1. **Stating the problem:** We are given a right triangle ABC with a right angle at A. Point D lies on segment CB such that AD is perpendicular to CB. The area of triangle ADC is 160 cm². We need to find the area of triangle ADB. 2. **Understanding the figure and given data:** - Triangle ABC is right-angled at A. - D lies on CB, dividing it into segments CD = 4k and DB = 3k. - AD is perpendicular to CB. - Area of triangle ADC = 160 cm². 3. **Formula for the area of a triangle:** The area of a triangle is given by $$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$$ 4. **Applying the formula to triangles ADC and ADB:** Since AD is perpendicular to CB, AD is the height for both triangles ADC and ADB, and CD and DB are their respective bases. 5. **Expressing areas in terms of k and AD:** - Area of triangle ADC: $$160 = \frac{1}{2} \times 4k \times AD = 2k \times AD$$ - Area of triangle ADB: $$\text{Area}_{ADB} = \frac{1}{2} \times 3k \times AD = \frac{3}{2} k \times AD$$ 6. **Finding AD in terms of k:** From the first equation: $$160 = 2k \times AD \implies AD = \frac{160}{2k} = \frac{80}{k}$$ 7. **Calculating area of triangle ADB:** Substitute $AD = \frac{80}{k}$ into the area formula for ADB: $$\text{Area}_{ADB} = \frac{3}{2} k \times \frac{80}{k} = \frac{3}{2} \times 80 = 120$$ **Final answer:** The area of triangle ADB is **120 cm²**. Hence, the correct choice is (c) 120.