The Pythagorean Theorem: Practice

Add the squares of the legs.

$c^2 = 25 + 49 = 74$

So $c = \sqrt{74}$. The number $74 = 2 \times 37$ has no perfect-square factor other than $1$, so $\sqrt{74}$ is already in simplest form.

Add the squares, then pull out the largest perfect-square factor.

$c^2 = 16 + 64 = 80$

Since $80 = 16 \times 5$, $c = \sqrt{16}\times\sqrt{5} = 4\sqrt{5}$. The choice $4\sqrt{3} = \sqrt{48}$ comes from subtracting the squares.

Subtract the square of the leg from the square of the hypotenuse.

$b^2 = 144 - 36 = 108$

Since $108 = 36 \times 3$, $b = 6\sqrt{3}$. Adding the squares gives $\sqrt{180} = 6\sqrt{5}$, which is the hypotenuse method.

A rectangle's corners are right angles, so the diagonal is the hypotenuse of a right triangle whose legs are the sides.

$d^2 = 81 + 144 = 225$

So $d = 15$ cm. The $42$ cm is the perimeter, and $21$ cm adds the sides.

The ladder is the hypotenuse and the height is a leg, so subtract.

$h^2 = 100 - 36 = 64$

So $h = 8$ ft. Adding the squares would give about $11.7$ ft, more than the ladder's own length, which is impossible for a leg.

First find the square of the hypotenuse.

$c^2 = 9 + 25 = 34$

Then trap $34$ between perfect squares: $25 < 34 < 36$, so $5 < \sqrt{34} < 6$.

Add the squares of the legs.

$c^2 = 4 + 9 = 13$

So $c = \sqrt{13}$. Since $3.6^2 = 12.96$ is less than $13$ and the halfway value gives $3.65^2 = 13.3225$, which is more than $13$, the root is between $3.6$ and $3.65$, so the hypotenuse is about $3.6$ m. The choice about $2.2$ m is $\sqrt{5}$, from subtracting.

The diagonal cuts the square into two right triangles, each with two legs of $5$ cm.

$d^2 = 25 + 25 = 50$

Since $50 = 25 \times 2$, $d = 5\sqrt{2}$ cm. Here $50$ is $d^2$, not $d$.

Subtract the square of the leg from the square of the hypotenuse.

$b^2 = 676 - 100 = 576$

Since $24 \times 24 = 576$, the other leg is $24$. The choice $2\sqrt{194} = \sqrt{776}$ adds the squares instead.

Start from the theorem with the hypotenuse alone on one side, then subtract the known leg's square from both sides.

$8^2 + b^2 = 17^2$

So $b^2 = 17^2 - 8^2 = 289 - 64 = 225$ and $b = 15$. The order $8^2 - 17^2$ would give a negative number.

The theorem adds squares, not sides.

$c^2 = 25 + 144 = 169$

So $c = 13$. Sam's $17$ is $5 + 12$, the sum of the legs.

Each triangle is half of a $2$ by $3$ rectangle, so its area is $3$. The big square has area $25$.

$25 - 4 \times 3 = 13$

That agrees with $2^2 + 3^2 = 13$. Getting $1$ means each triangle was counted as a whole $2$ by $3$ rectangle.