Question

1.)Prove that the perimeter of a triangle is greater than the sum of its 3 medians
2.) Show that the sum of three altitudes of a triangle is less than sum of three sides of triangle .
3.) Prove that any two sides of triangle are together greater than twice the median drawn to the third side .

These questions are for class 9 . Kindly help ..!!

Answer 1

Figure is in the attachment

1)
Let AD,BE & CF be the three medians of a ∆ABC.

WE KNOW THAT THE SUM OF ANY TWO SIDES OF A TRIANGLE IS GREATER THAN TWICE THE MEDIAN DRAWN TO THE THIRD SIDE.

AB+AC>2AD;. AB+BC>2BE & BC+AC>2CF.

Adding We get,
2(AB+BC+AC) >2(AD+BE+CF)

(AB+BC+AC) >(AD+BE+CF)

Hence, the perimeter of a triangle is greater than the sum of its three medians.

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2)

Use the result that the perpendicular drawn from a point( outside the line ) to a line is shorter( in length) than a line segment drawn from that point to the line and then add all three cases.
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Solution:

Consider ABC is a triangle and AL, BM and CN are the altitudes.

To Prove:
AL+BM+CN
Proof:

We know that the perpendicular AL drawn from the point A to the line BC is shorter than the line segment AB drawn from the point A to the line BC.

ALBMCN
On adding equation 1,2,3
AL+BM+CN

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3)

Given: A ∆ABC in which AD is median.

To Prove:
AC+AB>2AD

Construction:
Produce AD to E, such that AD=DE.
Join EC.

Proof:

In ∆ADB & ∆ EDC
AD=ED (by construction)
angleADB=angleEDC. (vertically opposite angle)
BD=CD. (D midpoint of BC)


∆ADB congruent ∆ EDC (by SAS)

AB=EC (by CPCT)

Now ,in ∆AEC, we have AC+EC>AE

[SINCE, SUM OF ANY TWO SIDES OF A TRIANGLE IS GREATER THAN THE THIRD SIDE]

AC+EC>AD+DE. (AE=AD+DE)

AC+AB>2AD (AD=ED & EC=AB)

Thus, the sum of any two sides of a triangle is greater than twice the median with respect to the third side.
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Hope this will help you....

Answer 2

Answer:

In the triangle ABC, D,E and F are the midpoints of sides BC,CA and AB respectively.

We know that the sum of two sides of a triangle is greater than twice the median bisecting the third side,

Hence in triangle ABD, AD is a median

⇒AB+AC>2(AD)

Similarly, we get

⇒BC+AC>2(CF)

⇒BC+AB>2(BE)

On adding the above inequations, we get

(AB+AC)+(BC+AC)+(BC+AB)>2AD+2CF+2BE

2(AB+BC+AC)>2(AD+BE+CF)

∴AB+BC+AC>AD+BE+CF

Then perimeter of a triangle is greater than sum of its three medians

Step-by-step explanation: