In `Delta LMN, NT bot LM.` Area of `Delta LMN =72 sq. m and NT = 9 cm. ` If area of `Delta NTL, ` area of `Delta NTM= 1:3,` find LT and TM.

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript.

### Step 1: Understand the problem
We have triangle \( \Delta LMN \) with a perpendicular \( NT \) from point \( N \) to line \( LM \). The area of triangle \( LMN \) is given as \( 72 \, \text{sq. m} \) and the height \( NT = 9 \, \text{cm} \). The areas of triangles \( NTL \) and \( NTM \) are in the ratio \( 1:3 \).

### Step 2: Calculate the base \( LM \)
The area of triangle \( LMN \) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Substituting the known values:
\[ 72 = \frac{1}{2} \times LM \times 9 \]
To find \( LM \), we rearrange the equation:
\[ 72 = \frac{9}{2} \times LM \]
Multiplying both sides by \( 2 \):
\[ 144 = 9 \times LM \]
Now, divide by \( 9 \):
\[ LM = \frac{144}{9} = 16 \, \text{cm} \]

### Step 3: Set up the areas of triangles \( NTL \) and \( NTM \)
Let \( LT = x \) and \( TM = y \). From the problem, we know:
\[ x + y = LM = 16 \, \text{cm} \quad \text{(Equation 1)} \]
The areas of triangles \( NTL \) and \( NTM \) can be expressed as:
\[ \text{Area of } NTL = \frac{1}{2} \times LT \times NT = \frac{1}{2} \times x \times 9 = \frac{9x}{2} \]
\[ \text{Area of } NTM = \frac{1}{2} \times TM \times NT = \frac{1}{2} \times y \times 9 = \frac{9y}{2} \]

### Step 4: Set up the ratio of areas
According to the problem, the ratio of the areas of triangles \( NTL \) and \( NTM \) is \( 1:3 \):
\[ \frac{\frac{9x}{2}}{\frac{9y}{2}} = \frac{1}{3} \]
This simplifies to:
\[ \frac{x}{y} = \frac{1}{3} \]
From this, we can express \( y \) in terms of \( x \):
\[ y = 3x \quad \text{(Equation 2)} \]

### Step 5: Substitute Equation 2 into Equation 1
Substituting \( y = 3x \) into \( x + y = 16 \):
\[ x + 3x = 16 \]
\[ 4x = 16 \]
Dividing both sides by \( 4 \):
\[ x = 4 \, \text{cm} \]

### Step 6: Find \( y \)
Now, substituting \( x \) back into Equation 2 to find \( y \):
\[ y = 3x = 3 \times 4 = 12 \, \text{cm} \]

### Final Answer:
Thus, the lengths are:
- \( LT = 4 \, \text{cm} \)
- \( TM = 12 \, \text{cm} \)