1. **Problem Statement:** We have a cyclic quadrilateral ABCD inscribed in a circle with given sides: $AB=22$, $BC=35$, $CD=24$, and diagonal $AC=40$. We need to find: (a) length $AD$, (b) length $BD$, (c) the area of the quadrilateral. 2. **Key Formula (Ptolemy's Theorem):** For a cyclic quadrilateral, Ptolemy's theorem states: $$AC \times BD = AB \times CD + BC \times AD$$ This relates the sides and diagonals. 3. **Step (a) and (b) - Find $AD$ and $BD$:** We know $AC=40$, $AB=22$, $BC=35$, $CD=24$. Let $AD = x$ and $BD = y$. From Ptolemy's theorem: $$40y = 22 \times 24 + 35x$$ $$40y = 528 + 35x \quad (1)$$ 4. **Using Law of Cosines in triangles $ABC$ and $ADC$ to relate $x$ and $y$: ** Since $ABCD$ is cyclic, angles subtended by the same chord are equal. We use the fact that triangles $ABC$ and $ADC$ share diagonal $AC$. In triangle $ABC$: $$BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(\angle BAC)$$ $$35^2 = 22^2 + 40^2 - 2 \times 22 \times 40 \times \cos(\angle BAC)$$ $$1225 = 484 + 1600 - 1760 \cos(\angle BAC)$$ $$1225 = 2084 - 1760 \cos(\angle BAC)$$ $$1760 \cos(\angle BAC) = 2084 - 1225 = 859$$ $$\cos(\angle BAC) = \frac{859}{1760}$$ In triangle $ADC$: $$CD^2 = AD^2 + AC^2 - 2 \times AD \times AC \times \cos(\angle DAC)$$ Since $\angle DAC = \angle BAC$ (angles subtended by chord $AC$), $$24^2 = x^2 + 40^2 - 2 \times x \times 40 \times \frac{859}{1760}$$ $$576 = x^2 + 1600 - \frac{2 \times 40 \times 859}{1760} x$$ Calculate coefficient: $$\frac{2 \times 40 \times 859}{1760} = \frac{68720}{1760} = 39$$ So: $$576 = x^2 + 1600 - 39x$$ Rearranged: $$x^2 - 39x + 1600 - 576 = 0$$ $$x^2 - 39x + 1024 = 0$$ 5. **Solve quadratic for $x$ (AD):** $$x = \frac{39 \pm \sqrt{39^2 - 4 \times 1024}}{2} = \frac{39 \pm \sqrt{1521 - 4096}}{2}$$ Discriminant: $$1521 - 4096 = -2575 < 0$$ No real solution here, so check calculations. Recalculate coefficient carefully: $$\frac{2 \times 40 \times 859}{1760} = \frac{68720}{1760}$$ Divide numerator and denominator by 80: $$\frac{859}{22} \approx 39.045$$ So coefficient is approximately 39.045. Rewrite equation: $$576 = x^2 + 1600 - 39.045 x$$ $$x^2 - 39.045 x + 1024 = 0$$ Discriminant: $$39.045^2 - 4 \times 1024 = 1524.5 - 4096 = -2571.5$$ Still negative. This suggests an error in approach. Instead, use Ptolemy's theorem to express $y$ in terms of $x$ and then use Law of Cosines in triangle $BCD$ or $ABD$. 6. **Alternative approach: Use Ptolemy's theorem and Law of Cosines in triangle $BCD$ to find $y$ and $x$.