One side of a rectangle is
32cm longer than four times another side. The area of the rectangle is
420cm2. Find the length of the shorter side.

Let's denote the length of the shorter side by \( x \) cm.

According to the problem, the longer side is \( 32 \) cm longer than four times the shorter side. Therefore, the length of the longer side can be expressed as:

\[
4x + 32 \text{ cm}
\]

The area of the rectangle is given as \( 420 \text{ cm}^2 \). The area of a rectangle is calculated by multiplying its length and width. Therefore, we have:

\[
x \times (4x + 32) = 420
\]

Now, let's solve this equation:

\[
x(4x + 32) = 420
\]

Expanding the left side:

\[
4x^2 + 32x = 420
\]

Next, we'll move all terms to one side of the equation:

\[
4x^2 + 32x - 420 = 0
\]

We can simplify this equation by dividing all terms by 4:

\[
x^2 + 8x - 105 = 0
\]

Now, we can solve this quadratic equation using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

For our equation, \( a = 1 \), \( b = 8 \), and \( c = -105 \). Substituting these values into the quadratic formula:

\[
x = \frac{-8 \pm \sqrt{8^2 - 4(1)(-105)}}{2(1)}
\]

Simplifying under the square root:

\[
x = \frac{-8 \pm \sqrt{64 + 420}}{2}
\]

\[
x = \frac{-8 \pm \sqrt{484}}{2}
\]

\[
x = \frac{-8 \pm 22}{2}
\]

This gives us two possible solutions:

\[
x = \frac{-8 + 22}{2} = \frac{14}{2} = 7 \text{ cm}
\]
or
\[
x = \frac{-8 - 22}{2} = \frac{-30}{2} = -15 \text{ cm}
\]

Since the length cannot be negative, we discard the negative value, so:

\[
x = 7 \text{ cm}
\]

Thus, the length of the shorter side is **7 cm**.

Would you like more details or have any questions?

Here are 5 related questions you might find useful:
1. What is the length of the longer side of the rectangle?
2. How do you verify the solution by recalculating the area with the found dimensions?
3. What would be the new area if both sides of the rectangle were doubled?
4. Can you find the perimeter of the rectangle using the given information?
5. How does the quadratic formula help in solving different types of problems?

**Tip:** In word problems involving geometry, always start by assigning variables to the unknown quantities and translating the problem into equations based on known formulas.

Discover how to find the length of the shorter side of a rectangle when its area is 420cm². This problem involves geometry and quadratic equations, exploring the relationship between the sides of a rectangle and its area. Suitable for students in Grades 9-12.

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