Construct a quadrilateral having its sides QR = 5 cm, RS = 5 cm, PS = 5.7 cm, PQ = 6 cm and PR = 5.6 cm and a triangle ∆RQT equal in area to the quadrilateral.

To solve this problem, we need to go through two main steps:

1. **Construct the quadrilateral \(PQRS\)** with the given side lengths.
2. **Construct a triangle \(\Delta RQT\)** that is equal in area to quadrilateral \(PQRS\).

### Step 1: Construct the Quadrilateral \(PQRS\)

Given the sides:
- \(PQ = 6\) cm
- \(QR = 5\) cm
- \(RS = 5\) cm
- \(PS = 5.7\) cm
- \(PR = 5.6\) cm

The construction steps for quadrilateral \(PQRS\) are as follows:

1. **Draw \(PQ = 6\) cm.**
2. **Using a compass, draw an arc of radius \(QR = 5\) cm from point \(Q\).**
3. **From point \(P\), draw an arc of radius \(PR = 5.6\) cm.** The intersection of these two arcs will be point \(R\).
4. **Draw line segments \(QR\) and \(PR\).**
5. **From point \(R\), draw an arc of radius \(RS = 5\) cm.**
6. **From point \(P\), draw an arc of radius \(PS = 5.7\) cm.** The intersection of these arcs will be point \(S\).
7. **Draw line segments \(RS\) and \(PS\).**

Now, you have constructed quadrilateral \(PQRS\).

### Step 2: Construct a Triangle \(\Delta RQT\) Equal in Area to Quadrilateral \(PQRS\)

1. **Calculate the Area of Quadrilateral \(PQRS\):**

   We can divide the quadrilateral into two triangles, \( \triangle PQR \) and \( \triangle PRS \), using diagonal \(PR\). The area of quadrilateral \(PQRS\) is the sum of the areas of these two triangles.

   - **Area of \(\triangle PQR\)** using Heron's formula:
     \[
     \text{Semi-perimeter } s_1 = \frac{PQ + QR + PR}{2} = \frac{6 + 5 + 5.6}{2} = 8.3 \text{ cm}
     \]
     \[
     \text{Area of } \triangle PQR = \sqrt{s_1(s_1 - PQ)(s_1 - QR)(s_1 - PR)}
     \]
     \[
     = \sqrt{8.3 \times (8.3 - 6) \times (8.3 - 5) \times (8.3 - 5.6)}
     \]
     \[
     = \sqrt{8.3 \times 2.3 \times 3.3 \times 2.7}
     \]

   - **Area of \(\triangle PRS\)** using Heron's formula:
     \[
     \text{Semi-perimeter } s_2 = \frac{PR + RS + PS}{2} = \frac{5.6 + 5 + 5.7}{2} = 8.15 \text{ cm}
     \]
     \[
     \text{Area of } \triangle PRS = \sqrt{s_2(s_2 - PR)(s_2 - RS)(s_2 - PS)}
     \]
     \[
     = \sqrt{8.15 \times (8.15 - 5.6) \times (8.15 - 5) \times (8.15 - 5.7)}
     \]
     \[
     = \sqrt{8.15 \times 2.55 \times 3.15 \times 2.45}
     \]

   - **Total area of \(PQRS\):**
     \[
     \text{Area of } PQRS = \text{Area of } \triangle PQR + \text{Area of } \triangle PRS
     \]

2. **Construct Triangle \(\Delta RQT\):**

   To make a triangle \(\Delta RQT\) that has the same area as quadrilateral \(PQRS\), choose points \(R\) and \(Q\) on the line as before.

   - Select any point \(T\) such that \( \triangle RQT \) has an area equal to the area of quadrilateral \(PQRS\).

   To determine the exact location of point \(T\), you can use dynamic geometry software like GeoGebra or a manual trial method with a compass and protractor to adjust the location until the area condition is satisfied.

### Summary of the Process

1. Construct the quadrilateral \(PQRS\) as per the steps outlined.
2. Calculate the area of \(PQRS\) using Heron's formula for triangles \(PQR\) and \(PRS\).
3. Construct triangle \(RQT\) with an area equal to the calculated area of quadrilateral \(PQRS\).

Would you like detailed calculations for each area, or have any other questions?

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**Related Questions:**

1. What is Heron's formula, and how is it used to calculate the area of a triangle?
2. How can you determine if a quadrilateral is cyclic using its side lengths?
3. What are some other methods to construct a quadrilateral given its side lengths?
4. How can you use coordinate geometry to find the area of a quadrilateral?
5. What are the properties of a quadrilateral that make it a parallelogram?

**Tip:** When constructing geometric shapes, always double-check each step for accuracy to ensure all sides and angles match the required specifications.

Learn how to construct a quadrilateral and triangle with equal areas using given side lengths. This geometry problem involves applying Heron's formula to calculate areas accurately. Suitable for high school students studying geometry.

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