How-could-5-be-drawn-on-numbered-line-using-scale-and-compass-only-Exactly-5-not-its-decimal-approximation-

Question Number 3249 by Rasheed Soomro last updated on 08/Dec/15

H o w c o u l d \sqrt{5} b e d r a w n o n n u m b e r e d l i n e u s i n g

s c a l e a n d c o m p a s s o n l y ? (E x a c t l y \sqrt{5} n o t i t s d e c i m a l a p p r o x i m a t i o n .)

Answered by prakash jain last updated on 08/Dec/15

For \sqrt{5}

draw a horizonal line AB of length 5 + 1 = 6

mark a point at 5 (C) .

bisect AB and get center point D .

draw a semicircle of radius AD .

draw ⊥^{r} to AB at C . Let it intersect semicircle

at E

length of CE = \sqrt{5}

DCE is a right angled triangle

DE = 3

CD = 2

{DE}^{2} = {CD}^{2} + {CE}^{2}

{CE}^{2} = 5 or CE = \sqrt{5}

Commented by RasheedAhmad last updated on 08/Dec/15

N i c e!