What is the area of rhombus ABCD ?

The area is 36 units squared.

You have three ways you can solve this question.

Method 1:

Split the rhombus into two equal triangles.
You will get triangle ACD and triangle ACB.

Count the base length and height length of each triangle using the diagram.
You will get base = 6 units and height = 6 units. Plug this into the area of a triangle and then multiply it by 2.

A = bh/2 * 2
A = 6(6) / 2 * 2
A = 36 / 2 * 2
A = 18 * 2
A = 36

Method 2:

Calculate the length of DP.
The length of line segment DP is √28.8

Calculate the length of DC.
The length of line segment DC is 3√5

Put into equation A = bh.

A = bh
A = 3√5(√28.8)
A = 36

Method 3:

Calculate the length of each diagonal and put into formula A = 1/2(d1 * d2)

Diagonal DB = 12 units
Diagonal AC = 6 units

A = 1/2(d1 * d2)
A = 1/2(12 * 6)
A = 1/2(72)
A = 36

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AkhileshT AkhileshT · Answer 2 · 2019-09-04T11:42:14+02:00

The area of the rhombus ABCD is [tex]\boxed{\bf 36\text{ \bf square units}}[/tex].

Further explanation:

Formula used:

The area [tex](A)[/tex] of the rhombus is calculated as follows:

[tex]\boxed{A=\dfrac{xy}{2}}[/tex]

Here, [tex]x[/tex] and [tex]y[/tex] is the length of the diagonals.

The distance [tex]d[/tex] between the two points [tex](x_{1},y_{1})[/tex] and [tex](x_{2},y_{2})[/tex] is calculated as follows:

[tex]\boxed{d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}}[/tex]

Calculation:

The coordinates of the corner points of rhombus ABCD are A [tex](-1,0)[/tex],B [tex](5,-3)[/tex], C [tex](-1,-6)[/tex] and D [tex](-7,-3)[/tex].

The diagonals of the rhombus ABCD are AC and BD as shown in attached Figure 1.

The distance between the point A [tex](-1,0)[/tex] and C [tex](-1,-6)[/tex] is calculated as follows:

[tex]\begin{aligned}\text{AC}&=\sqrt{(-1-(-1))^{2}+(-6-0)^{2}}\\&=\sqrt{(-1+1)^{2}+36}\\&=\sqrt{0+36}\\&=\sqrt{36}\\&=6\end{aligned}[/tex]

Therefore, the length AC is [tex]\boxed{6{\text{ unit}}}[/tex].

The distance between the point B [tex](5,-3)[/tex] and D [tex](-7,-3)[/tex] is calculated as follows:

[tex]\begin{aligned}\text{BD}&=\sqrt{(-7-5)^{2}+(-3-(-3))^{2}}\\&=\sqrt{(-12)^{2}+(-3+3)^{2}}\\&=\sqrt{144}\\&=12\end{aligned}[/tex]

Therefore, the distance BD is [tex]\boxed{12\text{ unit}}[/tex].

The area of the rhombus ABCD is calculated as follows:

[tex]\begin{aligned}A&=\dfrac{12\times 6}{2}\\&=6\times 6\\&=36\end{aligned}[/tex]

Therefore, the value of [tex]A[/tex] is [tex]36[/tex].

Thus, the area of the rhombus ABCD is [tex]\boxed{\bf 36\text{ \bf square units}}[/tex].

Learn more:

1. Learn more about problem on coordinate of the graph https://brainly.com/question/1286775

2. Learn more about problem on equations https://brainly.com/question/1473992

3. Learn more about problem on triangles https://brainly.com/question/7437053

Answer details:

Grade: Middle school

Subject: Mathematics

Chapter: Quadrilateral

Keywords: Area, rhombus, distance, coordinate, diagonal, length, quadrilateral, ABCD, distance formula, mensuration, mathematics, corner points.