The Pythagorean Theorem: Additional Practice (Set 1)

Square the decimals and add.

$c^2 = 0.36 + 0.64 = 1$

So $c = \sqrt{1} = 1$ m. The $1.4$ m adds the legs instead of their squares.

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

$b^2 = 225 - 81 = 144$

So $b = 12$. The choice $3\sqrt{34} = \sqrt{306}$ adds the squares instead.

Add the squares of the legs.

$c^2 = 100 + 100 = 200$

Since $200 = 100 \times 2$, $c = 10\sqrt{2}$ cm, about $14.1$ cm. The $200$ is $c^2$.

The string is the hypotenuse, and the height is at a right angle to the level ground.

$h^2 = 2500 - 900 = 1600$

So $h = 40$ m. Adding the squares gives about $58.3$ m, longer than the string itself.

The diagonal is the hypotenuse of a right triangle whose legs are the width and the height.

$h^2 = 2500 - 1600 = 900$

So the height is $30$ inches. Subtracting the lengths themselves gives $10$, which is not how the theorem works.

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

$c^2 = 9 + 36 = 45$

Since $45 = 9 \times 5$, $c = 3\sqrt{5}$. The choice $3\sqrt{3} = \sqrt{27}$ comes from subtracting the squares.

Square the hypotenuse: $(2\sqrt{10})^2 = 4 \times 10 = 40$. Then subtract the square of the leg.

$b^2 = 40 - 4 = 36$

So $b = 6$. Adding gives $\sqrt{44} = 2\sqrt{11}$, and subtracting the sides is not the theorem.

Square each leg. Since $(5\sqrt{3})^2 = 25 \times 3 = 75$,

$c^2 = 25 + 75 = 100$

So $c = 10$. The choice $5\sqrt{2} = \sqrt{50}$ comes from subtracting.

Equal sides are not enough: a squashed shape can have four equal sides. The corners must be right angles too. At each mark, the two triangles bring their two acute angles, and these add to $90^\circ$.

$180^\circ - 90^\circ = 90^\circ$

So every corner is a right angle. Saying the area is $c^2$ assumes the very thing being shown.

The gate's corners are right angles, so the brace is the hypotenuse.

$d^2 = 1.44 + 0.81 = 2.25$

Since $1.5 \times 1.5 = 2.25$, the brace is $1.5$ m. The $2.25$ is $d^2$, and $2.1$ m adds the sides.

The two parts of her walk meet at a right angle, so the straight path back is the hypotenuse.

$d^2 = 81 + 144 = 225$

So $d = 15$ km. The $21$ km is the distance she walked, not the distance back.

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

$b^2 = 6.25 - 2.25 = 4$

So $b = 2$ cm. Adding the squares gives about $2.9$ cm, longer than the hypotenuse, so it cannot be a leg.