Trapezium Angles Lengths

1. **State the problem:** We have trapezium ABCD with given sides AB=4 cm, AD=5 cm, BD=7 cm, and angle at C = 110°. We need to find (a)(i) angle \(\angle ABD\), (a)(ii) length BC, and (b)(ii) area of triangle AA'D where A' is on line BA extended such that AD = A'D. 2. **Find \(\angle ABD\):** - Triangle ABD has sides AB=4 cm, AD=5 cm, BD=7 cm. - Use the Law of Cosines to find \(\angle ABD\) opposite side AD: $$\cos(\angle ABD) = \frac{AB^2 + BD^2 - AD^2}{2 \times AB \times BD} = \frac{4^2 + 7^2 - 5^2}{2 \times 4 \times 7} = \frac{16 + 49 - 25}{56} = \frac{40}{56} = \frac{5}{7} \approx 0.7143$$ - Therefore, $$\angle ABD = \cos^{-1}(0.7143) \approx 44.42^\circ$$ 3. **Find length BC:** - Since ABCD is a trapezium with AB parallel to DC, and angle at C is 110°, we can use the Law of Cosines in triangle BCD. - We know BD=7 cm, DC = a cm (unknown), and angle at C = 110°. - To find BC, we need to find DC first. 4. **Find DC:** - Using trapezium properties, AB is parallel to DC, so angles at B and C are supplementary. - Angle at B is \(\angle ABC = 70^\circ\) because \(\angle ABD = 44.42^\circ\) and \(\angle ABC = 180^\circ - 110^\circ = 70^\circ\). - Using triangle BCD with angle C = 110°, BD=7 cm, and angle B = 70°, apply Law of Cosines or Law of Sines. 5. **Use Law of Cosines in triangle BCD to find BC:** - Let BC = x. - Using Law of Cosines: $$BD^2 = BC^2 + DC^2 - 2 \times BC \times DC \times \cos(110^\circ)$$ - But DC is unknown, so use Law of Sines instead: $$\frac{BD}{\sin(\angle BCD)} = \frac{BC}{\sin(\angle BDC)} = \frac{DC}{\sin(\angle B)}$$ - Angles in triangle BCD sum to 180°, so \(\angle BDC = 180^\circ - 110^\circ - 70^\circ = 0^\circ\), which is impossible. 6. **Reconsider approach:** - Since angle at C is 110°, and AB is parallel to DC, angle at B is 70°. - Triangle BCD has sides BD=7 cm, angle C=110°, angle B=70°, so angle D=0°, which is impossible. - Therefore, BC can be found by considering coordinates or vector approach. 7. **Coordinate geometry approach:** - Place point D at origin (0,0). - Since AD=5 cm and AB=4 cm, and angle at C=110°, assign coordinates: - Let D = (0,0), C = (a,0) since DC is horizontal. - Point A is at (0,5) because AD=5 cm vertical. - Point B is at (4,5) since AB=4 cm horizontal. - Calculate BD distance: $$BD = \sqrt{(4-0)^2 + (5-0)^2} = \sqrt{16 + 25} = \sqrt{41} \approx 6.4 \neq 7$$ - Given BD=7, adjust coordinates accordingly. 8. **Use Law of Cosines in triangle ABD to find angle at A:** - Using sides AB=4, AD=5, BD=7: $$BD^2 = AB^2 + AD^2 - 2 \times AB \times AD \times \cos(\angle BAD)$$ $$49 = 16 + 25 - 40 \cos(\angle BAD)$$ $$49 = 41 - 40 \cos(\angle BAD)$$ $$40 \cos(\angle BAD) = 41 - 49 = -8$$ $$\cos(\angle BAD) = -\frac{8}{40} = -0.2$$ $$\angle BAD = \cos^{-1}(-0.2) \approx 101.54^\circ$$ 9. **Calculate length BC:** - Since angle at C is 110°, and DC is unknown, use Law of Cosines in triangle BCD: - Let BC = x, DC = a. - Using Law of Cosines: $$BD^2 = BC^2 + DC^2 - 2 \times BC \times DC \times \cos(110^\circ)$$ - We know BD=7, angle C=110°, but two unknowns BC and DC. 10. **Use trapezium property:** - Since AB is parallel to DC, and AB=4 cm, DC = 4 cm. - So DC = 4 cm. 11. **Calculate BC:** $$7^2 = BC^2 + 4^2 - 2 \times BC \times 4 \times \cos(110^\circ)$$ $$49 = BC^2 + 16 - 8 BC \times \cos(110^\circ)$$ - \(\cos(110^\circ) = -0.3420\) $$49 = BC^2 + 16 + 8 BC \times 0.3420$$ $$49 = BC^2 + 16 + 2.736 BC$$ $$BC^2 + 2.736 BC + 16 - 49 = 0$$ $$BC^2 + 2.736 BC - 33 = 0$$ 12. **Solve quadratic for BC:** $$BC = \frac{-2.736 \pm \sqrt{2.736^2 - 4 \times 1 \times (-33)}}{2}$$ $$= \frac{-2.736 \pm \sqrt{7.487 + 132}}{2} = \frac{-2.736 \pm \sqrt{139.487}}{2}$$ $$= \frac{-2.736 \pm 11.81}{2}$$ - Positive root: $$BC = \frac{-2.736 + 11.81}{2} = \frac{9.074}{2} = 4.537 \text{ cm}$$ 13. **Find point A' such that AD = A'D:** - Since A' lies on line BA extended beyond A, and AD=5 cm, A'D=5 cm. - Vector BA has length 4 cm, so extend BA by 5 cm beyond A. 14. **Calculate area of triangle AA'D:** - Triangle AA'D is isosceles with sides AD = A'D = 5 cm. - Base AA' = length of extension = 5 cm. - Height can be found using Pythagoras: $$\text{height} = \sqrt{5^2 - \left(\frac{5}{2}\right)^2} = \sqrt{25 - 6.25} = \sqrt{18.75} = 4.33 \text{ cm}$$ - Area: $$\frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5 \times 4.33 = 10.83 \text{ cm}^2$$ **Final answers:** - (a)(i) \(\angle ABD \approx 44.42^\circ\) - (a)(ii) BC \(\approx 4.54\) cm - (b)(ii) Area of \(\triangle AA'D \approx 10.83\) cm\(^2\)