1. **State the problem:** We have points $A(0,135)$ and $C(110,80)$, and point $P$ lies on segment $AC$ dividing it in the ratio $4:7$. Line $AC$ is perpendicular to line $PB$. We want to find the area of triangle $\triangle ABP$. 2. **Find coordinates of point $P$:** Since $P$ divides $AC$ in ratio $4:7$, use section formula: $$P = \left(\frac{4 \times 110 + 7 \times 0}{4+7}, \frac{4 \times 80 + 7 \times 135}{4+7}\right) = \left(\frac{440}{11}, \frac{320 + 945}{11}\right) = (40, 117)$$ 3. **Find slope of $AC$:** $$m_{AC} = \frac{80 - 135}{110 - 0} = \frac{-55}{110} = -\frac{1}{2}$$ 4. **Since $AC \perp PB$, slope of $PB$ is negative reciprocal of $m_{AC}$:** $$m_{PB} = 2$$ 5. **Find equation of line $PB$ passing through $P(40,117)$:** $$y - 117 = 2(x - 40) \implies y = 2x + 37$$ 6. **Find point $B$ as intersection of line $PB$ and vertical line through $A$ (since $A$ is at $x=0$):** At $x=0$, $$y = 2(0) + 37 = 37$$ So, $B = (0,37)$. 7. **Calculate lengths $BP$ and $AP$ for area formula:** - $BP = |117 - 37| = 80$ (vertical distance) - $AP = $ distance between $A(0,135)$ and $P(40,117)$: $$AP = \sqrt{(40-0)^2 + (117-135)^2} = \sqrt{1600 + 324} = \sqrt{1924} = 2\sqrt{481}$$ 8. **Calculate area of $\triangle ABP$ using formula:** $$\text{Area} = \frac{1}{2} \times BP \times AP = \frac{1}{2} \times 80 \times 2\sqrt{481} = 80\sqrt{481}$$ 9. **Approximate numeric value:** $$\sqrt{481} \approx 21.93 \implies \text{Area} \approx 80 \times 21.93 = 1754.4$$ **Final answer:** The area of the park $\triangle ABP$ is $80\sqrt{481} \approx 1754.4$ square units.