Question

show that the perimeter of triangle is greater than the sum of the lengths of the three altitudes of the triangle.​

Answer 1

Answer:

Perimeter of the ∆ > sum of lengths of the three altitudes of that same triangle.

Step-by-step explanation:

Let us consider a ∆ABC. And, AD, BE and CF be the altitudes of the triangle from point A, B and C respectively.

From the properties of right angled triangles :

• In a right angled triangle, length of hypotenuse is always greater than the length of any other sides.

Thus, in ∆ADB,

= > AB must be greater than DB

= > AB > AD

Similarly,

= > BC > BE & AC > CF

From above, ( adding all )

= > AB + BC + AC > AD + BE + CF

= > Perimeter of the ∆ > sum of lengths of the three altitudes of that same triangle.

Hence, proved.

Answer 2

Answer:

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