In triangle ABC, AD is the internal angle bisector of angle ∠A, and the sides satisfy AB : AC = 3 : 4. If the area of triangle ABC is 350 cm², what is the area of triangle ABD in square centimetres?

Introduction / Context:
This problem tests the angle bisector theorem and how it relates to areas of sub triangles within a larger triangle. When an internal angle bisector is drawn from a vertex to the opposite side, it divides that side in the ratio of the adjacent sides. Because triangles sharing the same altitude have areas proportional to their bases, this side ratio directly translates into an area ratio. Given the overall area of triangle ABC, we can use these relationships to find the area of triangle ABD.

Given Data / Assumptions:
- Triangle ABC is any triangle. - AD is the internal angle bisector of angle ∠A. - Side lengths adjacent to angle A satisfy AB : AC = 3 : 4. - The total area of triangle ABC is 350 cm². - We need the area of triangle ABD.

Concept / Approach:
The angle bisector theorem states that if AD is the internal bisector of angle A meeting BC at D, then BD / DC = AB / AC. Given AB : AC = 3 : 4, it follows that BD : DC = 3 : 4. Triangles ABD and ACD share the same altitude drawn from A to side BC, so their areas are proportional to the lengths of their bases BD and DC. Therefore, area(ABD) : area(ACD) = BD : DC = 3 : 4. Since the total area is the sum of these two areas, we can split 350 cm² in the ratio 3 : 4 to obtain the required area.

Step-by-Step Solution:
Step 1: By the angle bisector theorem, BD / DC = AB / AC. Step 2: Given AB : AC = 3 : 4, we get BD : DC = 3 : 4. Step 3: Triangles ABD and ACD share the altitude from vertex A to side BC, so their heights to BC are equal. Step 4: Therefore, their areas are proportional to their bases on BC: area(ABD) : area(ACD) = BD : DC = 3 : 4. Step 5: Let area(ABD) = 3k and area(ACD) = 4k for some positive k. Step 6: The total area of triangle ABC is 350 cm², so 3k + 4k = 7k = 350. Step 7: Solve for k: k = 350 / 7 = 50. Step 8: Compute area(ABD) = 3k = 3 * 50 = 150 cm².

Verification / Alternative check:
As a check, compute area(ACD) = 4k = 4 * 50 = 200 cm². The total area is then 150 + 200 = 350 cm², which matches the given area of triangle ABC. Additionally, if we imagine a concrete example with AB = 3 units and AC = 4 units and choose BC such that the triangle is non degenerate, the angle bisector theorem would again produce side segments BD and DC in ratio 3 : 4, and the corresponding areas would share the same ratio. This conceptual consistency supports the result.

Why Other Options Are Wrong:
Option 200 is actually the area of triangle ACD, not ABD, if the ratio 3 : 4 is applied correctly. Option 210 would require the areas to be split in a different ratio, which contradicts AB : AC = 3 : 4. Option 240 also does not satisfy the ratio 3 : 4 when combined with the remaining area, since 240 : 110 does not match 3 : 4.

Common Pitfalls:
A frequent error is to assume that an angle bisector also bisects the opposite side, which is only true in special cases such as isosceles triangles with two equal adjacent sides. Another mistake is to try to use trigonometric area formulas rather than exploiting the simpler base height relationship and the angle bisector theorem. Students sometimes also invert the ratio AB : AC when applying it to BD : DC, leading to incorrect area splits. Carefully matching the side adjacent to AB with the corresponding base segment BD helps avoid this mistake.

Final Answer:
The area of triangle ABD is 150 square centimetres.