1. **Problem Statement:** Construct quadrilateral ABCD with given sides $AB=4.5$ cm, $AC=CD=5$ cm, $AD=6$ cm, and angle $\angle BAC=60^\circ$. Then construct triangle PBC such that its area equals the area of quadrilateral ABCD. 2. **Step 1: Construct triangle ABC** - Draw segment $AB=4.5$ cm. - At point A, construct angle $\angle BAC=60^\circ$. - On the ray from A forming this angle, mark point C such that $AC=5$ cm. 3. **Step 2: Locate point D** - Since $CD=5$ cm and $AD=6$ cm, point D lies at the intersection of two circles: - Circle centered at C with radius 5 cm. - Circle centered at A with radius 6 cm. - Find intersection point D of these two circles. 4. **Step 3: Quadrilateral ABCD is formed by points A, B, C, D.** 5. **Step 4: Calculate area of quadrilateral ABCD** - Divide ABCD into triangles ABC and ACD. - Area of $\triangle ABC$ using formula: $$\text{Area} = \frac{1}{2} AB \times AC \times \sin(\angle BAC) = \frac{1}{2} \times 4.5 \times 5 \times \sin 60^\circ = \frac{1}{2} \times 4.5 \times 5 \times \frac{\sqrt{3}}{2} = 9.74 \text{ cm}^2$$ - Area of $\triangle ACD$ using Heron's formula: - Sides: $AC=5$, $CD=5$, $AD=6$ - Semi-perimeter $s = \frac{5+5+6}{2} = 8$ - Area: $$\sqrt{s(s-AC)(s-CD)(s-AD)} = \sqrt{8(8-5)(8-5)(8-6)} = \sqrt{8 \times 3 \times 3 \times 2} = \sqrt{144} = 12 \text{ cm}^2$$ - Total area of quadrilateral ABCD: $$9.74 + 12 = 21.74 \text{ cm}^2$$ 6. **Step 5: Construct triangle PBC with area equal to quadrilateral ABCD** - Base $BC$ is known from points B and C. - Calculate length $BC$ using coordinates or law of cosines. - Use formula for area of triangle: $$\text{Area} = \frac{1}{2} \times BC \times h = 21.74$$ - Solve for height $h$: $$h = \frac{2 \times 21.74}{BC}$$ - Construct point P on the line perpendicular to $BC$ at B or C at distance $h$ to form triangle PBC. 7. **Visualization:** - Quadrilateral ABCD and triangle PBC share side BC. - Triangle PBC has area equal to quadrilateral ABCD. Final answer: The quadrilateral ABCD has area approximately $21.74$ cm$^2$, and triangle PBC is constructed with base $BC$ and height $h=\frac{2 \times 21.74}{BC}$ to have the same area.