle: $$5 \times 12 = 60\text{ m}^2$$ 3. Each triangle area: right triangle with angle 30°, height 5 m, base calculated by $$b = h \tan 30^\circ = 5 \times \frac{1}{\sqrt{3}} = \frac{5}{\sqrt{3}}$$ Area each triangle: $$\frac{1}{2} \times 5 \times \frac{5}{\sqrt{3}} = \frac{25}{2\sqrt{3}}$$ 4. Total removed area: $$2 \times \frac{25}{2\sqrt{3}} = \frac{25}{\sqrt{3}}$$ 5. Area colored: $$60 - \frac{25}{\sqrt{3}}$$ 6. Perimeter: original $$2(5+12) = 34$$ m Remove two vertical edges of length $$5$$ m each, add hypotenuse of triangles: Hypotenuse $$= \frac{5}{\cos 30^\circ} = \frac{5}{\sqrt{3}/2} = \frac{10}{\sqrt{3}}$$ Total hypotenuse length $$2 \times \frac{10}{\sqrt{3}} = \frac{20}{\sqrt{3}}$$ Perimeter colored: $$P = 34 - 10 + \frac{20}{\sqrt{3}} = 24 + \frac{20}{\sqrt{3}}$$ --- 1. Problem d: Square side 8.4 cm with two intersecting circle segments removed forming a pointed shape. 2. Area square: $$8.4^2 = 70.56\text{ cm}^2$$ 3. Without exact radius or segment details, assume each circle segment is a quarter circle with radius 8.4 cm. Area each quarter circle: $$\frac{1}{4} \pi (8.4)^2 = \frac{1}{4} \pi \times 70.56 = 17.64\pi$$ Two segments area: $$2 \times 17.64\pi = 35.28\pi$$ 4. Area colored: $$70.56 - 35.28\pi$$ 5. Perimeter: original $$4 \times 8.4 = 33.6$$ cm Replace two edges of length 8.4 cm each with two quarter circle arcs of length $$\frac{1}{4} 2\pi r = \frac{\pi r}{2} = \frac{8.4\pi}{2} = 4.2\pi$$ each Total arc length $$2 \times 4.2\pi = 8.4\pi$$ Perimeter colored: $$P = 33.6 - 16.8 + 8.4\pi = 16.8 + 8.4\pi$$ --- 1. Problem e: Square side 18 m with four quarter circle segments removed from middle of each side forming a star shape. 2. Area square: $$18^2 = 324\text{ m}^2$$ 3. Each quarter circle radius is half side: $$9$$ m Area each quarter circle: $$\frac{1}{4} \pi 9^2 = \frac{81\pi}{4}$$ Four quarter circles total area: $$4 \times \frac{81\pi}{4} = 81\pi$$ 4. Area colored: $$324 - 81\pi$$ 5. Perimeter: original $$4 \times 18 = 72$$ m Replace four straight segments of length 9 m each (total 36 m) with four quarter circle arcs of length $$\frac{1}{4} 2\pi r = \frac{\pi r}{2} = \frac{9\pi}{2}$$ each Total arc length $$4 \times \frac{9\pi}{2} = 18\pi$$ Perimeter colored: $$P = 72 - 36 + 18\pi = 36 + 18\pi$$