** From (1): $$40y = 528 + 35x \Rightarrow y = \frac{528 + 35x}{40}$$ Triangle $BCD$ has sides $BC=35$, $CD=24$, and diagonal $BD=y$. Use Law of Cosines in triangle $BCD$: $$BD^2 = BC^2 + CD^2 - 2 \times BC \times CD \times \cos(\angle BCD)$$ Similarly, triangle $ABD$ has sides $AB=22$, $AD=x$, and diagonal $BD=y$. Use Law of Cosines in triangle $ABD$: $$BD^2 = AB^2 + AD^2 - 2 \times AB \times AD \times \cos(\angle BAD)$$ Since $ABCD$ is cyclic, $\angle BCD = 180^\circ - \angle BAD$, so $$\cos(\angle BCD) = -\cos(\angle BAD)$$ Set the two expressions for $BD^2$ equal: $$BC^2 + CD^2 - 2 BC CD \cos(\angle BCD) = AB^2 + AD^2 - 2 AB AD \cos(\angle BAD)$$ Substitute $\cos(\angle BCD) = -\cos(\angle BAD)$: $$BC^2 + CD^2 + 2 BC CD \cos(\angle BAD) = AB^2 + AD^2 - 2 AB AD \cos(\angle BAD)$$ Group terms: $$2 \cos(\angle BAD)(BC CD + AB AD) = AB^2 + AD^2 - BC^2 - CD^2$$ Plug in values: $$2 \cos(\angle BAD)(35 \times 24 + 22 x) = 22^2 + x^2 - 35^2 - 24^2$$ Calculate constants: $$2 \cos(\angle BAD)(840 + 22x) = 484 + x^2 - 1225 - 576$$ $$2 \cos(\angle BAD)(840 + 22x) = x^2 - 1317$$ 7. **Use Law of Cosines in triangle $ABD$ to express $\cos(\angle BAD)$:** $$y^2 = 22^2 + x^2 - 2 \times 22 \times x \times \cos(\angle BAD)$$ $$y^2 = 484 + x^2 - 44 x \cos(\angle BAD)$$ Rearranged: $$\cos(\angle BAD) = \frac{484 + x^2 - y^2}{44 x}$$ 8. **Substitute $\cos(\angle BAD)$ into previous equation:** $$2 \times \frac{484 + x^2 - y^2}{44 x} (840 + 22 x) = x^2 - 1317$$ Simplify left side: $$\frac{2 (484 + x^2 - y^2)(840 + 22 x)}{44 x} = x^2 - 1317$$ Multiply both sides by $44 x$: $$2 (484 + x^2 - y^2)(840 + 22 x) = 44 x (x^2 - 1317)$$ 9. **Recall from (1):** $$y = \frac{528 + 35 x}{40}$$ So: $$y^2 = \left(\frac{528 + 35 x}{40}\right)^2 = \frac{(528 + 35 x)^2}{1600}$$ 10. **Substitute $y^2$ and expand:** $$2 (484 + x^2 - \frac{(528 + 35 x)^2}{1600})(840 + 22 x) = 44 x (x^2 - 1317)$$ This is a complicated algebraic equation in $x$. To simplify, solve numerically. 11. **Numerical solution for $x$ (AD):** Try approximate values near 30: - For $x=30$, compute left and right sides. Calculate $y$: $$y = \frac{528 + 35 \times 30}{40} = \frac{528 + 1050}{40} = \frac{1578}{40} = 39.45$$ Calculate left side: $$484 + 900 - \frac{(528 + 1050)^2}{1600} = 1384 - \frac{1578^2}{1600}$$ $$1578^2 = 2,490,084$$ $$\frac{2,490,084}{1600} = 1556.3$$ $$1384 - 1556.3 = -172.3$$ $$2 \times (-172.3) \times (840 + 660) = -344.6 \times 1500 = -516,900$$ Right side: $$44 \times 30 \times (900 - 1317) = 1320 \times (-417) = -550,440$$ Close values, so $x \approx 30$. 12. **Calculate $AD \approx 30$ m and $BD \approx 39.45$ m.** 13. **Step (c) - Area of quadrilateral ABCD:** Use Brahmagupta's formula for cyclic quadrilateral area: $$s = \frac{AB + BC + CD + AD}{2} = \frac{22 + 35 + 24 + 30}{2} = \frac{111}{2} = 55.5$$ Area: $$\sqrt{(s - AB)(s - BC)(s - CD)(s - AD)} = \sqrt{(55.5 - 22)(55.5 - 35)(55.5 - 24)(55.5 - 30)}$$ $$= \sqrt{33.5 \times 20.5 \times 31.5 \times 25.5}$$ Calculate product: $$33.5 \times 20.5 = 686.75$$ $$31.5 \times 25.5 = 803.25$$ $$686.75 \times 803.25 \approx 551,500$$ Area $\approx \sqrt{551,500} \approx 742.6$ m$^2$. **Final answers:** (a) $AD \approx 30$ m (b) $BD \approx 39.45$ m (c) Area $\approx 742.6$ m$^2$