Question

what is the value of 'x' in the shaft

Answer 1

Answer:

x = ( 10√2 - 9 ) mm

Step-by-step explanation:

Given: Radius of circle with centre O = 15mm and in attached figure

            ED = 6mm , AC = 10mm

To find: Value of x i.e., EB

Construction: Join OA, AC and draw radius OD such that it is perpendicular

                        with AC.

OA = OD = OC = 15mm (radius of circle)

AC is chord and OB is perpendicular then

$AB=CB=\frac{AC}{2}=5\,mm$ (because perpendicular from center bisect the chord)

In ΔOBC,

By Pythagoras Theorem

$OC^2=OB^2+CB^2\\\implies 15^2=OB^2+5^2\\\implies OB^2=15^2-5^2\\\implies OB^2=200\\\implies OB=\sqrt{200}=10\sqrt{2}\,mm$

Now we find value of BD,

BD = OD - OB

⇒ BD = ( 15 - 10√2 ) mm

∴ EB = ED - BD

⇒ x = 6 - ( 15 - 10√2 )

⇒ x = ( 10√2 - 9 ) mm