Question

prove that the perimeter of the triangle is greater than the sum of its altitutes​

Answer 1

given a ∆ABC and altitudes AF, BD AND CE

to prove :-alt. AF+BD+CE<the perimeter of the ∆ABC (perimeter of ∆ABC= AB+BC+CA)

Proof:- in right ∆AFC

angle AFC=90°, thus angle AFC > angle ACF

THUS, AF<AC ------(1)

(THEOREM:- in ∆, the side opposite to the larger angle is longest. or we can also say AF <AC Because AC is the hypotenuse of the right ∆ AFC)

SIMILARLY, in right ∆BDA

BD<AB ------(2)

and in right ∆CED,

CE<AC------ (3)

Now adding 1,2, and 3,we have

AF+BD+CE<AC+AB+AC

hence proofed.

(perimeter of a triangle =sum of all the sides of the triangle)

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