PQRS is a parallelogram with ST perpendicular to PQ. Given area of `Delta` PST = 54 sq.cm, QS = 13 cm, and TQ = 5 cm, find the perimeter of the parallelogram.

Answer

To find the perimeter of the parallelogram PQRS given the area of triangle PST, QS, and TQ, we can follow these steps:

### Step 1: Understand the Given Information
We know:
- Area of triangle PST = 54 sq.cm
- QS = 13 cm
- TQ = 5 cm
- ST is perpendicular to PQ.

### Step 2: Use the Pythagorean Theorem to Find ST
In triangle SQT, we can apply the Pythagorean theorem since ST is perpendicular to PQ.

\[ SQ^2 = ST^2 + TQ^2 \]

Substituting the values we know:
\[ 13^2 = ST^2 + 5^2 \]
\[ 169 = ST^2 + 25 \]
\[ ST^2 = 169 - 25 = 144 \]
\[ ST = \sqrt{144} = 12 \text{ cm} \]

### Step 3: Calculate the Base PT of Triangle PST
Using the area formula for triangle PST:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base is PT and the height is ST.

Substituting the known values:
\[ 54 = \frac{1}{2} \times PT \times 12 \]

Solving for PT:
\[ 54 = 6 \times PT \]
\[ PT = \frac{54}{6} = 9 \text{ cm} \]

### Step 4: Find the Length of PQ
Since PQ = PT + TQ:
\[ PQ = PT + TQ = 9 + 5 = 14 \text{ cm} \]

### Step 5: Find the Length of PS
Using the Pythagorean theorem again in triangle PST:
\[ PS^2 = ST^2 + PT^2 \]
Substituting the values:
\[ PS^2 = 12^2 + 9^2 \]
\[ PS^2 = 144 + 81 = 225 \]
\[ PS = \sqrt{225} = 15 \text{ cm} \]

### Step 6: Find the Lengths of the Other Sides
Since opposite sides of a parallelogram are equal:
- QR = PS = 15 cm
- RS = PQ = 14 cm

### Step 7: Calculate the Perimeter of the Parallelogram
The perimeter (P) of the parallelogram is given by:
\[ P = PQ + QR + RS + SP \]
Substituting the values:
\[ P = 14 + 15 + 14 + 15 = 58 \text{ cm} \]

### Final Answer
The perimeter of the parallelogram PQRS is **58 cm**.

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