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

1. The area of the rhombus can be found by the formula

$A=\dfrac{d_1\cdot d_2}{2},$ where $d_1,\ d_2$ are rhombus's diagonals.

Note that $d_1=4.5\ dm=45\ cm,$ then

$540=\dfrac{45\cdot d_2}{2},\\ \\540\cdot 2=45d_2,\\ \\d_2=24\ cm.$

2. The diagonals of rhombus are perpendicular and are bisectors of each other. Then the triangle formed with halfs of diagonals is right triangles with legs

$\dfrac{d_1}{2}=22.5\ cm,\ \dfrac{d_2}{2}=12\ cm.$

The hypotenuse of this triangle is the rhombus's side. By the Pythagorean theorem

$\text{rhombus's side}^2=(22.5)^2+12^2=506.25+144=650.25,\\ \\\text{rhombus's side}=25.5\ cm.$

3. The distance between the point of intersection of the diagonals and the side of the rhombus is the height of right triangle considered above.

Use twice the Pythagorean theorem to find this height:

$\left\{\begin{array}{l}x^2+h^2=12^2\\(25.5-x)^2+h^2=22.5^2,\end{array}\right.$

where x is projection of leg 12 cm and h is height.

Subtract the first equation from the second:

$(25.5-x)^2+h^2-x^2-h^2=22.5^2-12^2,\\ \\650.25-51x=506.25-144,\\ \\51x=650.25-362.25=288,\\ \\x=\dfrac{96}{17}\ cm.$

Then

$h^2=144-\left(\dfrac{96}{17}\right)^2=144-\dfrac{9216}{289}=\dfrac{32400}{289},\\ \\h=\dfrac{180}{17}\ cm.$

Answer: $h=\dfrac{180}{17}\ cm.$

Answer 2

¹⁸⁰/₁₇

Further explanation

Given:

• The area of the rhombus is 540 cm².
• The length of one of its diagonals is 4.5 dm.

Question:

What is the distance between the point of intersection of the diagonals and the side of the rhombus?

The Process:

Step-1: calculate the length of the second diagonal

Let us call the formula for calculating the area of rhombus.

$\boxed{ \ Area = \frac{1}{2} \times d_1 \times d_2 \ }$

The data is as follows:

• Area = 540 cm²
• Diagonal-1 = 4.5 dm = 45 cm.

Let us calculate the length of the second diagonal (d₂).

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

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

$\boxed{ \ d_2 = 2 \times 12 \ }$

∴ $\boxed{\boxed{ \ d_2 = 24 \ cm \ }}$

Step-2: calculate the length of the hypotenuse of the rhombus

In one of the four right triangle sections, we prepare the lengths of the half diagonal sides.

$\boxed{ \ \frac{1}{2}d_1 \rightarrow \frac{1}{2} \times 45 = 22.5 \ cm \ }$

$\boxed{ \ \frac{1}{2}d_2 \rightarrow \frac{1}{2} \times 24 = 12 \ cm \ }$

The relationship between the two perpendicular sides and the hypotenuse is given in the Pythagorean theorem.

$\boxed{ \ (hypotenuse)^2 = 12^2 + (22.5)^2\ }$

$\boxed{ \ hypotenuse = \sqrt{144 + 506.25} \ }$

$\boxed{ \ hypotenuse = \sqrt{650.25} \ }$

∴ $\boxed{ \ hypotenuse = 25.5 \ cm \ }$

Step-3: calculate the distance between the points of intersection of the diagonals and the side of the rhombus

Let's look at the attached picture. We will find out the length of x.

We can use the principle of congruence from the area of a triangle.

$\boxed{ \ b_1 \times h_1 = b_2 \times h_2 \ }$

where b = base and h = height.

Let us say,

• b₁ = 12 cm
• h₁ = 22.5 cm
• b₂ = 25.5 cm
• h₂ = x

$\boxed{ \ 12 \cdot 22.5 = 25.5 \cdot x \ }$

$\boxed{ \ x = \frac{12 \cdot 22.5}{25.5} \ }$

$\boxed{ \ x = \frac{12 \cdot 15}{17} \ }$

$\boxed{\boxed{ \ x = \frac{180}{17} \ }}$

Thus, the distance between the point of intersection of the diagonals and the side of the rhombus is ¹⁸⁰/₁₇ cm.

Learn more

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