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question 1 of 4, step 1 of 1 a rectangle has a length of 33 yards less …

Question

question 1 of 4, step 1 of 1 a rectangle has a length of 33 yards less than 6 times its width. if the area of the rectangle is 6195 square yards, find the length of the rectangle.

Explanation:

Step1: Define variables

Let the width of the rectangle be $w$ yards. Then the length $l = 6w - 33$ yards.

Step2: Use area formula

The area of a rectangle $A=l\times w$. Given $A = 6195$ square - yards. So, $(6w - 33)\times w=6195$.

Step3: Expand the equation

$6w^{2}-33w - 6195 = 0$. Divide through by 3 to simplify: $2w^{2}-11w - 2065 = 0$.

Step4: Solve the quadratic equation

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 2$, $b=-11$, $c = - 2065$), the quadratic formula is $w=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-11)^{2}-4\times2\times(-2065)=121 + 16520=16641$.
Then $w=\frac{11\pm\sqrt{16641}}{4}=\frac{11\pm129}{4}$.
We get two solutions for $w$: $w_1=\frac{11 + 129}{4}=\frac{140}{4}=35$ and $w_2=\frac{11-129}{4}=\frac{-118}{4}=-29.5$. Since width cannot be negative, $w = 35$ yards.

Step5: Find the length

Substitute $w = 35$ into the length formula $l=6w - 33$. So, $l=6\times35-33=210 - 33=177$ yards.

Answer:

177