The Pythagorean Theorem: Additional Practice (Set 2)

Add the squares of the legs.

$c^2 = 36 + 36 = 72$

Since $72 = 36 \times 2$, $c = 6\sqrt{2}$. The $72$ is $c^2$, and $12$ adds the legs.

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

$b^2 = 1681 - 81 = 1600$

So $b = 40$. Adding the squares gives $\sqrt{1762}$, which is longer than the hypotenuse.

The diagonal is the hypotenuse of a right triangle with two equal legs $s$, so $s^2 + s^2 = 64$.

$s^2 = 64 \div 2 = 32$

So $s = \sqrt{16 \times 2} = 4\sqrt{2}$ cm. Halving the diagonal to get $4$ would give a diagonal of only $4\sqrt{2}$.

The width is a leg of the right triangle that has the diagonal as its hypotenuse.

$w^2 = 625 - 576 = 49$

So $w = 7$ cm, and the perimeter is $2 \times (24 + 7) = 62$ cm. The $168$ is the area, and $98$ uses the diagonal as a side.

First find the square of the hypotenuse.

$c^2 = 1 + 4 = 5$

Then trap $5$ between perfect squares: $4 < 5 < 9$, so $2 < \sqrt{5} < 3$.

The ladder is the hypotenuse, so subtract the square of the height.

$d^2 = 25 - 23.04 = 1.96$

Since $1.4 \times 1.4 = 1.96$, the foot is $1.4$ m from the wall. The $1.96$ is $d^2$, and $0.2$ m subtracts the lengths.

Square each leg: $(\sqrt{5})^2 = 5$ and $2^2 = 4$.

$c^2 = 5 + 4 = 9$

So $c = 3$. Subtracting the squares gives $1$, and $9$ is $c^2$.

For a leg, subtract rather than add.

$b^2 = 400 - 144 = 256$

So $b = 16$. Mia's $\sqrt{544}$ is longer than the hypotenuse, which a leg can never be.

The diagonal is the hypotenuse of a right triangle with the pool's sides as legs.

$d^2 = 625 + 144 = 769$

Since $27.7^2 = 767.29$ is less than $769$ and the halfway value gives $27.75^2 = 770.0625$, which is more, $\sqrt{769}$ lies between $27.7$ and $27.75$, so it is about $27.7$ m. The choice about $21.9$ m comes from subtracting.

The square on the hypotenuse matches the two squares on the legs together.

$49 + 64 = 113$

So its area is $113$ square units. The $225$ is $(7 + 8)^2$, which squares the sum instead of adding the squares.

With a right angle at the corner, the gap is the hypotenuse.

$2.25 + 4 = 6.25$

Since $2.5 \times 2.5 = 6.25$, the marks should be $2.5$ m apart. Any other reading means the corner is not square.

The height cuts the triangle into two right triangles, each with a hypotenuse of $10$ cm and a leg of $6$ cm.

$h^2 = 100 - 36 = 64$

So $h = 8$ cm, and the area is $\tfrac12 \times 12 \times 8 = 48 \text{ cm}^2$. Using the slanted side as the height gives $60$.