Question

FIll in the blank with the most appropriate answer.
The sum of the three sides of a triangle is greater than the sum of its three ___.

• A. interior angles
• B. exterior angles
• C. sides
• D. medians

Answer 1

The correct option is D medians


Let ABC be the triangle and D, E and F the midpoints of BC, CA and AB respectively.
Recall that the sum of two sides of a triangle is greater than twice the median bisecting the third side,
Hence in $\Delta ABD$, AD is a median
$\Rightarrow AB+AC>2AD$
Similarly, we get,
$BC+AC>2CF$
$BC+AB>2BE$
On adding the above inequations, we get
$(AB+AC)+(BC+AC)+(BC+AB)>(2AD+2CD+2BE)$
$2(AB+BC+AC)>2(AD+BE+CF)$
$\therefore AB+BC+AC>AD+BE+CF$
Hence, we can say that the perimeter of a triangle is greater than the sum of the medians.