JEE Mathematics 2025 Question with Solution

Given: In tetrahedron $ABCD$, the edges $AB$, $AC$ and $AD$ are mutually perpendicular. The areas of $\triangle ABC$, $\triangle ACD$ and $\triangle ADB$ are $5$, $6$ and $7$ square units respectively.

Find: The area of the face $\triangle BCD$.

Use De Gua's theorem for a tetrahedron with a right-angled vertex:

$$(A_{BCD})^2 = (A_{ABC})^2 + (A_{ACD})^2 + (A_{ADB})^2$$

Substituting the given areas,

$$(A_{BCD})^2 = 5^2 + 6^2 + 7^2$$ $$(A_{BCD})^2 = 25 + 36 + 49 = 110$$

Therefore,

$$A_{BCD} = \sqrt{110}$$

So the area of $\triangle BCD$ is $\sqrt{110}$ square units.

The solution concludes $\sqrt{110}$, but this does not match any option. The solution's marks the correct option as C, so the recorded answer is C despite the discrepancy.

Given: Vertex $A$ is a right-angled vertex because $AB \perp AC$, $AC \perp AD$, and $AD \perp AB$. The three faces meeting at $A$ are pairwise perpendicular.

Find: The area opposite $A$, namely $\triangle BCD$.

For such a tetrahedron, the area of the opposite face satisfies the three-dimensional analogue of the Pythagorean theorem:

$$\text{Area}(\triangle BCD)^2 = \text{Area}(\triangle ABC)^2 + \text{Area}(\triangle ACD)^2 + \text{Area}(\triangle ADB)^2$$

Now insert the values:

$$\text{Area}(\triangle BCD)^2 = 5^2 + 6^2 + 7^2$$ $$\text{Area}(\triangle BCD)^2 = 110$$

Hence,

$$\text{Area}(\triangle BCD) = \sqrt{110}$$

Therefore, the computed face area is $\sqrt{110}$ square units, which is inconsistent with the listed options and the marked option C.