Maths Q5 – JEE-MAIN 2025

Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular. Let the areas of the triangles ABC, ACD and ADB be 5, 6 and 7 square units respectively. Then the area (in square units) of the

△

BCD is equal to :

Clue 1 of 3: Geometric Interpretation

Visualize the tetrahedron with three mutually perpendicular edges meeting at a vertex as a corner of a cuboid.

🧠 Full Solution Path

● Video Walkthrough

Step 1: Set up the coordinate system and define variables✦ Active

Place vertex A at the origin $(0, 0, 0)$ . Since edges AB, AC, and AD are mutually perpendicular, align them with the coordinate axes. Let their lengths be $A B = x$ , $A C = y$ , and $A D = z$ . The coordinates of the vertices are $A (0, 0, 0)$ , $B (x, 0, 0)$ , $C (0, y, 0)$ , and $D (0, 0, z)$ . The given areas of the right-angled triangles are:

A_{A B C} = \frac{1}{2} x y = 5 ⟹ x y = 10 (1)

A_{A C D} = \frac{1}{2} y z = 6 ⟹ y z = 12 (2)

A_{A D B} = \frac{1}{2} z x = 7 ⟹ z x = 14 (3)

Step 2: Calculate the area of

△ B C D

using vector cross product○ Expand

To find the area of $△ B C D$ , we can use the vector cross product. Consider two vectors forming sides of the triangle, for example, $\vec{B C}$ and $\vec{B D}$ :

\vec{B C} = C - B = (0 - x, y - 0, 0 - 0) = (- x, y, 0)

\vec{B D} = D - B = (0 - x, 0 - 0, z - 0) = (- x, 0, z)

\vec{B C} \times \vec{B D} = | \begin{matrix} i & j & k \\ - x & y & 0 \\ - x & 0 & z \end{matrix} | = y z i + x z j + x y k

| \vec{B C} \times \vec{B D} | = \sqrt{(y z)^{2} + (x z)^{2} + (x y)^{2}}

A_{B C D} = \frac{1}{2} | \vec{B C} \times \vec{B D} | = \frac{1}{2} \sqrt{(y z)^{2} + (x z)^{2} + (x y)^{2}}

💡 Teacher's Secret Hint

Remember that the area of a triangle formed by vectors $\vec{u}$ and $\vec{v}$ is $\frac{1}{2} | \vec{u} \times \vec{v} |$ . This formula is particularly useful in 3D geometry.

Step 3: Substitute known values and compute the final area○ Expand

Substitute the values of $x y$ , $y z$ , and $z x$ from equations (1), (2), and (3) into the area formula for $△ B C D$ :

A_{B C D} = \frac{1}{2} \sqrt{(12)^{2} + (14)^{2} + (10)^{2}}

A_{B C D} = \frac{1}{2} \sqrt{144 + 196 + 100}

A_{B C D} = \frac{1}{2} \sqrt{440}

A_{B C D} = \frac{1}{2} \sqrt{4 \times 110}

A_{B C D} = \frac{1}{2} \times 2 \sqrt{110}

A_{B C D} = \sqrt{110} square units

💡 Teacher's Secret Hint

Simplify the square root carefully by factoring out perfect squares.

✦ STEM Console utilizes AI models to generate step-by-step explanations and math clues. AI can make mistakes.

Maths Question 5 – JEE-MAIN 2025

Clue 1 of 3: Geometric Interpretation

🧠 Full Solution Path