Triangle ABC is a right angled triangle with angle A = 90°, AB = 5 cm and AC = 12 cm. The triangle is rotated about its hypotenuse BC to form a double cone (two right circular cones with a common base). What is the approximate total volume (in cubic cm) of this double cone?

Introduction / Context:
This is an advanced geometry and mensuration problem that combines right triangles, rotation in three dimensions and the volume formula for cones. When a right angled triangle is rotated about its hypotenuse, it generates a solid known as a double cone (two right circular cones joined at a common base). The problem requires careful use of right triangle properties and cone volume formulas.

Given Data / Assumptions:

• Triangle ABC is right angled at A.
• AB = 5 cm and AC = 12 cm are the legs of the right triangle.
• Hypotenuse BC is the axis of rotation.
• On rotation of triangle ABC around BC, a double cone is formed.
• We use π = 22 / 7 for approximate volume calculation.
• We must find the total volume of the two cones formed.

Concept / Approach:
First, we find the length of the hypotenuse BC using Pythagoras theorem. When the triangle rotates about BC, the right angle vertex A traces a circle whose radius equals the perpendicular distance from A to BC (the altitude from A to BC). The point where this altitude meets BC splits BC into two segments; each segment is the height of one cone, and the common base radius is this altitude. The total volume of the double cone is: V_total = (1 / 3) * π * r^2 * h1 + (1 / 3) * π * r^2 * h2 = (1 / 3) * π * r^2 * (h1 + h2) Here h1 + h2 equals the hypotenuse BC. The altitude r from the right angle to the hypotenuse can be expressed in terms of the legs AB and AC.

Step-by-Step Solution:
Step 1: Use Pythagoras theorem to find BC: BC^2 = AB^2 + AC^2 = 5^2 + 12^2 = 25 + 144 = 169, so BC = 13 cm. Step 2: The altitude from A to BC, call it h, can be found using the area of the triangle. Step 3: Area of triangle ABC = (1 / 2) * AB * AC = (1 / 2) * 5 * 12 = 30 square cm. Step 4: The same area is also (1 / 2) * BC * h, so 30 = (1 / 2) * 13 * h. Step 5: Solve for h: h = (30 * 2) / 13 = 60 / 13 cm. Step 6: When rotated about BC, point A traces a circle of radius r = h = 60 / 13 cm, which becomes the radius of the common circular base of the double cone. Step 7: The two cones have heights equal to the two segments into which BC is divided, but their sum is BC = 13 cm, so h1 + h2 = 13. Step 8: Therefore total volume is V_total = (1 / 3) * π * r^2 * (h1 + h2) = (1 / 3) * π * (60 / 13)^2 * 13. Step 9: Simplify r^2: (60 / 13)^2 = 3600 / 169. Step 10: So V_total = (1 / 3) * π * (3600 / 169) * 13 = (1 / 3) * π * (3600 / 13). Step 11: Thus V_total = (1200 / 13) * π cubic cm. Step 12: Using π ≈ 22 / 7, V_total ≈ (1200 / 13) * (22 / 7) cubic cm. Step 13: Compute approximately: 1200 * 22 = 26400, and 13 * 7 = 91, so V_total ≈ 26400 / 91 ≈ 290 cubic cm (approximately).

Verification / Alternative check:
We can also compute the lengths of the two segments of BC and volumes of each cone separately using the relations AB^2 = (segment1) * BC and AC^2 = (segment2) * BC. This leads back to the same total volume expression, confirming that the shortcut formula V_total = (1 / 3) * π * AB^2 * AC^2 / BC is correct and yields the same numerical result near 290 cubic cm.

Why Other Options Are Wrong:
145 cubic cm: This is roughly half the correct value and would correspond to using only one cone volume instead of both.
435 cubic cm and 580 cubic cm: These values are significantly larger than the correct volume and would arise from miscalculations such as squaring the hypotenuse again or adding extra factors of π.
260 cubic cm: This underestimates the volume and does not correspond to any consistent application of the formula.

Common Pitfalls:
A key difficulty is visualizing the double cone and identifying the correct radius and heights. Another common mistake is to treat one of the legs as the radius instead of the altitude to the hypotenuse. Errors in using the area relation or mishandling the fractions also occur. Keeping the geometric picture clear and systematically using area and Pythagoras relations helps avoid these issues.

Final Answer:
The approximate total volume of the double cone is 290 cubic cm.