Pythagorean theorem — proof (by similar triangles, rearrangement), converse: Exercises

This page is a self-test. Each problem is stated cleanly, then a full worked solution hides inside a collapsible callout. Read the problem, try it yourself on paper, then open the solution. Difficulty climbs from L1 Recognition (just spot the pattern) up to L5 Mastery (chain several ideas together).

Everything here rests on the one identity from the parent note, the Pythagorean theorem:

Before we start, one picture to fix every word — leg, hypotenuse, right angle — in your mind.

Figure — Pythagorean theorem — proof (by similar triangles, rearrangement), converse

Level 1 — Recognition

Here you only need to recognise which rule applies and plug in. No algebra tricks.

Problem 1.1

A right triangle has legs $a = 6$ and $b = 8$ . Find the hypotenuse $c$ .

Recall Solution 1.1

What we want: the longest side $c$ . Why the theorem applies: we are told it is a right triangle, and $6, 8$ are the two legs (the sides at the right angle). $c^{2} = a^{2} + b^{2} = 6^{2} + 8^{2} = 36 + 64 = 100$ Now finish — take the square root (the theorem gives $c^{2}$ , not $c$ ): $c = 100 = 10$ This is the classic $6$ - $8$ - $10$ triangle, a scaled $3$ - $4$ - $5$ . See Pythagorean Triples.

Problem 1.2

A right triangle has hypotenuse $c = 13$ and one leg $b = 5$ . Find the other leg $a$ .

Recall Solution 1.2

Watch out: $13$ is the hypotenuse (given), so it goes on the big side of the equation. We rearrange to isolate the missing leg: $a^{2} = c^{2} - b^{2} = 1 3^{2} - 5^{2} = 169 - 25 = 144$ $a = 144 = 12$ A $5$ - $12$ - $13$ triangle.

Level 2 — Application

Now you apply the theorem inside a small real-world or coordinate setup.

Problem 2.1

Find the distance between the points $P = (1, 2)$ and $Q = (4, 6)$ in the plane.

Recall Solution 2.1

Why Pythagoras? The horizontal gap and the vertical gap between the points meet at a right angle — they are the two legs of a right triangle, and the straight-line distance $P Q$ is the hypotenuse (look at figure s02). $Δ x = 4 - 1 = 3, Δ y = 6 - 2 = 4$ $P Q = (Δ x)^{2} + (Δ y)^{2} = 3^{2} + 4^{2} = 9 + 16 = 25 = 5$ This is the Distance Formula in Coordinate Geometry — it is Pythagoras wearing coordinates.

Problem 2.2

A ladder $10 m$ long leans against a vertical wall, its foot $6 m$ from the base of the wall. How high up the wall does it reach?

Recall Solution 2.2

The right angle is where wall meets ground. So the wall-height $h$ and the ground-distance $6$ are the legs; the ladder $10$ is the hypotenuse. $h^{2} = 1 0^{2} - 6^{2} = 100 - 36 = 64$ $h = 64 = 8 m$

Level 3 — Analysis

Here you must decide whether the theorem applies, or dig out a hidden right triangle.

Problem 3.1

A triangle has sides $9$ , $12$ , and $16$ . Is it a right triangle? If not, is its largest angle acute or obtuse?

Recall Solution 3.1

Use the converse as a test. Take the longest side as the candidate hypotenuse: $c = 16$ . Compare $a^{2} + b^{2}$ against $c^{2}$ : $a^{2} + b^{2} = 9^{2} + 1 2^{2} = 81 + 144 = 225$ $c^{2} = 1 6^{2} = 256$ Since $225 \neq = 256$ , it is not a right triangle. Which way is it off? Because $a^{2} + b^{2} = 225 < 256 = c^{2}$ , the side opposite $c$ is "too long" for a right angle — the angle there has been stretched open past $9 0^{\circ}$ , so the largest angle is obtuse. (This is exactly what the Law of Cosines $c^{2} = a^{2} + b^{2} - 2 ab cos C$ says: $c^{2} > a^{2} + b^{2}$ forces $cos C < 0$ , i.e. $C > 9 0^{\circ}$ .)

Problem 3.2

An isosceles triangle has two equal sides of length $13$ and a base of length $10$ . Find its height (the perpendicular from the apex to the base) and its area.

Recall Solution 3.2

Find the hidden right triangle. The height drops from the apex straight down to the base and, because the triangle is isosceles, it lands exactly at the midpoint of the base. That splits the base into two halves of $5$ each and creates a right triangle with:

hypotenuse $= 13$ (an equal side),
one leg $= 5$ (half the base),
other leg $= h$ (the height we want). $h^{2} = 1 3^{2} - 5^{2} = 169 - 25 = 144 \Rightarrow h = 12$ Area of the whole triangle: $Area = \frac{1}{2} \times base \times height = \frac{1}{2} \times 10 \times 12 = 60$

Level 4 — Synthesis

Combine two or more ideas in one problem.

