PQRS is a rectangle. T is a point on PQ such that RTQ is an isosceles triangle and PT = 5 QT.If the area of triangle RTQ is `12 sqrt3`sq.cm, then the area of the rectangle PQRS is:

Options

A. `142 sqrt 3 sq. cm`
B. `134 sqrt3 sq. cm`
C. `144 sqrt3 sq. cm`
D. `142 sq. cm`

Answer

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

### Step 1: Define Variables
Let QT = x. Since PT = 5QT, we have:
- PT = 5x
- QT = x
- QR = QT = x (because triangle RTQ is isosceles)

### Step 2: Find PQ
Since T is a point on PQ, we can express PQ as:
\[ PQ = PT + QT = 5x + x = 6x \]

### Step 3: Identify the Height of Triangle RTQ
In triangle RTQ, RT is the height from R to QT. Since QT = x, we can assume that RT is also x (as the triangle is isosceles).

### Step 4: Calculate the Area of Triangle RTQ
The area of triangle RTQ can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Here, the base QT = x and the height RT = x. Therefore:
\[ \text{Area} = \frac{1}{2} \times x \times x = \frac{x^2}{2} \]

### Step 5: Set Up the Equation for Area
We know from the problem that the area of triangle RTQ is \( 12\sqrt{3} \) sq. cm. Thus, we can set up the equation:
\[ \frac{x^2}{2} = 12\sqrt{3} \]

### Step 6: Solve for x^2
To solve for \( x^2 \), we multiply both sides by 2:
\[ x^2 = 24\sqrt{3} \]

### Step 7: Calculate the Area of Rectangle PQRS
The area of rectangle PQRS can be calculated as:
\[ \text{Area} = \text{length} \times \text{breadth} \]
Here, the length PQ = 6x and the breadth QR = x. Thus:
\[ \text{Area} = 6x \times x = 6x^2 \]

### Step 8: Substitute x^2 into the Area Formula
Now, substitute \( x^2 = 24\sqrt{3} \) into the area formula:
\[ \text{Area} = 6 \times 24\sqrt{3} = 144\sqrt{3} \text{ sq. cm} \]

### Final Answer
The area of rectangle PQRS is \( 144\sqrt{3} \) sq. cm.

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