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’ *5 Rec-tan ole 1 8 1' Rectangle 2 h - Rectangle 3 5 n . T - .22; - surface area of the figure ‘

5 Rectan ole 1 8 1 Rectangle 2 h Rectangle 3 5 n T 22 surface area of the figure class=

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I don't quite understand your topic sentence however to my best knowledge This is what i've come up with:
Rectangle 1 and 3 are the same. The only known side is the width at 5 inches whilst the other is unknown so it will be h.
Rectangle 1 and 3 area are "5h"

Rectangle 2 has a width of 8 and a length of h. Its area is "8h"

Without triangles our TSA for the rectangles is:   2 X 5H + 8H = TSA

The triangles are both assumed to be the same are. To get the Are of one it is [tex] \frac{1}{2}*(B*H) [/tex]
Which = 12

Ans: TSA = 18H + 24

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