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.

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)#
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Answer 2
  1. To prove that l=dmin24+dmaj24 l = \sqrt{\frac{d_{\text{min}}^2}{4} + \frac{d_{\text{maj}}^2}{4}} in a rhombus where dmin d_{\text{min}} is the length of the minor diagonal and dmaj d_{\text{maj}} is the length of the major diagonal:

In a rhombus, the diagonals bisect each other at right angles. This divides the rhombus into four congruent right triangles.

Using the Pythagorean theorem in one of these triangles: l2=(dmin2)2+(dmaj2)2l^2 = \left(\frac{d_{\text{min}}}{2}\right)^2 + \left(\frac{d_{\text{maj}}}{2}\right)^2

Solving for l l : l=(dmin2)2+(dmaj2)2l = \sqrt{\left(\frac{d_{\text{min}}}{2}\right)^2 + \left(\frac{d_{\text{maj}}}{2}\right)^2} l=dmin24+dmaj24l = \sqrt{\frac{d_{\text{min}}^2}{4} + \frac{d_{\text{maj}}^2}{4}}

  1. To prove that r=dmindmaj4l r = \frac{d_{\text{min}}d_{\text{maj}}}{4l} in a rhombus where l l is the side length of the rhombus and r r is the radius of the inscribed circle:

The radius of the inscribed circle (the circle tangent to all four sides of the rhombus) is the distance from the center of the rhombus to any of its vertices.

In a rhombus, the diagonals bisect each other at right angles, forming four congruent right triangles.

The radius of the inscribed circle is the height of one of these triangles, and the side length of the rhombus is the base.

Using the formula for the area of a triangle (A=12bh A = \frac{1}{2}bh ), and substituting l l for b b and r r for h h : A=12lrA = \frac{1}{2} \cdot l \cdot r

Since the area of a rhombus can also be expressed as half the product of its diagonals: A=dmindmaj2A = \frac{d_{\text{min}} \cdot d_{\text{maj}}}{2}

Equating the two expressions for the area: 12lr=dmindmaj2\frac{1}{2} \cdot l \cdot r = \frac{d_{\text{min}} \cdot d_{\text{maj}}}{2}

Solving for r r : r=dmindmaj2lr = \frac{d_{\text{min}} \cdot d_{\text{maj}}}{2l}

Since l=dmin2+dmaj22 l = \frac{\sqrt{d_{\text{min}}^2 + d_{\text{maj}}^2}}{2} : r=dmindmaj4lr = \frac{d_{\text{min}} \cdot d_{\text{maj}}}{4l}

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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