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The three sides of a right-angled triangle have lengths $a$, $b$ and $c$, where $a$, $b$, $c \in \mathbb{Z}$ - AQA - A-Level Maths: Pure - Question 6 - 2019 - Paper 3

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The three sides of a right-angled triangle have lengths $a$, $b$ and $c$, where $a$, $b$, $c \in \mathbb{Z}$. 6 (a) State an example where $a$, $b$ and $c$ are all ... show full transcript

Worked Solution & Example Answer:The three sides of a right-angled triangle have lengths $a$, $b$ and $c$, where $a$, $b$, $c \in \mathbb{Z}$ - AQA - A-Level Maths: Pure - Question 6 - 2019 - Paper 3

Step 1

State an example where $a$, $b$ and $c$ are all even.

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Answer

An example of a right-angled triangle where all sides aa, bb, and cc are even can be:

  • a=6a = 6,
  • b=8b = 8,
  • c=10c = 10.

This set of lengths satisfies the Pythagorean theorem since 62+82=36+64=100=1026^2 + 8^2 = 36 + 64 = 100 = 10^2.

Step 2

Prove that it is not possible for all $a$, $b$ and $c$ to be odd.

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Answer

Assume that aa and bb are both odd. We can express them as:

  • a=2m+1a = 2m + 1,
  • b=2n+1b = 2n + 1

for some integers mm and nn. According to the Pythagorean theorem, we have:

c2=a2+b2c^2 = a^2 + b^2

Substituting aa and bb gives:

c2=(2m+1)2+(2n+1)2c^2 = (2m + 1)^2 + (2n + 1)^2

Expanding this results in:

c2=(4m2+4m+1)+(4n2+4n+1)c^2 = (4m^2 + 4m + 1) + (4n^2 + 4n + 1)

which simplifies to:

c2=4m2+4m+4n2+4n+2c^2 = 4m^2 + 4m + 4n^2 + 4n + 2

This can be rewritten as:

c2=2(2m2+2m+2n2+2n+1)c^2 = 2(2m^2 + 2m + 2n^2 + 2n + 1)

Since c2c^2 is even, it follows that cc must be even (as the square of an odd number is odd). This leads to a contradiction, as we started with the assumption that aa, bb, and cc are all odd. Thus, it is not possible for all of aa, bb, and cc to be odd.

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