Question

O is the origin.
ABCDEF is a regular hexagon and O is the midpoint of AD
OA= a and OC= c.
Find, in terms of a and c, in their simplest form
a) BE
b) DB
Check the file attached.

543c034b1f2c71b328023967cd127df9.docx

Answer 1

Answer:

b - DB

Step-by-step explanation:

E O is the origin ABCDEF is a regular hexagon and is the midpoint of AD. x = a and occ Find, in terms of a and c, in their simplest form (a) BE Answerda) BE- [2] (b) DR Answery) DB- [2] (c) the position vector of Answery) [2]

Answer 2

Answer:

BE is 2a and DB is $\sqrt{3}c$

Step-by-step explanation:

Given OA =a , OC=c ,O is midpoint of AD that means O is the center of Hexagon ABCDEF.

Then OD =a  

And ABCDEF is a regular hexagon that means it has all side of same length and same interior angle.

(a) By joining OB we get equilateral ΔAOB

OB = a  ( equilateral  triangle have same side length)

now , by joining OD and OE we get equilateral ΔEOD

∵ OE = a  ( equilateral  triangle have same side length)

now ,  BE = OB + OE

          BE=2a

(b) By joining DB it intersect OC at G make 90° angle with OC

In Δ COD

DG is height of the Δ

∵ height of a equilateral ΔCOD

  h =$\frac{\sqrt{3} }{2} base$

  h=  $\frac{\sqrt{3} c}{2}$           ( length of the base of ΔCOD is c )

DG =  $\frac{\sqrt{3} c}{2}$    

similarly in ΔBOD

BG =  $\frac{\sqrt{3} c}{2}$      ( height of BOD is BG )

   

Now DB= DG+BG

        DB=  $\frac{\sqrt{3} c}{2}$   +   $\frac{\sqrt{3} c}{2}$  

         DB=  $\sqrt{3}c$