Question

the length of a rectangle is 10 yd less than three times the width, and the area of the rectangle is 77 yd². find the dimensions of the rectangle.
length : \boxed{} yd
width : \boxed{} yd

Explanation:

Step1: Define variables

Let the width of the rectangle be \( w \) yards. Then the length \( l \) is \( 3w - 10 \) yards (since length is 10 yd less than three times the width).

Step2: Use area formula

The area of a rectangle is \( A = l \times w \). We know \( A = 77 \) yd², so substitute \( l = 3w - 10 \) into the area formula:
\[
(3w - 10)w = 77
\]
Expand the left side:
\[
3w^2 - 10w = 77
\]
Rearrange into standard quadratic form:
\[
3w^2 - 10w - 77 = 0
\]

Step3: Solve the quadratic equation

We can solve \( 3w^2 - 10w - 77 = 0 \) using the quadratic formula \( w=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \), where \( a = 3 \), \( b = -10 \), and \( c = -77 \).

First, calculate the discriminant \( D = b^2 - 4ac \):
\[
D=(-10)^2-4\times3\times(-77)=100 + 924 = 1024
\]
Then, find \( w \):
\[
w=\frac{10\pm\sqrt{1024}}{2\times3}=\frac{10\pm32}{6}
\]
We have two solutions:

• \( w=\frac{10 + 32}{6}=\frac{42}{6}=7 \)
• \( w=\frac{10 - 32}{6}=\frac{-22}{6}=-\frac{11}{3} \) (discard since width can't be negative)

Step4: Find the length

Now that \( w = 7 \), substitute into the length formula \( l = 3w - 10 \):
\[
l=3\times7 - 10=21 - 10 = 11
\]

Answer:

Length: \( 11 \) yd, Width: \( 7 \) yd