Answer :
To find the length and width of a rectangle where the area is 210 square meters and the length and width are consecutive integers, let's follow these steps:
1. Define the variables:
- Let [tex]\( w \)[/tex] be the width of the rectangle.
- Since the length is a consecutive integer, it will be [tex]\( w + 1 \)[/tex].
2. Set up the equation for the area:
- The area of a rectangle is given by the product of its length and width.
- Therefore, the equation for the area is: [tex]\( w \cdot (w + 1) = 210 \)[/tex].
3. Simplify and solve the equation:
- Expand the equation: [tex]\( w^2 + w = 210 \)[/tex].
- Rewrite it as a standard quadratic equation: [tex]\( w^2 + w - 210 = 0 \)[/tex].
4. Solve the quadratic equation:
- We can solve this quadratic equation using the quadratic formula: [tex]\( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( ax^2 + bx + c = 0 \)[/tex].
- Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -210 \)[/tex].
Plugging these values into the quadratic formula:
[tex]\[ w = \frac{-1 \pm \sqrt{1 + 840}}{2} \\ w = \frac{-1 \pm \sqrt{841}}{2} \\ w = \frac{-1 \pm 29}{2} \][/tex]
5. Calculate the possible values for [tex]\( w \)[/tex]:
- [tex]\( w = \frac{28}{2} = 14 \)[/tex]
- [tex]\( w = \frac{-30}{2} = -15 \)[/tex] (This value is not possible as width cannot be negative)
Therefore, [tex]\( w = 14 \)[/tex].
6. Find the length:
- The length is [tex]\( w + 1 \)[/tex], so: [tex]\( 14 + 1 = 15 \)[/tex].
Thus, the width of the rectangle is 14 meters, and the length of the rectangle is 15 meters.
Final Answer:
- Width = 14 meters
- Length = 15 meters
1. Define the variables:
- Let [tex]\( w \)[/tex] be the width of the rectangle.
- Since the length is a consecutive integer, it will be [tex]\( w + 1 \)[/tex].
2. Set up the equation for the area:
- The area of a rectangle is given by the product of its length and width.
- Therefore, the equation for the area is: [tex]\( w \cdot (w + 1) = 210 \)[/tex].
3. Simplify and solve the equation:
- Expand the equation: [tex]\( w^2 + w = 210 \)[/tex].
- Rewrite it as a standard quadratic equation: [tex]\( w^2 + w - 210 = 0 \)[/tex].
4. Solve the quadratic equation:
- We can solve this quadratic equation using the quadratic formula: [tex]\( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( ax^2 + bx + c = 0 \)[/tex].
- Here, [tex]\( a = 1 \)[/tex], [tex]\( b = 1 \)[/tex], and [tex]\( c = -210 \)[/tex].
Plugging these values into the quadratic formula:
[tex]\[ w = \frac{-1 \pm \sqrt{1 + 840}}{2} \\ w = \frac{-1 \pm \sqrt{841}}{2} \\ w = \frac{-1 \pm 29}{2} \][/tex]
5. Calculate the possible values for [tex]\( w \)[/tex]:
- [tex]\( w = \frac{28}{2} = 14 \)[/tex]
- [tex]\( w = \frac{-30}{2} = -15 \)[/tex] (This value is not possible as width cannot be negative)
Therefore, [tex]\( w = 14 \)[/tex].
6. Find the length:
- The length is [tex]\( w + 1 \)[/tex], so: [tex]\( 14 + 1 = 15 \)[/tex].
Thus, the width of the rectangle is 14 meters, and the length of the rectangle is 15 meters.
Final Answer:
- Width = 14 meters
- Length = 15 meters