Question

Let ABCD be a tetrahedron such that the edges AB, AC and AD are mutually perpendicular,
area of triangles ABC, ACD and ADB be 3, 4 and 5 sq. units respectively. Then the area of
triangle BCD, is -​

Answer 1

area of triangle BCD, is   5 √2 sq units where ABCD  a tetrahedron where area of ABC, ACD and ADB be 3, 4 and 5 sq. units

Step-by-step explanation:

Area of Δ BCD

= (1/2) √ ((bc)² + (cd)²  + (db)²)

area of triangles ABC = bc/2 = 3

=> bc = 6

area of triangles ACD = cd/2 = 4

=> cd = 8

area of triangles ABD = bd/2 = 5

=> bd = 10

Area of Δ BCD

= (1/2) √ (6² + 8²  + 10²)

= √200 / 2

= 10 √2 /2

= 5 √2

area of triangle BCD, is   5 √2 sq units

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