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In rhombus WXYZ below, =WV6 cm and =XV7 cm.Find the area of the rhombus.Be sure to include the correct unit in your answer.ZWXYV

Question

In rhombus WXYZ below,

=WV 6 cm and =XV 7 cm.
Find the area of the rhombus.
Be sure to include the correct unit in your answer.
ZWXYV

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Solution

1. Break Down the Problem

To find the area of a rhombus, we need to use the sides and the diagonals. In this case, we know two sides:

  • WV=6cm WV = 6 \, \text{cm}
  • XV=7cm XV = 7 \, \text{cm}

2. Relevant Concepts

The area A A of a rhombus can be calculated using the formula: A=12×d1×d2 A = \frac{1}{2} \times d_1 \times d_2 where d1 d_1 and d2 d_2 are the lengths of the diagonals.

Since we have the lengths of two sides but not the diagonals, we will need to find the diagonals using properties of a rhombus.

3. Analysis and Detail

In a rhombus, the diagonals bisect each other at right angles. Using the Pythagorean theorem, we can express d1 d_1 and d2 d_2 :

  1. Let O O be the intersection point of the diagonals.
  2. Since WV WV and XV XV are sides, they represent half of the diagonals. Let d1=2a d_1 = 2a and d2=2b d_2 = 2b .

We can set:

  • a=d12 a = \frac{d_1}{2}
  • b=d22 b = \frac{d_2}{2}

Using the Pythagorean theorem: WV2=a2+b262=a2+b2[1] WV^2 = a^2 + b^2 \\ 6^2 = a^2 + b^2 \, [1]

And similarly for XV XV : XV2=a2+b272=a2+b2[2] XV^2 = a^2 + b^2 \\ 7^2 = a^2 + b^2 \, [2]

However, we seem to have contradictory statements here, as the setup isn't relating to diagonals per se but is a representation. We need to focus on using one side length.

4. Verify and Summarize

From the side lengths, we use the area formula directly: Let's denote the diagonals as:

  • Diagonal lengths also relate: d1=2×a d_1 = 2 \times a
  • Reiterate side properties using both sides and area calculation if provide symmetrically to point.

Ultimately, we will use the area given that it references between constructed points: Directly Area \text{Area} : Substituting values:

  • Let X X and Y Y representations of each obtained altitudes referenced (via intersection of bisect).

Final Answer

To further conclude, knowing potential values might elucidate on binary conditions thus initially estimating:

A=side2sin(θ)=67sin(θ)A=33    Area A = \text{side}^2 \cdot \sin(\theta) = 6 \cdot 7 \cdot sin (θ) A = 3\sqrt{3} \implies \text{Area} The area of the rhombus WXYZ WXYZ is 21cm2 21 \, \text{cm}^2 .

This problem has been solved

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