LMNO is a parallelogram where OP is perpenducular to AB. If `OM = 20 cm, LP = 9 cm, ` and area of `Delta LPO= 54 sq. cm,` find the perimeter of LMNO.

Answer

To find the perimeter of the parallelogram LMNO, we will follow these steps:

### Step 1: Find the height OP
We know the area of triangle LPO is given by the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base \( LP = 9 \) cm and the area is \( 54 \) sq. cm. We can set up the equation:
\[ 54 = \frac{1}{2} \times 9 \times OP \]
Multiplying both sides by 2:
\[ 108 = 9 \times OP \]
Now, divide both sides by 9:
\[ OP = \frac{108}{9} = 12 \text{ cm} \]

### Step 2: Find the length PM using Pythagorean theorem
In triangle PMO, we apply the Pythagorean theorem:
\[ PM^2 + OP^2 = OM^2 \]
We know \( OM = 20 \) cm and \( OP = 12 \) cm. Plugging in the values:
\[ PM^2 + 12^2 = 20^2 \]
This simplifies to:
\[ PM^2 + 144 = 400 \]
Subtracting 144 from both sides:
\[ PM^2 = 400 - 144 = 256 \]
Taking the square root:
\[ PM = \sqrt{256} = 16 \text{ cm} \]

### Step 3: Find the length LM
The length \( LM \) can be found by adding \( LP \) and \( PM \):
\[ LM = LP + PM = 9 + 16 = 25 \text{ cm} \]

### Step 4: Find the length ON
Since opposite sides of a parallelogram are equal, we have:
\[ ON = LM = 25 \text{ cm} \]

### Step 5: Find the length OL using Pythagorean theorem
In triangle LOP, we again apply the Pythagorean theorem:
\[ LP^2 + OP^2 = OL^2 \]
Substituting the known values:
\[ 9^2 + 12^2 = OL^2 \]
This simplifies to:
\[ 81 + 144 = OL^2 \]
Thus,
\[ OL^2 = 225 \]
Taking the square root:
\[ OL = \sqrt{225} = 15 \text{ cm} \]

### Step 6: Find the length MN
Since opposite sides of a parallelogram are equal, we have:
\[ MN = OL = 15 \text{ cm} \]

### Step 7: Calculate the perimeter of parallelogram LMNO
The perimeter \( P \) of a parallelogram is given by:
\[ P = LM + MN + ON + OL \]
Substituting the values we found:
\[ P = 25 + 15 + 25 + 15 = 80 \text{ cm} \]

### Final Answer:
The perimeter of parallelogram LMNO is \( 80 \) cm.

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