The Pythagorean Theorem: Practice

Call each leg $a$. The theorem gives $a^2 + a^2 = 100$, so two equal squares make $100$.

$a^2 = 100 \div 2 = 50$

So $a = \sqrt{25 \times 2} = 5\sqrt{2}$. Halving the hypotenuse to get $5$ would make $5^2 + 5^2 = 50$, not $100$.

The diagonal and two sides make a right triangle with the diagonal as hypotenuse. Find the length first.

$l^2 = 289 - 64 = 225$

So $l = 15$ cm and the area is $8 \times 15 = 120 \text{ cm}^2$. Multiplying by the diagonal gives $136$, and $46$ is the perimeter in centimeters.

The ramp is the hypotenuse of a right triangle with legs $1$ m and $2.4$ m.

$c^2 = 1 + 5.76 = 6.76$

Since $2.6 \times 2.6 = 6.76$, the ramp is $2.6$ m long. The $6.76$ is $c^2$, and about $2.2$ m comes from subtracting.

First find the shared side.

$6^2 + 8^2 = 100$

So it is $10$. Then use $10$ and $24$ as the legs of the second triangle.

$10^2 + 24^2 = 676$

Since $26 \times 26 = 676$, the answer is $26$. The choice $34$ adds $10$ and $24$ instead of their squares.

The diagonal is the hypotenuse of a right triangle with legs $40$ and $40$.

$d^2 = 1600 + 1600 = 3200$

So $d = 40\sqrt{2}$, about $56.6$ m. Two sides make $80$ m, and $80 - 56.6$ is about $23$ m. The choice about $57$ m is the diagonal itself.

Square each leg. Since $(2\sqrt{3})^2 = 4 \times 3 = 12$,

$c^2 = 12 + 4 = 16$

So $c = 4$. Subtracting gives $\sqrt{8} = 2\sqrt{2}$, and adding the legs gives $2 + 2\sqrt{3}$.

The square of $\sqrt{50}$ is $50$, so subtract the square of the known leg.

$b^2 = 50 - 25 = 25$

So $b = 5$, and the two legs are equal. Adding the squares gives $\sqrt{75} = 5\sqrt{3}$.

Find the hypotenuse first.

$c^2 = 81 + 1600 = 1681$

Since $41 \times 41 = 1681$, the hypotenuse is $41$ cm. The perimeter adds all three sides: $9 + 40 + 41 = 90$ cm. The $180$ is the area, in square centimeters.

The hypotenuse is always the longest side of a right triangle, so a leg of $6$ cannot sit with a hypotenuse of $5$. The subtraction shows the same thing.

$5^2 - 6^2 = 25 - 36 = -11$

No length has a negative square, so no right triangle has those sides.

If the corner were a right angle, the gap would be the hypotenuse.

$60^2 + 80^2 = 10{,}000$

So a right angle would put the marks exactly $\sqrt{10{,}000} = 100$ cm apart. The gap is $98$ cm, so the corner is not a right angle.

The height cuts the triangle into two right triangles. Each has a hypotenuse of $13$ cm and a leg of half the base, $5$ cm.

$h^2 = 169 - 25 = 144$

So $h = 12$ cm. The choice $\sqrt{269}$ makes two mistakes at once: it uses the whole base of $10$ and adds the squares, $13^2 + 10^2 = 269$. Subtracting the sides, $13 - 10$, gives $3$.

The two squares on the legs together match the square on the hypotenuse, so the total is two copies of the hypotenuse's square.

$2c^2 = 200, \text{ so } c^2 = 100$

So $c = 10$ cm. Taking $\sqrt{200} = 10\sqrt{2}$ treats the whole total as one square.