Area of rhombus is `96 cm^(2)` and one of its diagonals is 12 cm . Find the side of the rhombus .
The following steps are involved in solving the above problem . Arrange them in sequential order .
(A) Let PQRS be a rhombus ,PR = 12 cm , and the diagonals PR and QS intersect at T .
(B) Area of the rhombus = `1/2 xxPR xx QR = 96 `
(C ) ` rArr QS = 16 cm `
(D ) PTQ is a right triangle and `PQ^(2) = 6^(2) +8^(2) rArr PQ = 10 cm ` .
(E ) PT = `(PR)/2 = 6 cm and QT = (QS) /2 = 8` cm

Options

A. ABCDE
B. ABECD
C. ABCED
D. ACEBD

Answer

To find the side of the rhombus given the area and one of its diagonals, we can follow these steps:

### Step-by-Step Solution:

**Step 1: Identify the given values.**
Let PQRS be a rhombus. We know that the area of the rhombus is \(96 \, cm^2\) and one of its diagonals, \(PR\), is \(12 \, cm\).

**Step 2: Use the area formula for a rhombus.**
The area \(A\) of a rhombus can be calculated using the formula:
\[ A = \frac{1}{2} \times d_1 \times d_2 \]
where \(d_1\) and \(d_2\) are the lengths of the diagonals. Here, \(d_1 = PR = 12 \, cm\) and we need to find \(d_2 = QS\).

**Step 3: Set up the equation.**
Substituting the known values into the area formula:
\[ 96 = \frac{1}{2} \times 12 \times QS \]
This simplifies to:
\[ 96 = 6 \times QS \]

**Step 4: Solve for \(QS\).**
Now, divide both sides by \(6\):
\[ QS = \frac{96}{6} = 16 \, cm \]

**Step 5: Find the lengths of the halves of the diagonals.**
Since the diagonals bisect each other at point \(T\):
\[ PT = \frac{PR}{2} = \frac{12}{2} = 6 \, cm \]
\[ QT = \frac{QS}{2} = \frac{16}{2} = 8 \, cm \]

**Step 6: Apply the Pythagorean theorem.**
In right triangle \(PTQ\), we can use the Pythagorean theorem to find the length of side \(PQ\):
\[ PQ^2 = PT^2 + QT^2 \]
Substituting the known values:
\[ PQ^2 = 6^2 + 8^2 \]
Calculating the squares:
\[ PQ^2 = 36 + 64 = 100 \]

**Step 7: Solve for \(PQ\).**
Taking the square root of both sides:
\[ PQ = \sqrt{100} = 10 \, cm \]

Thus, the length of each side of the rhombus is \(10 \, cm\).