Question

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?

Answer 1

Answer:

Given: Area of rhombus is 540 square cm and the length of one of its diagonals is 4.5 dm

To find the other diagonal of the rhombus:

Using formula of Area of rhombus:

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

Use conversion:

1 dm = 10 cm

then;

4.5 dm = $4.5 \times 10 = 45$ cm

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

$540 =\frac{1}{2} \times 45 \times d_2$

or

$540 \times 2 = 45 \times d_2$

$1080 = 45d_2$

Divide by 45 both sides we get;

$d_2 = 24 cm$

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 $\frac{d_1}{2}$  and $\frac{d_2}{2}$ can be found from:

$\frac{1}{x^2} =\frac{1}{(\frac{d_1}{2})^2} +\frac{1}{(\frac{d_2}{2})^2}$

then:

$\frac{1}{x^2} =\frac{1}{(\frac{45}{2})^2} +\frac{1}{(\frac{24}{2})^2}$

$\frac{1}{x^2} =\frac{1}{(22.5)^2} +\frac{1}{(12)^2}$

$\frac{1}{x^2} =\frac{1}{(22.5)^2} +\frac{1}{(12)^2}$

$\frac{1}{x^2} =\frac{1}{506.25} +\frac{1}{144}$

$\frac{1}{x^2} =\frac{1}{506.25} +\frac{1}{144}$

or

$\frac{1}{x^2} =\frac{506.25+144}{72900}$

$\frac{1}{x^2} =\frac{650.25}{72900}$

or

$x^2 = 112.110727$

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