Geometry — Pythagorean Theorem

1. Using the Pythagorean theorem $a^2 + b^2 = c^2$:

$5^2 + 12^2 = c^2$ $25 + 144 = c^2$ $169 = c^2$ $c = \sqrt{169} = 13$

The hypotenuse is $13$ cm. (This is the 5-12-13 Pythagorean triple.)

2. The wall, ground, and ladder form a right triangle. The ladder is the hypotenuse ($c = 10$) and the ground distance is one leg ($a = 6$). We need the other leg $b$ (height on the wall):

$a^2 + b^2 = c^2$ $6^2 + b^2 = 10^2$ $36 + b^2 = 100$ $b^2 = 64$ $b = \sqrt{64} = 8$

The ladder reaches $8$ feet up the wall. (This is the 6-8-10 triple, a multiple of 3-4-5.)

3. Using the distance formula (derived from the Pythagorean theorem):

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ $d = \sqrt{(4 - 1)^2 + (6 - 2)^2}$ $d = \sqrt{3^2 + 4^2}$ $d = \sqrt{9 + 16}$ $d = \sqrt{25} = 5$

The distance between $A$ and $B$ is $5$ units. (This is the 3-4-5 Pythagorean triple.)

4. The diagonal of the rectangle is the hypotenuse of a right triangle with legs $24$ m and $7$ m:

$d^2 = 24^2 + 7^2$ $d^2 = 576 + 49$ $d^2 = 625$ $d = \sqrt{625} = 25$

The diagonal path is $25$ meters. (This is the 7-24-25 Pythagorean triple.)

5. First, find the side lengths using the distance formula.

$PQ$: from $(-2, 1)$ to $(6, 1)$ — same $y$-coordinate, so this is horizontal:

$PQ = |6 - (-2)| = 8$

$QR$: from $(6, 1)$ to $(6, 16)$ — same $x$-coordinate, so this is vertical:

$QR = |16 - 1| = 15$

Since $PQ$ is horizontal and $QR$ is vertical, they meet at a right angle at $Q$. So triangle $PQR$ is a right triangle with the right angle at $Q$.

The hypotenuse is $PR$:

$PR = \sqrt{(6 - (-2))^2 + (16 - 1)^2}$ $PR = \sqrt{8^2 + 15^2}$ $PR = \sqrt{64 + 225}$ $PR = \sqrt{289} = 17$

The hypotenuse $PR$ is $17$ units. (This is the 8-15-17 Pythagorean triple.)