The area of the rhombus is 540 cm2; the length of one of its diagonals is 4.5 dm. What is the distance between the point of intersection of the diagonals and the side of the rhombus?

Area of rhombus(A) =[tex]\frac{1}{2} d_1 \times d_2[/tex] ......[1]; where [tex]d_1 ,d_2[/tex] are the diagonals and A is the area of rhombus respectively.

Use conversion:

1 dm = 10 cm

then;

4.5 dm = [tex]4.5 \times 10 = 45[/tex] cm

Substitute the value of A = 540 square cm and [tex]d_1 = 45cm[/tex] in [1] we have;

[tex]540 =\frac{1}{2} \times 45 \times d_2[/tex]

or

[tex]540 \times 2 = 45 \times d_2[/tex]

[tex]1080 = 45d_2[/tex]

Divide by 45 both sides we get;

[tex]d_2 = 24 cm[/tex]

The distance between the point of intersection of the diagonals and the side of the rhombus :

The line perpendicular to the rhombus side i,e x the altitude to the hypotenuse of one of those right triangles.

The length x of such an altitude in a right triangle with leg lengths [tex]\frac{d_1}{2}[/tex] and [tex]\frac{d_2}{2}[/tex] can be found from:

[tex]\frac{1}{x^2} =\frac{1}{(\frac{d_1}{2})^2} +\frac{1}{(\frac{d_2}{2})^2}[/tex]

then:

[tex]\frac{1}{x^2} =\frac{1}{(\frac{45}{2})^2} +\frac{1}{(\frac{24}{2})^2}[/tex]

[tex]\frac{1}{x^2} =\frac{1}{(22.5)^2} +\frac{1}{(12)^2}[/tex]

[tex]\frac{1}{x^2} =\frac{1}{506.25} +\frac{1}{144}[/tex]

or

[tex]\frac{1}{x^2} =\frac{506.25+144}{72900}[/tex]

[tex]\frac{1}{x^2} =\frac{650.25}{72900}[/tex]

or

[tex]x^2 = 112.110727[/tex]

Simplify:

x ≈ 10.59

Therefore, the distance between the point of intersection of the diagonals and the side of the rhombus is, 10.59 cm