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We are interested in the dimensions of a certain square. A rectangle has a length 5 units more than the side of this square and a width half the side of this square. If the two areas are equal, what are the square's dimensions?

(Note: Let the side of the square be denoted as [tex]\( s \)[/tex].)


Sagot :

Let's solve this problem step-by-step:

1. Define Variables:
- Let [tex]\( s \)[/tex] represent the side length of the square.

2. Area of the Square:
- The area of the square is given by:
[tex]\[ \text{Area of Square} = s^2 \][/tex]

3. Dimensions of the Rectangle:
- The length of the rectangle is 5 units more than the side of the square, which can be expressed as [tex]\( s + 5 \)[/tex].
- The width of the rectangle is half the side of the square, which can be expressed as [tex]\( \frac{s}{2} \)[/tex].

4. Area of the Rectangle:
- The area of the rectangle is given by:
[tex]\[ \text{Area of Rectangle} = \text{Length} \times \text{Width} = (s + 5) \times \frac{s}{2} \][/tex]

5. Set Up the Equation:
- We are given that the area of the square is equal to the area of the rectangle. Therefore, we equate the two areas:
[tex]\[ s^2 = (s + 5) \times \frac{s}{2} \][/tex]

6. Simplify the Equation:
- Simplifying the right side of the equation:
[tex]\[ s^2 = s \times \frac{s + 5}{2} = \frac{s(s + 5)}{2} \][/tex]
- Multiply both sides by 2 to clear the fraction:
[tex]\[ 2s^2 = s(s + 5) \][/tex]

7. Expand and Solve the Equation:
- Expand the right side:
[tex]\[ 2s^2 = s^2 + 5s \][/tex]
- Bring all terms to one side of the equation to set it to zero:
[tex]\[ 2s^2 - s^2 - 5s = 0 \implies s^2 - 5s = 0 \][/tex]

8. Factor the Equation:
- Factor the quadratic equation:
[tex]\[ s(s - 5) = 0 \][/tex]

9. Solve for [tex]\( s \)[/tex]:
- Set each factor to zero:
[tex]\[ s = 0 \quad \text{or} \quad s - 5 = 0 \implies s = 5 \][/tex]
- Therefore, the solutions are [tex]\( s = 0 \)[/tex] and [tex]\( s = 5 \)[/tex].

10. Interpret the Solutions:
- The side length [tex]\( s = 0 \)[/tex] is not meaningful in this context because it would imply no square.
- The meaningful solution is [tex]\( s = 5 \)[/tex].

Therefore, the side length of the square is 5 units. The dimensions of the square are [tex]\( 5 \times 5 \)[/tex].