Understanding the Rectangle Problem

This problem involves a rectangle where the relationship between its sides, area, and perimeter is given. We need to find the length of the longer side.

Setting Up the Equations

Let's define the variables:

• Let the length of the longer side be $L$ cm.
• Let the length of the shorter side be $S$ cm.

From the problem statement, we know:

• The shorter side is 15 cm less than the longer side: $S = L - 15$.
• The area ($A$) of the rectangle is $A = L \times S$. Substituting $S$, we get $A = L(L - 15)$.
• The perimeter ($P$) of the rectangle is $P = 2(L + S)$. Substituting $S$, we get $P = 2(L + (L - 15)) = 2(2L - 15)$.
• The numerical value of the area is equal to 5 times the numerical value of the perimeter: $A = 5P$.

Solving the Algebraic Equation

Now, we substitute the expressions for $A$ and $P$ into the equation $A = 5P$:

$L(L - 15) = 5 \times [2(2L - 15)]$

Simplify the equation:

$L^2 - 15L = 10(2L - 15)$

$L^2 - 15L = 20L - 150$

Rearrange the terms to form a standard quadratic equation ($ax^2 + bx + c = 0$):

$L^2 - 15L - 20L + 150 = 0$

$L^2 - 35L + 150 = 0$

Finding the Roots of the Quadratic Equation

We need to solve the quadratic equation $L^2 - 35L + 150 = 0$. We can do this by factoring. We are looking for two numbers that multiply to 150 and add up to -35.

The numbers are -5 and -30, since $(-5) \times (-30) = 150$ and $(-5) + (-30) = -35$.

So, we can factor the equation as:

$(L - 5)(L - 30) = 0$

This gives two possible solutions for $L$:

• $L - 5 = 0 \implies L = 5$
• $L - 30 = 0 \implies L = 30$

Validating the Solutions

We must check if these values for $L$ are valid in the context of the problem.

• If $L = 5$ cm:
  The shorter side would be $S = L - 15 = 5 - 15 = -10$ cm. Since a side length cannot be negative, this solution is not valid.
• If $L = 30$ cm:
  The longer side is $L = 30$ cm.
  The shorter side is $S = L - 15 = 30 - 15 = 15$ cm. Both dimensions are positive, which is physically possible.
  
  Let's verify the area and perimeter condition:
  Area $A = L \times S = 30 \times 15 = 450$ cm$^2$.
  Perimeter $P = 2(L + S) = 2(30 + 15) = 2(45) = 90$ cm.
  Check if $A = 5P$: $450 = 5 \times 90$, which is $450 = 450$. This condition is satisfied.

Conclusion

The value $L = 30$ cm is the only valid solution. Therefore, the length of the longer side of the rectangle is 30 cm.