Pythagoras’ Theorem

Pythagoras relates the squares of the sides: the square of the hypotenuse equals the sum of the squares of the two shorter sides, $c^2 = a^2 + b^2$. ($c^2 = (a+b)^2$ squares the sum: $(5+12)^2 = 289$, but $5^2 + 12^2 = 169$.)

The hypotenuse is the side opposite the right angle, and it is always the longest side. Here that is the 10 cm side. (The 6 cm and 8 cm sides form the right angle.)

$c^2 = 5^2 + 12^2 = 25 + 144 = 169$. (17 adds the sides; 289 squares the sum $(5+12)^2$.)

Test with $c^2 = a^2 + b^2$ using the longest side as $c$. For $5, 12, 13$: $5^2 + 12^2 = 25 + 144 = 169 = 13^2$, so it is right-angled. ($6, 8, 9$ gives $100 \ne 81$; the triple is $6, 8, 10$.)

$c^2 = 3^2 + 4^2 = 9 + 16 = 25$, so $c = \sqrt{25} = 5$ cm. This is the $(3, 4, 5)$ triple. (25 forgets the square root; 7 adds the sides.)

On the map the straight line is the hypotenuse: $\sqrt{3^2 + 4^2} = \sqrt{25} = 5$ cm. Each cm is 100 m, so the real distance is $5 \times 100 = 500$ m. (Adding the legs, $(3+4)\times 100 = 700$ m, gives the walking distance, not the straight line.)

First find the other leg with Pythagoras: $b^2 = 13^2 - 5^2 = 169 - 25 = 144$, so $b = 12$ cm. The two legs are the base and height, so area $= \tfrac{1}{2}\times 5 \times 12 = 30$ cm². (Forgetting the $\tfrac{1}{2}$ gives 60; using the hypotenuse as a side gives 32.5.)

$c^2 = 5^2 + 7^2 = 25 + 49 = 74$, so $c = \sqrt{74}$. Since $74 = 2 \times 37$ has no perfect-square factor, it does not simplify (about 8.60). ($\sqrt{24}$ subtracts inside the root.)

$c^2 = 6^2 + 6^2 = 72$, so $c = \sqrt{72}$. Since $72 = 36 \times 2$, $\sqrt{72} = \sqrt{36}\times\sqrt{2} = 6\sqrt{2}$ cm. ($\sqrt{72}$ is correct but not simplest form; 8.49 is the decimal.)

Finding a shorter side, so subtract: $a^2 = 13^2 - 5^2 = 169 - 25 = 144$, giving $a = \sqrt{144} = 12$ cm. $(5, 12, 13)$ triple. (8 subtracts the lengths $13 - 5$ instead of the squares.)

Drop a perpendicular to the base; it splits the base into two 7 cm halves. The height satisfies $h^2 = 25^2 - 7^2 = 625 - 49 = 576$, so $h = 24$ cm. Area $= \tfrac{1}{2}\times 14 \times 24 = 168$ cm². (Using the slant side 25 as the height gives 175.)

$a^2 = 25^2 - 7^2 = 625 - 49 = 576$, so $a = \sqrt{576} = 24$ cm. $(7, 24, 25)$ triple.

The diagonals cut each other in half, forming a right-angled triangle with legs $\tfrac{12}{2} = 6$ cm and $\tfrac{16}{2} = 8$ cm. Each side is $\sqrt{6^2 + 8^2} = \sqrt{100} = 10$ cm. (Using the full diagonals gives $\sqrt{12^2 + 16^2} = 20$, which is a diagonal, not a side.)

$a^2 = 8^2 - 5^2 = 64 - 25 = 39$. Since 39 is not a perfect square, $a = \sqrt{39}$ cm (about 6.24). ($\sqrt{89}$ adds inside the root; 3 subtracts the lengths.)

The ladder is the hypotenuse: $h^2 = 5^2 - 3^2 = 25 - 9 = 16$, so $h = \sqrt{16} = 4$ m. $(3, 4, 5)$ triple.

The diagonal is the hypotenuse: $d^2 = 8^2 + 6^2 = 64 + 36 = 100$, so $d = 10$ cm. (48 is the area $8 \times 6$.)

Horizontal gap $4 - 1 = 3$, vertical gap $6 - 2 = 4$. Distance: $d^2 = 3^2 + 4^2 = 25$, so $d = 5$ units. (7 is the L-shaped walking distance.)

$d^2 = 48^2 + 27^2 = 2304 + 729 = 3033$, so $d = \sqrt{3033} \approx 55$ cm. (1296 is the area $48 \times 27$.)

The diagonal is the hypotenuse of a right triangle with two equal legs of 6: $d^2 = 6^2 + 6^2 = 72$, so $d = \sqrt{72} \approx 8.49$ cm.

By the converse, $10^2 + 24^2 = 100 + 576 = 676$ and $26^2 = 676$. They are equal, so yes, it is right-angled. $(10, 24, 26)$ is $(5, 12, 13)$ doubled. (Adding $10 + 24$ tests the wrong thing.)

By the converse, $4^2 + 5^2 = 16 + 25 = 41$ but $6^2 = 36$. Since $41 \ne 36$, it is not right-angled.

First the hypotenuse: $c^2 = 9^2 + 12^2 = 225$, so $c = 15$ cm. Perimeter $= 9 + 12 + 15 = 36$ cm. (21 forgets the hypotenuse; 54 is the area.)

The walks are the two shorter sides: $c^2 = 9^2 + 12^2 = 81 + 144 = 225$, so $c = \sqrt{225} = 15$ m. (21 is the total distance walked.)

With side $s$: $s^2 + s^2 = 10^2$, so $2s^2 = 100$, $s^2 = 50$ and $s = \sqrt{50} \approx 7$ cm. (5 just halves the diagonal.)

Let the sides be $3x$ and $4x$: $(3x)^2 + (4x)^2 = 25^2$, so $9x^2 + 16x^2 = 625$, giving $25x^2 = 625$, $x^2 = 25$ and $x = 5$. The shortest side is $3x = 15$ cm. (Check: $15^2 + 20^2 = 625$.)