The plane `x/a+y/b+z/c=1` meets the co-ordinate axes in the point A, B, C respectively. Find the
area `triangle ABC.`

Explanation

Let the given plane `x/a + y/b+ z/c =` meet the coordinate axes in the point A,(a, 0, 0), B(0, b, 0) and C(0, 0, c).
Let `triangle xy, triangleyz, trianglezx` be the projection of the area of trangle
ABC on `x- y, y - z and z - x `plane. Then
`triangle^(2) = triangle_(xy)^(2)+ triangle_(yz)^(2)+triangle_(zx)^(2)`
Now `trianglexy =1/2[a(b-0)+0(0-0)+0]=(ab)/2`
Similarly `triangleyz = 1/2bc and triangle zx=1/2ca.`
Hence `triangle^(2) = triangle_(xy)^(2)+ triangle_(yz)^(2)+triangle_(zx)^(2)= 1/4 (a^(2)b^(2)+b^(2)c^(2)+c^(2)a^(2))`
`rArr triangle = 1/2sqrt(a^(2)b^(2)+b^(2)c^(2)+c^(2)a^(2))`.