To find the length and width of the playground, let's use algebra. Let the length of the playground be \( L \) yards. According to the problem, the width is 5 yards longer than the length, so we can express the width as \( L + 5 \) yards.
The area of the playground is given as 336 square yards, which can be expressed using the formula for the area of a rectangle:
\[ \text{Area} = \text{Length} \times \text{Width} \]
Substitute the given values into the equation:
\[ L \times (L + 5) = 336 \]
Now we have a quadratic equation to solve:
\[ L^2 + 5L - 336 = 0 \]
To solve this quadratic equation, we can use the quadratic formula where \( a = 1 \), \( b = 5 \), and \( c = -336 \):
\[ L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Substitute the values into the formula:
\[ L = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot (-336)}}{2 \cdot 1} \]
\[ L = \frac{-5 \pm \sqrt{25 + 1344}}{2} \]
\[ L = \frac{-5 \pm \sqrt{1369}}{2} \]
\[ L = \frac{-5 \pm 37}{2} \]
This gives us two possible solutions for \( L \):
\[ L = \frac{-5 + 37}{2} = \frac{32}{2} = 16 \]
\[ L = \frac{-5 - 37}{2} = \frac{-42}{2} = -21 \]
Since a length cannot be negative, we discard \( L = -21 \). So, the length \( L \) is 16 yards. The width is:
\[ L + 5 = 16 + 5 = 21 \text{ yards} \]
Thus, the correct dimensions of the playground are:
\[ \text{Length} = 16 \text{ yards}, \text{Width} = 21 \text{ yards} \]
So the correct answer is:
\[ \text{Length} = 16 \text{ yd}, \text{Width} = 21 \text{ yd} \]
18.
The area of a playground is 336 yd2. The width of the playground is 5 yd longer than its length. Find the length and width of the playground.
length = 21 yd, width = 16 yd
length = 21 yd, width = 26 yd
length = 16 yd, width = 21 yd
length = 26 yd, width = 21 yd
1 answer