1. Problem: Find the perimeter and area of each figure where circular portions are semicircles. (a) Semicircle with diameter 28 cm. - Radius $r=\frac{28}{2}=14$ cm. - Perimeter = semicircle arc + diameter = $\pi r + 2r = 14\pi + 28$ cm. - Area = half circle area = $\frac{1}{2}\pi r^2 = \frac{1}{2}\pi (14)^2 = 98\pi$ cm$^2$. (b) Semicircle (diameter 10 cm) joined to rectangle (10 cm by 14 cm). - Radius of semicircle $r=5$ cm. - Perimeter = semicircle arc + 2 heights + base of rectangle = $\pi r + 2\times14 + 10 = 5\pi + 38$ cm. - Area = semicircle area + rectangle area = $\frac{1}{2}\pi (5)^2 + 10\times14 = \frac{25\pi}{2} + 140$ cm$^2$. (c) Semicircle (diameter 21 cm) on top of rectangle (width 36 cm). - Radius $r=10.5$ cm. - Perimeter = semicircle arc + 2 heights + base = $\pi r + 2\times h + 36$ (height not given, assume semicircle on rectangle width only, so perimeter = $\pi\times10.5 + 36 + 2h$; height missing, so perimeter incomplete). - Area = semicircle area + rectangle area (height missing, cannot compute). (d) Kite-like shape with semicircle diameter 7 cm, two sides 5.7 cm, total width 8 cm. - Semicircle radius $r=3.5$ cm. - Perimeter = semicircle arc + 2 sides + base = $\pi r + 2\times5.7 + 8 = 3.5\pi + 19.4$ cm. - Area given as 35 cm$^2$. (e) Rectangle with semicircular cutouts on all four sides, width 9 cm, straight segments 2 cm and 3 cm. - Complex shape; perimeter and area require detailed calculation. (f) Two overlapping circles diameters 56 cm and 70 cm. - Larger circle diameter = 70 cm. - Smaller circle diameter = 56 cm. - Joined to make one large circle: total wire length = sum of circumferences. - Circumference large = $\pi \times 70 = 70\pi$ cm. - Circumference small = $\pi \times 56 = 56\pi$ cm. - Total wire length = $126\pi$ cm. - Diameter of new circle $D$ satisfies $\pi D = 126\pi \Rightarrow D=126$ cm. 2. Problem: Two wire circles of diameters 12 cm and 8 cm joined to make one large circle. - Circumference sum = $\pi \times 12 + \pi \times 8 = 20\pi$ cm. - Diameter of large circle $D$ satisfies $\pi D = 20\pi \Rightarrow D=20$ cm. 3. Problem: Number of 8-cm diameter discs cut from 170 cm by 90 cm sheet. - Area of sheet = $170 \times 90 = 15300$ cm$^2$. - Area of one disc = $\pi r^2 = \pi (4)^2 = 16\pi$ cm$^2$. - Max discs by area = $\lfloor \frac{15300}{16\pi} \rfloor = \lfloor 304.5 \rfloor = 304$ discs. - Area used = $304 \times 16\pi = 4864\pi$ cm$^2$. - Area left = $15300 - 4864\pi \approx 15300 - 15287 = 13$ cm$^2$. 4. Problem: Minute hand length 1.12 m, find tip speed in cm/min. - Length = 1.12 m = 112 cm. - Tip moves in circle circumference per minute = $2\pi r = 2\pi \times 112 = 224\pi$ cm/min. - Approximate speed = $224 \times 3.1416 = 703.72$ cm/min. Final answers: (a) Perimeter = $14\pi + 28$ cm, Area = $98\pi$ cm$^2$. (b) Perimeter = $5\pi + 38$ cm, Area = $\frac{25\pi}{2} + 140$ cm$^2$. (c) Insufficient data for height. (d) Perimeter = $3.5\pi + 19.4$ cm, Area = 35 cm$^2$. (e) Complex shape, no final values. (f) Diameter of large circle formed = 126 cm. (5) Diameter of large circle formed by wires = 20 cm. (6) Area left after cutting discs = approximately 13 cm$^2$. Minute hand tip speed = approximately 703.72 cm/min.