Two rectangles have the exact same area.  The measurement of the first rectangle is 3k by 3. And the measurement of the second rectangle is k+3 by 6. Calculate the value of k.

The areas of the two rectangles are equal, so we can start by writing expressions for the areas of both rectangles and setting them equal to each other.

- The area of the first rectangle is:
\[
A_1 = 3k \times 3 = 9k
\]

- The area of the second rectangle is:
\[
A_2 = (k + 3) \times 6 = 6(k + 3)
\]

Since the areas are the same, we set these two expressions equal to each other:
\[
9k = 6(k + 3)
\]

Now, solve for \(k\):

1. Expand the right-hand side:
\[
9k = 6k + 18
\]

2. Subtract \(6k\) from both sides:
\[
9k - 6k = 18
\]
\[
3k = 18
\]

3. Divide both sides by 3:
\[
k = 6
\]

Thus, the value of \(k\) is \(6\).

Would you like more details on any step, or do you have any other questions?

Here are 5 related questions for practice:
1. What is the perimeter of the first rectangle if \(k = 6\)?
2. What is the area of the second rectangle after solving for \(k\)?
3. If the width of the second rectangle is doubled, how would that affect the area?
4. What would the dimensions of the first rectangle be if \(k = 10\)?
5. How would you solve if the areas of the rectangles were not equal?

**Tip:** Always double-check your work by substituting the calculated values back into the original equations!

Two rectangles have the exact same area. The measurement of the first rectangle is 3k by 3. And the measurement of the second rectangle is k+3 by 6. Calculate the value of k.

Learn how to solve a problem where two rectangles have equal areas. The first rectangle measures 3k by 3, and the second rectangle measures k+3 by 6. This problem involves algebraic equations and understanding area formulas. Suitable for students in Grades 6-8.

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