Question

Show that in rhombus the side, 1) `=>" "l = sqrt(d_min^2/4 + d_(maj)^2/4)` Where `d_min = "minor diagonal" " " d_(maj) = "major diagonal"` 2) `=>" "r = (d_mind_(maj))/(4l)` Where `l and r` are the side of rhombus and radius of inscribed circle?

Answer 1

Please see below for proof.

Explanation:

In a rhombus, diagonals bisect each other at right angles. Hence, all the four triangles, to which diagonals of a rhombus divide it, are congruent.

(1) As hypotenuse of each triangle is equal to side of the rhombus and each side of triangles are half of the diagonals and using Pythagoras theorem

`l^2=(d_(maj)/2)^2+(d_(min)/2)^2`

or `l=sqrt((d_(maj)/2)^2+(d_(min)/2)^2)=sqrt(d_(maj)^2/4+d_(min)^2/4`

(2) Area of each triangle can be calculated in two ways. Note that `r` is perpendicular to hypotenuse, hence area of each triangle can be written as

either `1/2xxrxxl` or `1/2xxd_(min)/2xxd_(maj)/2`

Therefore `1/2xxrxxl=1/2xxd_(min)/2xxd_(maj)/2`

or `rl=(d_(min)d_(maj))/4`

and `r=(d_(min)d_(maj))/(4l)`