Question

quadrilatè fcde a sanble ak quadrilatè jghi a. jwenn valè x.
options: x=4, x=5, x=11, x=4.5

Explanation:

Step1: Identify the shapes

FCDE is a square with side 6m, so its area is \(6\times6 = 36\) \(m^2\). JGHI has one side 4m and the other \(x\), so its area is \(4\times x\) \(m^2\).

Step2: Set up the area equation

Since the areas are equal, \(4x=36\).

Step3: Solve for \(x\)

Divide both sides by 4: \(x = \frac{36}{4}=9\)? Wait, no, wait the options. Wait maybe the problem is about perimeters? Wait FCDE is a square, perimeter is \(4\times6 = 24\)m. JGHI is a rectangle with length 4m and width \(x\), perimeter is \(2(4 + x)\). If perimeters are equal: \(2(4 + x)=24\) → \(4 + x = 12\) → \(x = 8\)? No, the options are x=4,5,11,4.5. Wait maybe the problem is about similarity? Wait FCDE is a square (6x6), JGHI is a rectangle (4x x). If they are similar, then \(\frac{6}{4}=\frac{6}{x}\)? No, that would be \(x = 4\). Wait, maybe the problem is that FCDE and JGHI have the same area. Wait FCDE area: \(6\times6 = 36\). JGHI area: \(4\times x\). So \(4x = 36\) → \(x = 9\), but that's not an option. Wait maybe the problem is about side ratios. Wait the first square has side 6, the second has one side 4, and we need to find x such that maybe the other sides relate. Wait the options include x=4. Wait maybe I misread. Wait the first quadrilateral is FCDE, a square (6m each side). The second is JGHI, with one side 4m and the other x. If they are similar (since both are rectangles, square is a rectangle), then the ratio of sides should be equal. So \(\frac{6}{4}=\frac{6}{x}\)? No, that's not. Wait maybe the problem is that FCDE and JGHI have the same perimeter. Perimeter of FCDE: \(4\times6 = 24\). Perimeter of JGHI: \(2(4 + x)\). So \(2(4 + x)=24\) → \(4 + x = 12\) → \(x = 8\). Not an option. Wait the options are x=4,5,11,4.5. Wait maybe the problem is about area proportion. Wait maybe FCDE is 6x6=36, JGHI is 4x x. If 4x = 18 (half of 36), x=4.5. Ah! Maybe the area of JGHI is half? No, the problem says "a sanble ak" which might mean "equal to" in Creole. Wait maybe the original problem is that FCDE and JGHI have equal areas. Wait 66=36, 4x=36 → x=9. Not an option. Wait maybe the first figure is a square with side 6, the second is a rectangle with length 4 and width x, and maybe the problem is about something else. Wait the options include x=4.5. Let's check 44.5=18, 63=18? No. Wait maybe the problem is about perimeter. Wait 64=24, 2(4 + x)=24 → x=8. No. Wait maybe the problem is that FCDE has area 36, and JGHI has area 4x, and maybe there's a miscalculation. Wait the options have x=4. Let's see, if x=4, then JGHI is a square (4x4), area 16, FCDE is 36. No. Wait maybe the problem is not about area or perimeter. Wait the user's image shows options x=4,5,11,4.5. Wait maybe the correct answer is x=4.5? Wait 66=36, 4*4.5=18. No. Wait maybe the problem is about side lengths. Wait FCDE has sides 6, JGHI has sides 4 and x, and maybe the sum of sides? No. Wait I think I made a mistake. Wait let's re-express. The first quadrilateral is a square with side 6. The second is a rectangle with length 4 and width x. If they are similar (since square is a rectangle with equal sides), then the ratio of length to width should be equal. So for the square, length/width = 6/6 = 1. For the rectangle, 4/x = 1 → x=4. Ah! That makes sense. So if they are similar (both rectangles, square is a rectangle with l/w=1), then the rectangle must also have l/w=1, so x=4.

Answer:

x = 4 (the option with x=4)