Polygon Construction Edff06

1. **Problem:** Construct an irregular polygon ABCDE with given sides and angles, then measure the length of AE. 2. **Step 1: Understand the polygon construction** - AB is horizontal and $AB=35$mm. - BC is inclined at $120^\circ$ to AB and $BC=40$mm. - CD is perpendicular ($90^\circ$) to BC and $CD=80$mm. - DE is at $75^\circ$ to CD and $DE=33$mm. 3. **Step 2: Use vector or coordinate geometry to find coordinates of points** - Place $A$ at origin $(0,0)$. - $B$ at $(35,0)$ since AB is horizontal. - Calculate $C$ using $B$ and length and angle: $$C_x = 35 + 40 \cos 120^\circ = 35 + 40 \times (-0.5) = 35 - 20 = 15$$ $$C_y = 0 + 40 \sin 120^\circ = 0 + 40 \times 0.866 = 34.64$$ So, $C = (15, 34.64)$. - $D$ is $80$mm perpendicular to BC. Since BC is at $120^\circ$, CD is at $120^\circ + 90^\circ = 210^\circ$. $$D_x = 15 + 80 \cos 210^\circ = 15 + 80 \times (-0.866) = 15 - 69.28 = -54.28$$ $$D_y = 34.64 + 80 \sin 210^\circ = 34.64 + 80 \times (-0.5) = 34.64 - 40 = -5.36$$ So, $D = (-54.28, -5.36)$. - $E$ is $33$mm at $75^\circ$ to CD. Angle of CD is $210^\circ$, so DE is at $210^\circ + 75^\circ = 285^\circ$. $$E_x = -54.28 + 33 \cos 285^\circ = -54.28 + 33 \times 0.2588 = -54.28 + 8.54 = -45.74$$ $$E_y = -5.36 + 33 \sin 285^\circ = -5.36 + 33 \times (-0.9659) = -5.36 - 31.87 = -37.23$$ So, $E = (-45.74, -37.23)$. 4. **Step 3: Calculate length AE** $$AE = \sqrt{(x_E - x_A)^2 + (y_E - y_A)^2} = \sqrt{(-45.74 - 0)^2 + (-37.23 - 0)^2}$$ $$= \sqrt{(-45.74)^2 + (-37.23)^2} = \sqrt{2092.7 + 1386.1} = \sqrt{3478.8} = 59.0 \text{ mm}$$ --- 5. **Step 4: Construct a similar polygon on a new base 52mm** - Similar polygons have proportional sides. - Scale factor $k = \frac{52}{35} = 1.4857$. - Multiply all sides by $k$ to get new polygon. --- 6. **Step 5: Construct a triangle equal in area to the triangle in question 2** - Calculate area of triangle in question 2 (polygon with base 52mm). - Use coordinates scaled by $k$. - Construct a triangle with same area using standard methods. --- 7. **Step 6: Construct an equilateral triangle with side 80mm and a rectangle on the same base equal in area to the triangle** - Area of equilateral triangle: $$A = \frac{\sqrt{3}}{4} \times 80^2 = \frac{\sqrt{3}}{4} \times 6400 = 2771.3 \text{ mm}^2$$ - Rectangle with base 80mm and height $h$ equal to triangle area: $$80 \times h = 2771.3 \Rightarrow h = \frac{2771.3}{80} = 34.64 \text{ mm}$$ **Final answer:** - Length $AE = 59.0$ mm.