Explore geometric proofs of the Pythagorean theorem and its converse. Detailed solutions to practice problems are included to help you master right triangle geometry.
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.
Proof 2: Similar Triangles Method
Construct rectangle from right triangle, creating three similar triangles. From similarity of \(\triangle ABE\) and \(\triangle AED\): \[ \frac{a}{c} = \frac{x}{a} \implies a^2 = cx \] From similarity of \(\triangle ECD\) and \(\triangle AED\): \[ \frac{b}{c} = \frac{y}{b} \implies b^2 = cy \] Adding equations: \[ a^2 + b^2 = c(x + y) = c^2 \]
Converse of the Pythagorean Theorem
If triangle sides \(a\), \(b\), \(c\) satisfy \(a^2 + b^2 = c^2\) (with \(c\) being the longest side), then the triangle is right-angled with hypotenuse \(c\).
Practice Problems
- Find the area of a right triangle with hypotenuse 10 cm and one leg 6 cm.
- Which triplets are sides of a right triangle?
a) \((2, 3, 4)\) b) \((12, 16, 20)\) c) \((3\sqrt{2}, 3\sqrt{2}, 6)\) - Vertices \(A(0,2)\), \(B(-2,-1)\), and \(C(-1,k)\) form a right triangle with hypotenuse \(AB\). Find \(k\).
- One leg is 2 meters shorter than the hypotenuse, the other leg is 4 meters shorter than the hypotenuse. Find the triangle's perimeter.
- Find the perimeter of an equilateral triangle with height 60 cm.
- Calculate the area of a square with diagonal 100 meters.
- Find \(x\) in the figure below:
Detailed Solutions
Solution 1
Let the unknown leg be \(L\). By the Pythagorean theorem:
\[ L^2 + 6^2 = 10^2 \implies L = \sqrt{100 - 36} = 8 \, \text{cm} \]Area \(= \frac{1}{2} \times 6 \times 8 = 24 \, \text{cm}^2\)
Solution 2
Test using the converse theorem:
- a) \(2^2 + 3^2 = 13 \neq 16 = 4^2\) → Not a right triangle.
- b) \(12^2 + 16^2 = 144 + 256 = 400 = 20^2\) → Right triangle.
- c) \((3\sqrt{2})^2 + (3\sqrt{2})^2 = 18 + 18 = 36 = 6^2\) → Right isosceles triangle.
Solution 3
Distance squares:
\[ AB^2 = (-2-0)^2 + (-1-2)^2 = 4 + 9 = 13 \] \[ AC^2 = (-1-0)^2 + (k-2)^2 = 1 + k^2 - 4k + 4 = k^2 - 4k + 5 \] \[ BC^2 = (-1 - (-2))^2 + (k - (-1))^2 = 1^2 + (k+1)^2 = k^2 + 2k + 2 \]Using \(AB^2 = AC^2 + BC^2\):
\[ 13 = (k^2 - 4k + 5) + (k^2 + 2k + 2) \] \[ 2k^2 - 2k - 6 = 0 \implies k^2 - k - 3 = 0 \] \[ k = \frac{1 \pm \sqrt{13}}{2} \approx 2.30 \text{ or } -1.30 \]Two possible points: \(C(-1, 2.30)\) or \(C(-1, -1.30)\).
Solution 4
Let the hypotenuse be \(h\). Legs are \(h-2\) and \(h-4\). By the theorem:
\[ h^2 = (h-2)^2 + (h-4)^2 \] \[ h^2 = h^2 - 4h + 4 + h^2 - 8h + 16 \] \[ h^2 - 12h + 20 = 0 \implies (h-10)(h-2) = 0 \]Solutions: \(h = 2\) (invalid, since sides would be 0 and -2) or \(h = 10\).
Perimeter \(= 10 + 8 + 6 = 24\) meters.
Solution 5
Let the side be \(s\). The height splits the base into \(\frac{s}{2}\). By the theorem:
\[ s^2 = 60^2 + \left(\frac{s}{2}\right)^2 \] \[ s^2 = 3600 + \frac{s^2}{4} \implies \frac{3s^2}{4} = 3600 \implies s^2 = 4800 \] \[ s = \sqrt{4800} = 40\sqrt{3} \, \text{cm} \]Perimeter \(= 3s = 120\sqrt{3} \, \text{cm}\).
Solution 6
Let the side of the square be \(s\). The diagonal gives:
\[ s^2 + s^2 = 100^2 \implies 2s^2 = 10000 \implies s^2 = 5000 \]Area \(= s^2 = 5000 \, \text{m}^2\).
Solution 7
First, find \(CD\) from the lower right triangle \(\triangle ECD\):
\[ 5^2 = 3^2 + CD^2 \implies CD = \sqrt{25 - 9} = 4 \]Then \(FC = 6 - 4 = 2\). In the upper right triangle \(\triangle EFC\):
\[ x^2 = 3^2 + 2^2 = 9 + 4 = 13 \implies x = \sqrt{13} \]