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Prove that of all triangles on the same base and of the same area, the isosceles triangle has the least perimeter.

Prove that of all triangles on the same base and of the same area, the isosceles triangle has the least perimeter.

Grade:9

1 Answers

Arun
25763 Points
3 years ago
Dear Anirban
 
To prove that of all the triangles on the same base and of the same area, the isosceles triangle has the least perimeter.
 
If we drop a vertical line from the vertex of the triangle to the base (B), the length of that line is the height of the triangle (H), and the point where it intersects the base can be defined as the distance x from one end of the base.
 
For a given area A and base length, the height h must be constant, since the area of a triangle is equal to A = (1/2)* B *H
 
The distance from that point to the other end of the base is B-x.
 
Note that for an isoceles triangle the value of x is B/2. So the job here is to show that the length of the perimeter is a minimum when x = B/2.

The perimeter of the triangle is simply the sum of the lengths of the three sides:

P(x) = B + ( x2 + H2)1/2 + ( ( B-x)2 + H2))1/2

Taking  the derivative of P wrt x, and set it to zero - that will define the value for x where the length of the perimeter is either at a max or a min, we get 
 
dP/dx = [x / ( x2 + H2)1/2 ] - [(B-x) / ( B2 - 2Bx + x2 + H2 )1/2 ]

Now  substitute in x = B/2, we find for  dP/dx is zero.
Hence the perimeter is either at a max or a min for an isoceles triangle.
To prove it's a min, you could take the 2nd derivative of P(x) and show that it's negative at x = B/2.
To show that for x = + infinity or - infinity the value of P(x) goes to infinity . Hence the value of P at x = B/2 must be a min, and the isoceles traingle has the minimum length perimeter for a given base and area.
 
 
Regards
Arun (askIITians forum expert)

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