This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Alright Mercy — let's do this. Question 2: A cuboid carton is open at the top with length l = 50 cm, width w = 30 cm, and height h = 25 cm. Step 1: Identify the formula for the outer surface area of a cuboid open at the top. The total surface area of a closed cuboid is 2(lw + lh + wh). Since the top is open, we subtract the area of the top face (lw). The formula for the outer surface area (OSA) is: OSA = lw + 2(lh + wh) Step 2: Substitute the given values into the formula. OSA = (50 cm × 30 cm) + 2((50 cm × 25 cm) + (30 cm × 25 cm)) Step 3: Calculate the area. OSA = 1500 cm^2 + 2(1250 cm^2 + 750 cm^2) OSA = 1500 cm^2 + 2(2000 cm^2) OSA = 1500 cm^2 + 4000 cm^2 OSA = 5500 cm^2 The outer surface area of the carton is 5500 cm^2. --- Question 3: A company wants to make cuboid boxes each measuring 60 cm by 40 cm by 30 cm. We need to determine the surface area of each board (which implies the total surface area of one closed box). Step 1: Identify the formula for the total surface area of a closed cuboid. Given length L_b = 60 cm, width W_b = 40 cm, and height H_b = 30 cm. The formula for the total surface area (TSA) is: TSA = 2(L_b W_b + L_b H_b + W_b H_b) Step 2: Substitute the given values into the formula. TSA = 2((60 cm × 40 cm) + (60 cm × 30 cm) + (40 cm × 30 cm)) Step 3: Calculate the surface area. TSA = 2(2400 cm^2 + 1800 cm^2 + 1200 cm^2) TSA = 2(5400 cm^2) TSA = 10800 cm^2 The surface area of each board (for one box) is 10800 cm^2. --- Question 4: An open water-tank in the shape of a cuboid has a length l_t = 4.5 m, a width w_t = 2.5 m, and a depth (height) h_t = 1.0 m. We need to find the surface area of the tank that will be in contact with water when the tank is full. Step 1: Identify the formula for the surface area in contact with water for an open tank. When the tank is full, the water touches the base and the four side walls. It does not touch the top. The formula for the area in contact with water is: Area = l_t w_t + 2(l_t h_t + w_t h_t) Step 2: Substitute the given values into the formula. Area = (4.5 m × 2.5 m) + 2((4.5 m × 1.0 m) + (2.5 m × 1.0 m)) Step 3: Calculate the area. Area = 11.25 m^2 + 2(4.5 m^2 + 2.5 m^2) Area = 11.25 m^2 + 2(7.0 m^2) Area = 11.25 m^2 + 14.0 m^2 Area = 25.25 m^2 The surface area of the tank in contact with water is 25.25 m^2. --- Question 5: The surface area of a cuboid is 126 cm^2. The cuboid is 6 cm long and 3 cm tall. We need to find the width of the cuboid. Step 1: Identify the formula for the total surface area of a closed cuboid. Given total surface area (TSA) = 126 cm^2, length l_c = 6 cm, and height h_c = 3 cm. Let the width be w_c. The formula for the total surface area (TSA) is: TSA = 2(l_c w_c + l_c h_c + w_c h_c) Step 2: Substitute the known values into the formula. 126 cm^2 = 2((6 cm × w_c) + (6 cm × 3 cm) + (w_c × 3 cm)) Step 3: Simplify and solve for w_c. 126 = 2(6w_c + 18 + 3w_c) 126 = 2(9w_c + 18) Divide both sides by 2: (126)/(2) = 9w_c + 18 63 = 9w_c + 18 Subtract 18 from both sides: 63 - 18 = 9w_c 45 = 9w_c Divide by 9: w_c = (45)/(9) w_c = 5 cm The width of the cuboid is 5 cm. Send me the next one 📸