Two paths each 2m wide run across the middle of a rectangular field. If the area of the cross path is 196m and its length is 60m
1 answer:
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Answer:
Container B has smaller surface area.
Step-by-step explanation:
Given:
Container A
Radius = 60/2 = 30 mm
Height = 4 x 60 = 240 mm
Container B
Length = 120
Width = 120
Height = 60
Computation:
Surface area of container A (Cylinder) = 2πr[h+r]
Surface area of container A (Cylinder) = 2[22/7][60][120+60]
Surface area of container A (Cylinder) = 67,885.70 mm² (Approx)
Surface area of container B (Cuboid) = 2[lb+bh+hl]
Surface area of container B (Cuboid) = 2[(14,400)+(7,200)+(7,200)]
Surface area of container B (Cuboid) = 57,600 mm²
Container B has smaller surface area.
Answer:
(C)72.4 in
Step-by-step explanation:
Given an acute triangle in which the longest side measures 30 inches; and the other two sides are congruent.
Consider the attached diagram
AB=BC=x
However to be able to solve for x, we form a right triangle with endpoints A and C.
Since the hypotenuse is always the longest side in a right triangle
Hypotenuse, AC=30 Inches
Using Pythagoras Theorem
Therefore, the smallest possible perimeter of the triangle
Perimeter=2x+30
=2(21.21)+30
=42.42+30
=72.4 Inches (rounded to the nearest tenth)
