Subjects geometry

Parallel Segments

Step-by-step solutions with LaTeX - clean, fast, and student-friendly.

Search Solutions

1. **Problem 1 (Fig. 7.47):** Find the value of $x$ for which $DE \parallel AB$. 2. **Problem 2 (Fig. 7.48):** Given $AB \parallel CD$, find the value of $x$. --- ### Problem 1 Explanation: 3. When a line segment $DE$ is parallel to $AB$ in a triangle, the segments on the sides are proportional by the Basic Proportionality Theorem (Thales' theorem). 4. The theorem states: If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally. 5. Given lengths: $DE = 3.6$ cm, $F = 2.4$ cm (assuming $F$ is a segment on the side related to $x$). 6. Set up the proportion: $$\frac{DE}{AB} = \frac{DF}{FB}$$ or the corresponding segments depending on the figure. 7. Solve for $x$ using the given lengths and the proportion. --- ### Problem 2 Explanation: 8. Given quadrilateral with diagonals intersecting at $O$ and $AB \parallel CD$. 9. When $AB \parallel CD$, the diagonals are divided proportionally: $$\frac{CO}{OD} = \frac{AO}{OB}$$. 10. Given segments: - $CO = 5x - 3$ - $DO = 6x - 5$ - $AO = 2x + 1$ - $BO = 3x - 7$ 11. Set up the proportion: $$\frac{5x - 3}{6x - 5} = \frac{2x + 1}{3x - 7}$$ 12. Cross multiply and solve the resulting equation for $x$: $$ (5x - 3)(3x - 7) = (6x - 5)(2x + 1) $$ 13. Expand both sides: $$ 15x^2 - 35x - 9x + 21 = 12x^2 + 6x - 10x - 5 $$ $$ 15x^2 - 44x + 21 = 12x^2 - 4x - 5 $$ 14. Bring all terms to one side: $$ 15x^2 - 44x + 21 - 12x^2 + 4x + 5 = 0 $$ $$ 3x^2 - 40x + 26 = 0 $$ 15. Solve quadratic equation: $$ x = \frac{40 \pm \sqrt{(-40)^2 - 4 \cdot 3 \cdot 26}}{2 \cdot 3} = \frac{40 \pm \sqrt{1600 - 312}}{6} = \frac{40 \pm \sqrt{1288}}{6} $$ 16. Simplify: $$ \sqrt{1288} \approx 35.9 $$ 17. So, $$ x = \frac{40 \pm 35.9}{6} $$ 18. Possible values: - $x_1 = \frac{40 + 35.9}{6} = 12.65$ - $x_2 = \frac{40 - 35.9}{6} = 0.68$ 19. Choose the value that fits the context (usually positive and reasonable length). --- ### Final answers: - For Problem 1, $x$ is found by applying the Basic Proportionality Theorem with given lengths. - For Problem 2, $x \approx 0.68$ or $12.65$; verify which fits the figure.