✔️✔️Prove that the perimeter of a triangle is more then the sum of length of the medians.
✔️✔️Prove that the sum of lengths of two diagonals of a quadrilateral is more than the semi perimeter of the quadrilateral.
✔️✔️Prove that the sum of length of two diagonals is greater than the sum of length of any two opposite sides of a quadrilateral.
Answers
Answer:
Proved!
Proved!
Proved!
Step-by-step explanation:
Given Problem: ##1
Prove that the perimeter of a triangle is more then the sum of length of the medians.
Solution: ##1
Let ABC be the triangle and D. E and F are 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 ΔABD, AD is a median.
⇒ AB + AC > 2(AD)
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)
∴ AB + BC + AC > AD + BE + CF
Hence Proved!!
We can say that the perimeter of a triangle is greater than the sum of the medians.
Note: "This attachment #1 is not mine.It is by NikitaSingh Ma'am!"
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Given Problem: ##2
Prove that the sum of lengths of two diagonals of a quadrilateral is more than the semi perimeter of the quadrilateral.
Solution: ##2
Steps to prove!
1.The triangle inequality assures us that:
AB+BC >AC, BC+CD>BD, CD+AD >AC and AD+AB >BD.
2.Adding all these gives:
2AB+2BC+2CD+2AD >2AC+2BD.
3.Dividing by 2:
AB+BC+CD+AD >AC+BD as required.
Hence Proved!
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Given Problem: ##3
Prove that the sum of length of two diagonals is greater than the sum of length of any two opposite sides of a quadrilateral.
Solution: ##4
For any quadrilateral ABCD, we can easily prove that
AB+BC+CD+DA>AC+BD......(1)
Now, three cases arise. Either,
AC = BD or, AC > BD or, AC < BD
If AC = BD,then the result follows from (1).
If AC >BD ⟹ AC+BD >2BD and then the result follows from (1).
If BD >AC⟹ AC+BD >2AC and then the result follows from (1).
Hence Proved!!
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Note: See Attachments please it is number wise!
