a triangle is formed by joining mid points of alternate sides of a regular hexagon. what is the ratio of area of triangle so formed and area of regular hexagon

Answer :

we draw our figure , As  :



Here as we know ABCDEF is a regular hexagon, SO

AB  =  BC  =  CD  =  DE  =  EF  =  FA 

Now we divide it in 6 equal parts , after joining the diagonals, So

Area of $∆$ AOB  = Area of $∆$ BOC =  Area of $∆$ COD =  Area of $∆$ DOE  =  Area of $∆$ EOF =  Area of $∆$ FOA 

As there base is equal and their Apothem (  Height of every triangle is same )

And now we Divide each triangle into four equal parts , after joining there mid points  ,

Then we get Two rquired triangles , As $∆$ UQS  and $∆$ RTP  , as they formed by joining the mid point of alternate sides  ,

Area Area of each small triangle is same  ,  So

Area of $∆$ UQS or $∆$ RTP  =  9 ( Area of small triangle )  ( As we can see in both triangle we have 9 small triangle  .

And

Area of Hexagon ABCDEF  =  24 ( Area of small triangle )  (  As we can see that in hexagon we have 24 small triangle  .

SO,

$\frac{Area of ∆ UQS or ∆ RTP}{Area of Hexagon ABCDEF} = \frac{9 \left( Area of small triangle \right)}{24 \left( Area of small triangle \right)} = \frac{9}{24} = \frac{3}{8}$

SO,

Area of $∆$ UQS or $∆$ RTP  : Area of Hexagon ABCDEF  = 3  : 8  ( Ans )