Problem 4.1

A rectangular box has dimensions $3 \times 4 \times 12$ . Find the length of its longest internal diagonal (corner to opposite corner).

Recall Solution 4.1

Two Pythagoras steps stacked. First find the diagonal $d$ across the base rectangle ( $3 \times 4$ ): $d^{2} = 3^{2} + 4^{2} = 9 + 16 = 25 \Rightarrow d = 5$ That base-diagonal $d$ and the vertical height $12$ meet at a right angle, and their hypotenuse is the space diagonal $D$ : $D^{2} = d^{2} + 1 2^{2} = 25 + 144 = 169 \Rightarrow D = 13$ Combining in one line gives the 3D Distance Formula: $D = 3^{2} + 4^{2} + 1 2^{2} = 169 = 13$

Problem 4.2

Point $A = (2, 1)$ , point $B = (7, 1)$ , and point $C = (7, 13)$ . Show that triangle $A B C$ has a right angle, and state at which vertex.

Recall Solution 4.2

Step 1 — measure all three sides with the Distance Formula in Coordinate Geometry: $A B = (7 - 2)^{2} + (1 - 1)^{2} = 25 = 5$ $B C = (7 - 7)^{2} + (13 - 1)^{2} = 144 = 12$ $A C = (7 - 2)^{2} + (13 - 1)^{2} = 25 + 144 = 169 = 13$ Step 2 — apply the converse. The longest side is $A C = 13$ . Test: $A B^{2} + B C^{2} = 5^{2} + 1 2^{2} = 25 + 144 = 169 = 1 3^{2} = A C^{2} ✓$ Since $a^{2} + b^{2} = c^{2}$ holds, the triangle is right-angled, and the right angle is opposite the longest side $A C$ — that is at vertex $B = (7, 1)$ .

Level 5 — Mastery

One idea, pushed to its limits — including a degenerate/boundary case.

Problem 5.1

Find all integer values of the leg $b$ for which the triangle with legs $b$ and $b + 7$ has an integer hypotenuse, given the hypotenuse is $b + 8$ .

Recall Solution 5.1

Set up the requirement. We demand $b^{2} + (b + 7)^{2} = (b + 8)^{2}$ with $b$ a positive integer. Expand each square carefully: $b^{2} + (b^{2} + 14 b + 49) = b^{2} + 16 b + 64$ $2 b^{2} + 14 b + 49 = b^{2} + 16 b + 64$ Bring everything to one side (subtract the right side): $b^{2} - 2 b - 15 = 0$ Factor: $(b - 5) (b + 3) = 0$ , so $b = 5$ or $b = - 3$ . A side length must be positive, so $b = 5$ is the only value. Check: legs $5$ and $12$ , hypotenuse $13$ — the $5$ - $12$ - $13$ triple. The negative root $b = - 3$ is rejected: lengths cannot be negative.

Problem 5.2 (boundary case)

In a right triangle the two legs are equal, each of length $a$ . Express the hypotenuse in terms of $a$ , and find the exact ratio hypotenuse : leg. Then, as a degenerate check, describe what happens to the triangle if one leg shrinks to $0$ .

Recall Solution 5.2

Equal legs. Put $b = a$ into the theorem: $c^{2} = a^{2} + a^{2} = 2 a^{2} \Rightarrow c = 2 a^{2} = a 2$ So the ratio is $\frac{c}{a} = 2 \approx 1.41421.$ This is the diagonal-to-side ratio of a square, and the shape of a $4 5^{\circ} - 4 5^{\circ} - 9 0^{\circ}$ triangle — see Trigonometric Ratios. Degenerate case ( $b \to 0$ ): let one leg shrink toward $0$ while the other stays $a$ . Then $c = a^{2} + 0^{2} = a^{2} = a .$ The hypotenuse collapses onto the remaining leg: the "triangle" flattens into a single line segment of length $a$ . The theorem still gives a sensible answer at the boundary — it degrades gracefully to $c = a$ . This is why $a^{2} = ∣ a ∣$ (the magnitude), the same non-negativity idea behind Vectors and Magnitude and Complex Numbers.

Recall Quick self-check ladder

L1 answer for $6, 8$ ::: $c = 10$ L1 answer for hypotenuse $13$ , leg $5$ ::: other leg $= 12$ L2 distance from $(1, 2)$ to $(4, 6)$ ::: $5$ L2 ladder height ::: $8$ m L3 triangle $9, 12, 16$ ::: not right; largest angle is obtuse L3 isosceles $13, 13, 10$ height and area ::: $h = 12$ , area $= 60$ L4 box $3, 4, 12$ space diagonal ::: $13$ L4 right-angle vertex of $A (2, 1), B (7, 1), C (7, 13)$ ::: at $B$ L5 integer leg value ::: $b = 5$ L5 equal-leg ratio ::: $2 \approx 1.41421$