Challenging +1.2 This is a proof by contradiction requiring students to apply a given result about rectangles with fixed area to prove the converse statement about rectangles with fixed perimeter. While it requires understanding the logical structure of proof by contradiction and careful manipulation of the constraint (perimeter = 4L), the question explicitly provides the key fact needed and guides students to the method. The algebraic manipulation is straightforward once the contradiction setup is established, making this moderately above average difficulty but not requiring deep geometric insight.
15 It is given in lines \(31 - 32\) that the square has the smallest perimeter of all rectangles with the same area. Using this fact, prove by contradiction that among rectangles of a given perimeter, \(4 L\), the square with side \(L\) has the largest area.
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Suppose that for the given perimeter \(4L\), there is a rectangle which is larger in area than the square.
M1
Setting up a statement for contradiction
There is a square which has the same area as this rectangle but a smaller perimeter so its side is less than \(L\).
A1
Use of statement in line 31–32
The square with side \(L\) has perimeter \(4L\) and an area larger than the given rectangle. This is a contradiction so the square must have the largest area of all rectangles with given perimeter.
A1
Completion to correct conclusion, including contradiction
## Question 15:
| Answer | Mark | Guidance |
|--------|------|----------|
| Suppose that for the given perimeter $4L$, there is a rectangle which is larger in area than the square. | M1 | Setting up a statement for contradiction |
| There is a square which has the same area as this rectangle but a smaller perimeter so its side is less than $L$. | A1 | Use of statement in line 31–32 |
| The square with side $L$ has perimeter $4L$ and an area larger than the given rectangle. This is a contradiction so the square must have the largest area of all rectangles with given perimeter. | A1 | Completion to correct conclusion, including contradiction |
15 It is given in lines $31 - 32$ that the square has the smallest perimeter of all rectangles with the same area. Using this fact, prove by contradiction that among rectangles of a given perimeter, $4 L$, the square with side $L$ has the largest area.
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\hfill \mbox{\textit{OCR MEI Paper 3 2018 Q15 [3]}}