Geometry Basics

Pythagorean Theorem and Distance

Use a² + b² = c² for a 3-4-5 triangle and the distance formula for points (1,2) and (4,6).

Example

Use squares and square roots for a triangle and coordinate distance.

highlighted = computed this step

Step 1 — Triangle legs

Set up the given values.

a=3b=4a= 3 \quad b= 4
Right triangle with legs 3 and 4a = 3b = 4c = sqrt(25)The slanted side is the hypotenuse.

Step 2 — Square the legs

Square the legs: 3^2 = 9, 4^2 = 16; 9+16=25.

32+42=9+16=253^{2} + 4^{2} = \hl{9} + \hl{16} = \hl{25}
Square the legs: 9 + 16 = 25a = 3b = 4c = sqrt(25)3^2 = 94^2 = 169 + 16 = 25The slanted side is the hypotenuse.

Step 3 — Take the square root

Take the square root: sqrt(25) = 5.

c=25=5c= \sqrt{25} = \hl{5}
c = sqrt(25) = 5a = 3b = 4c = sqrt(25) = 53^2 = 94^2 = 169 + 16 = 25Keep the root symbolic until sqrt(25) = 5.

Step 4 — Distance points

Set up the given values.

(1,2)→(4,6)( 1 , 2 )\to( 4 , 6 )
Plot (1, 2) and (4, 6)01234501234567Connect the points, then make a right triangle.(1, 2)(4, 6)

Step 5 — Coordinate changes

Find changes: 4 - 1 = 3 and 6 - 2 = 4.

dx=4−1=3dy=6−2=4dx= 4 - 1 = \hl{3} \quad dy= 6 - 2 = \hl{4}
Horizontal change 3, vertical change 401234501234567dx = 3dy = 4Connect the points, then make a right triangle.(1, 2)(4, 6)

Step 6 — Distance formula

Use distance: sqrt(3^2+4^2) = sqrt(25) = 5.

d=25=5d= \sqrt{25} = \hl{5}
d = sqrt(25) = 501234501234567dx = 3dy = 4d = sqrt(25) = 53^2 + 4^2 = 25, so d = sqrt(25).(1, 2)(4, 6)

Step 7 — Result

Read the final result.

c=5d=5c= 5 \quad d= 5
Distance matches the 3-4-5 triangle01234501234567dx = 3dy = 4d = sqrt(25) = 53^2 + 4^2 = 25, so d = sqrt(25).(1, 2)(4, 6)
pythagorean-and-distance The Pythagorean theorem: a² + b² = c² - a and b are the two legs. - c is the hypotenuse (longest side). Distance between (x₁,y₁) and (x₂,y₂): - d = √((x₂-x₁)² + (y₂-y₁)²).