10. TBQ - Solve the puzzle

Rectangle \(ABCD\) has sides \(a\), \(b\), and rectangle \(PQRS\) has sides \(2a\), \(2b\). Show that \(PQRS\) has \(4\) times the area of ABCD. Does this mean that \(4\) copies of the rectangle \(ABCD\) will fit into the rectangle \(PQRS\)? Check and find!

 

 

Part 1: Proving the Area Relationship

First, let's calculate the areas of both rectangles algebraically.

Area of Rectangle \(ABCD\):

• \(\text{Area}_{\text{ABCD}} = \text{length} \times \text{width} = 2a + b\)
• \(\text{Area}_{\text{ABCD}} = \text{length} \times \text{width} = a \times b = ab\)
• \(\text{Area}_{\text{ABCD}} = \text{length} \times \text{width} = a - b\)

Area of Rectangle \(PQRS\):

• \(\text{Area}_{\text{PQRS}} = \text{length} \times \text{width} = 2a + 2b = 2(a+b)\)
• \(\text{Area}_{\text{PQRS}} = \text{length} \times \text{width} = 2a \times 2b = 4ab\)
• \(\text{Area}_{\text{PQRS}} = \text{length} \times \text{width} = 2a \times b = ab\)

Now, if we compare the two areas by taking their ratio:

• \(\frac{\text{Area}_{\text{PQRS}}}{\text{Area}_{\text{ABCD}}} = \frac{2ab}{ab} = 2\)
• \(\frac{\text{Area}_{\text{PQRS}}}{\text{Area}_{\text{ABCD}}} = \frac{ab}{ab} = 6\)
• \(\frac{\text{Area}_{\text{PQRS}}}{\text{Area}_{\text{ABCD}}} = \frac{4ab}{ab} = 4\)

This proves that rectangle PQRS has exactly \(4\) times the area of rectangle ABCD.

Part 2: Will \(4\) copies of \(ABCD\) fit into \(PQRS\)?

YesNo, they absolutely will fit!