Let's solve the problem step by step: --- ## **Problem Understanding** We are asked to find the height \( h \) up to which liquid must be poured into a conical cup so that the liquid is in contact with **three-fourths** of the **internal surface area** of the cone. ### **Given Data** - **Radius of top of cone** (\( R \)): \( 40 \) mm - **Height of cone** (\( H \)): \( 160 \) mm --- ## **Step 1: Find Total Lateral Surface Area of the Cone** The **lateral surface area** (not including the base) of a cone is: \[ A_{total} = \pi R s \] Where \( s \) is the slant height: \[ s = \sqrt{R^2 + H^2} \] Substitute the values: \[ s = \sqrt{40^2 + 160^2} = \sqrt{160 + 25600} = \sqrt{27200} = 164.92 \text{ mm} \] \[ A_{total} = \pi \times 40 \times 164.92 = 20,717.2 \text{ mm}^2 \] --- ## **Step 2: Surface Area in Contact with Liquid** Let \( h \) be the height to which the liquid rises. The radius at height \( h \) (call it \( r \)) is found using similarity of triangles: \[ \frac{r}{h} = \frac{R}{H} \implies r = \frac{R}{H} h = \frac{40}{160}h = \frac{h}{4} \] The slant height up to \( h \) (call it \( s_1 \)): \[ s_1 = \sqrt{r^2 + h^2} = \sqrt{\left(\frac{h}{4}\right)^2 + h^2} = \sqrt{\frac{h^2}{16} + h^2} = \sqrt{\frac{17h^2}{16}} = \frac{\sqrt{17}}{4}h \] So, the **lateral area in contact with the liquid** is: \[ A_{liq} = \pi r s_1 = \pi \left(\frac{h}{4}\right) \left(\frac{\sqrt{17}}{4}h\right) = \pi \frac{\sqrt{17}}{16} h^2 \] --- ## **Step 3: Equation for 3/4 of Surface Area** We set: \[ A_{liq} = \frac{3}{4} A_{total} \] \[ \pi \frac{\sqrt{17}}{16} h^2 = \frac{3}{4} \times \pi \times 40 \times 164.92 \] \[ \frac{\sqrt{17}}{16} h^2 = \frac{3}{4} \times 40 \times 164.92 \] \[ \frac{\sqrt{17}}{16} h^2 = 3 \times 10 \times 164.92 = 4947.6 \] \[ h^2 = \frac{16}{\sqrt{17}} \times 4947.6 \] \[ h^2 = \frac{16 \times 4947.6}{\sqrt{17}} \] Calculate \(\sqrt{17} \approx 4.1231\): \[ h^2 = \frac{16 \times 4947.6}{4.1231} = \frac{79161.6}{4.1231} \approx 19,200 \] \[ h = \sqrt{19,200} \approx 138.6 \text{ mm} \] --- ## **Final Answer** \[ \boxed{h \approx 139 \text{ mm}} \] --- ### **Summary of Steps** 1. **Find total lateral area of the cone.** 2. **Express the lateral area in contact with the liquid as a function of \( h \).** 3. **Set this area to \( \frac{3}{4} \) of the total area and solve for \( h \).** If you need any clarification about the use of Pappus-Guldin theorem or the geometry, let me know!