1. **State the problem:** We have rectangle ABDE and parallelogram BCDF. We need to find the perpendicular distance from point B to line DF. 2. **Identify given measurements:** - BF = 31 cm - FD = 62 cm (diagonal of parallelogram) - BD = 64 cm (common side) 3. **Analyze the shapes:** - Since ABDE is a rectangle, sides AB and DE are perpendicular to BD. - Parallelogram BCDF shares side BD with the rectangle. 4. **Using properties of parallelogram:** The diagonal FD splits it into two triangles: BFD and CFD. 5. **Find area of triangle BFD using sides BF, BD, and angle:** We do not have angles directly, so consider using the Law of Cosines in triangle BFD to find angle BFD. 6. **Apply Law of Cosines in triangle BFD:** $$BD^2 = BF^2 + FD^2 - 2 \cdot BF \cdot FD \cdot \cos(\angle BFD)$$ Plug in values: $$64^2 = 31^2 + 62^2 - 2 \cdot 31 \cdot 62 \cdot \cos(\angle BFD)$$ $$4096 = 961 + 3844 - 3844 \cdot \cos(\angle BFD)$$ $$4096 = 4805 - 3844 \cdot \cos(\angle BFD)$$ 7. **Solve for \(\cos(\angle BFD)\):** $$3844 \cdot \cos(\angle BFD) = 4805 - 4096 = 709$$ $$\cos(\angle BFD) = \frac{709}{3844} \approx 0.1845$$ 8. **Find sine of the angle:** $$\sin(\angle BFD) = \sqrt{1 - (0.1845)^2} = \sqrt{1 - 0.0340} = \sqrt{0.966} \approx 0.983$$ 9. **Calculate area of triangle BFD:** Area = \(\frac{1}{2} \times BF \times FD \times \sin(\angle BFD)\) $$= \frac{1}{2} \times 31 \times 62 \times 0.983 \approx 945.9\text{ cm}^2$$ 10. **Calculate the height (perpendicular distance) from B to line DF:** Area = \(\frac{1}{2} \times FD \times \text{height}\) So, $$945.9 = \frac{1}{2} \times 62 \times \text{height}$$ $$\text{height} = \frac{2 \times 945.9}{62} = \frac{1891.8}{62} \approx 30.51 \text{ cm}$$ **Final answer:** The perpendicular distance from B to DF is approximately $30.51$ cm.