(i) The area of the cross section ABCDEFG is 1771.6cm³
(ii) The volume of the concrete lab is 212592 cm³
(iii) the total surface area of the concrete slab is 30284 cm²
(iv) The mass of the concrete slab is 40746.8 kg/m³.
Given, dimensions are 120 cm by 60 cm by
40 cm.
(i) area of cross section = 40 × 60 ₋ π(60 ₋10 ₋ 10)/2 . 1/2
= 1771.6 cm³
(ii) the volume of the concrete slab = 40 × 60 × 120 ₋ 1/2 π((60 ₋ 10 ₋10)/2)² . 120
= 212592 cm³
(iii) The total surface area is:
=40 × 120 × 2 ₊ 60 × 120 ₊ 1771.6 × 2 ₊ 10 × 120 × 2 ₊ 1/2 π(60 ₋ 10 ₋10) × 120
= 30284 cm²
(iv) Mass of the concrete slab given density is 2300 kg/m³
Mass = Volume × density
Mass = 17.716 m³ × 2300
= 40746.8 kg/m³
Hence we get the mass as 40746.8 kg/m³.
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30•4=120, 8•2=16. 120+16=136
Answer: The volume is 20 cubic feet
The Width is 1/3 feet
Hope I helped
Step-by-step explanation:
Answer:
Step-by-step explanation:
Represent the width by W. Then, "The length of a rectangular field is 7 m less than 4 times the width" expressed symbolically is
L = 4W - 7 (dimensions in meters)
Recall that the perimeter formula in this case is P = 2L + 2W, and recognize that the perimeter value is 136 m. After substituting 4W - 7 for L, we get:
136 m = 2(4W - 7) + 2W, or
136 = 8W - 14 + 2W, or
150 = 10W These three equations are equivalent mathematical statements.
150 = 10W reduces to W = 15 (meters).
Part A: the independent variable is W, the width of the field.
Part B: The mathematical statement is 136 m = 2(4W - 7) + 2W, which after algebraic manipulation becomes 150 = 10W.
Part C: The above equation can be solved for W: W = 15 meters. This is the value of the independent variable.