Answer:
2.5 or 3.5
Explanation:
You have an x-cm square and a rectangle ℓ cm by 3 cm. Together, they have a total perimeter of 32 cm. You want the length ℓ and the area S in terms of x, and you want to know the values of x that make the area be 30.25 cm².
(i) Wire length
The length of the wire is 32 cm. This is the sum of the perimeters of the square and the rectangle.
P = 4s +2(l+w)
32 = 4x + 2(ℓ +3) . . . . . . . . substitute given values
(ii) Rectangle length
Solving the equation for ℓ, we have ...
16 = 2x +ℓ +3 . . . . . . . . . divide by 2
ℓ = 13 -2x . . . . . . . . . . . subtract 3+2x
(iii) Total area
The total area of the two figures is the sum of the area of the square and that of the rectangle:
S = x² +3ℓ
S = x² +3(13 -2x) . . . . . . substitute for ℓ
S = x² -6x +39 . . . . . . . simplify
(iv) Value of x
For an area of 30.25 cm, we require ...
30.25 = x² -6x +39
-8.75 = x² -6x . . . . . . . . . . subtract 39
0.25 = x² -6x +9 . . . . . . . complete the square
±0.5 = x -3 . . . . . . . . . . . . take the square root
x = 3 ± 0.5 = {2.5, 3.5}
The total area is 30.25 cm² for x = 2.5 or x = 3.5.
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