Pythagorean Theorem and Problems with Solutions

Explore geometric proofs of the Pythagorean theorem and its converse. Detailed solutions to practice problems are included.

Proofs of the Pythagorean Theorem

Proof 1: Geometric Area Method

Two squares with sides \(a + b\) and \(c\) are shown. The large square's area equals the small square's area plus four congruent right triangles: \[ (a + b)^2 = c^2 + 4 \cdot \frac{1}{2}ab \] Expanding and simplifying: \[ a^2 + b^2 + 2ab = c^2 + 2ab \] \[ a^2 + b^2 = c^2 \] where \(a\) and \(b\) are legs of the right triangle, and \(c\) is the hypotenuse.

Geometric proof of Pythagorean theorem

Proof 2: Similar Triangles Method

Construct rectangle from right triangle, creating three similar triangles. From similarity of \(\triangle ABE\) and \(\triangle AED\): \[ \frac{a}{c} = \frac{x}{a} \Rightarrow a^2 = cx \] From similarity of \(\triangle ECD\) and \(\triangle AED\): \[ \frac{b}{c} = \frac{y}{b} \Rightarrow b^2 = cy \] Adding equations: \[ a^2 + b^2 = c(x + y) = c^2 \]

Similar triangles proof

Converse of the Pythagorean Theorem

If triangle sides \(a\), \(b\), \(c\) satisfy \(a^2 + b^2 = c^2\) (with \(c\) longest), then the triangle is right-angled with hypotenuse \(c\).

Practice Problems

  1. Find the area of a right triangle with hypotenuse 10 cm and one leg 6 cm.
  2. Which triplets are sides of a right triangle?
    a) \((2, 3, 4)\)   b) \((12, 16, 20)\)   c) \((3\sqrt{2}, 3\sqrt{2}, 6)\)
  3. Vertices \(A(0,2)\), \(B(-2,-1)\), and \(C(-1,k)\) form a right triangle with hypotenuse \(AB\). Find \(k\).
  4. One leg is 2 meters shorter than the hypotenuse, the other leg is 4 meters shorter than the hypotenuse. Find the triangle's perimeter.
  5. Find the perimeter of an equilateral triangle with height 60 cm.
  6. Calculate the area of a square with diagonal 100 meters.
  7. Find \(x\) in the figure below: Nested right triangles problem

Detailed Solutions

  1. Solution: Let unknown leg be \(L\). By Pythagorean theorem: \[ L^2 + 6^2 = 10^2 \Rightarrow L = \sqrt{100 - 36} = 8 \text{ cm} \] Area \(= \frac{1}{2} \times 6 \times 8 = 24 \text{ cm}^2\)
  2. Solution: Test using converse theorem.
    a) \(2^2 + 3^2 = 13 \neq 16 = 4^2\) → Not right triangle.
    b) \(12^2 + 16^2 = 400 = 20^2\) → Right triangle.
    c) \((3\sqrt{2})^2 + (3\sqrt{2})^2 = 18 + 18 = 36 = 6^2\) → Right isosceles triangle.
  3. Solution: Distance squares: \[ AB^2 = (-2-0)^2 + (-1-2)^2 = 13 \] \[ AC^2 = (-1-0)^2 + (k-2)^2 = k^2 - 4k + 5 \] \[ BC^2 = (-1+2)^2 + (k+1)^2 = k^2 + 2k + 2 \] Using \(AB^2 = AC^2 + BC^2\): \[ 13 = (k^2 - 4k + 5) + (k^2 + 2k + 2) \] \[ 2k^2 - 2k - 6 = 0 \Rightarrow k^2 - k - 3 = 0 \] \[ k = \frac{1 \pm \sqrt{13}}{2} \approx 2.30 \text{ or } -1.30 \] Two possible points: \(C(-1, 2.30)\) or \(C(-1, -1.30)\).
  4. Solution: Let hypotenuse be \(h\). Legs: \(h-2\) and \(h-4\). By theorem: \[ h^2 = (h-2)^2 + (h-4)^2 \] \[ h^2 = h^2 - 4h + 4 + h^2 - 8h + 16 \] \[ h^2 - 12h + 20 = 0 \] Solutions: \(h = 2\) (invalid, sides would be 0 and -2) or \(h = 10\).
    Perimeter \(= 10 + 8 + 6 = 24\) meters.
  5. Solution: Let side be \(s\). Height splits base into \(s/2\). By theorem: \[ s^2 = 60^2 + \left(\frac{s}{2}\right)^2 \] \[ s^2 = 3600 + \frac{s^2}{4} \Rightarrow \frac{3s^2}{4} = 3600 \] \[ s = 40\sqrt{3} \text{ cm} \] Perimeter \(= 3s = 120\sqrt{3} \text{ cm}\).
  6. Solution: Let side be \(s\). Diagonal gives: \[ s^2 + s^2 = 100^2 \Rightarrow 2s^2 = 10000 \] Area \(= s^2 = 5000 \text{ m}^2\).
  7. Solution: First, find \(CD\) from \(\triangle ECD\): \[ 5^2 = 3^2 + CD^2 \Rightarrow CD = 4 \] Then \(FC = 6 - 4 = 2\). In \(\triangle EFC\): \[ x^2 = 3^2 + 2^2 = 13 \Rightarrow x = \sqrt{13} \]

Additional Resources