The area of a playground is { 221 yd^2 }, The width of the playground is 4 yd longer than its...
Question:
The area of a playground is {eq}221 yd^2 {/eq}, The width of the playground is 4 yds longer than its length. Find the length and width of the playground?
Rectangle and its properties
A Rectangle is a closed 2-dimensional planner 4-sided geometry which has a pair of equal and parallel sides.
And one of the sides is Rectangles Length(L) and another is Width(W).
All the angles of a rectangle are at the right angle {eq}\displaystyle 90^ \circ {/eq}.
Perimeter of the Rectangle
{eq}\displaystyle P = 2 \times (Length + Width) = 2 \times (L + W) {/eq}.
Area of the Rectangle
{eq}\displaystyle A = Length \times Width = L \times W {/eq}.
Answer and Explanation: 1
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View this answerAccording to the question, The playground is Rectangular because it has Length as well as Width.
And the Width of the playground is 4 yds longer than...
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