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analyze julia knows that a figure has a surface area of 40 square centi…

Question

analyze julia knows that a figure has a surface area of 40 square centimeters. the net shown has 5 - centimeters and 2 - centimeters edges. could the net represent the figure? justify your argument. the area of this net is square centimeters. need help?

Explanation:

Step1: Calculate the area of each rectangle

The formula for the area of a rectangle is \(A = l\times w\).
For the rectangles with \(l = 5\) cm and \(w=2\) cm, \(A_1=5\times2 = 10\) \(cm^{2}\).
There are 4 such rectangles.
For the rectangles with \(l = 5\) cm and \(w = 5\) cm (the two larger - like parts in the net, assume it's a mis - understanding of the net structure, actually, if we consider the net as a rectangular prism net: a rectangular prism has 6 faces. Let's assume the net has 2 faces of \(5\times5\) and 4 faces of \(5\times2\))
The area of a \(5\times5\) rectangle \(A_2 = 5\times5=25\) \(cm^{2}\), but no, wait, correct way: a rectangular prism net (assuming the given net is of a rectangular prism). The surface area formula of a rectangular prism \(S=2(lw + lh+wh)\). Here, assume \(l = 5\) cm, \(w = 2\) cm, \(h = 2\) cm (from the net structure, by counting the rectangles: there are 2 rectangles of \(5\times2\), 2 rectangles of \(2\times2\) (if we assume the correct net structure for a rectangular - like 3D figure, no, wait, another way: count the number of rectangles. The net has 6 rectangles. There are 2 rectangles with dimensions \(5\times2\) and 4 rectangles with dimensions \(2\times2\) (no, wrong. Let's re - count:
The net has 6 rectangles. If we assume the figure is a rectangular prism. Let's calculate the area of each rectangle in the net.
The net has 2 rectangles with area \(A_{big}=5\times2 = 10\) \(cm^{2}\) and 4 rectangles with area \(A_{small}=2\times2=4\) \(cm^{2}\) (this is wrong. Wait, no, the correct way:
The net (assuming it's a rectangular prism net) has 6 faces. Two faces have dimensions \(5\times2\), two faces have dimensions \(2\times2\) and two faces have dimensions \(5\times2\) (no, wait, no. Let's use the formula \(S = 2(lw+lh + wh)\). If \(l = 5\), \(w = 2\), \(h = 2\)
\(S=2(5\times2+5\times2 + 2\times2)\)
\(=2(10 + 10+4)\)
\(=2\times24=48\) \(cm^{2}\)

Another way: count the area of each rectangle in the net.
The net has 6 rectangles. There are 2 rectangles with area \(A_1=5\times2=10\) \(cm^{2}\), 2 rectangles with area \(A_2 = 2\times2 = 4\) \(cm^{2}\) and 2 rectangles with area \(A_3=5\times2=10\) \(cm^{2}\)
\(S=2\times10+2\times4 + 2\times10\)
\(=20 + 8+20\)
\(=48\) \(cm^{2}\)

Answer:

No, the area of this net is 48 square centimeters. Since \(48
eq40\), the net does not represent the figure.