ABCD is quadrilateral. Is AB + BC + CD + DA < 2(AC + BD)?
Solution:
This question is based on the triangle inequality theorem that the sum of lengths of two sides of a triangle is always greater than the third side.
Now visually identify that the quadrilateral ABCD is divided by diagonals AC and BD into four triangles.
O is the point where the two diagonals are intersecting.
Now, take each triangle separately that is the triangle AOB, COD, BOC, and AOD and apply the above property and then add L.H.S and R.H.S of the equation formed
In triangle AOB,
AB < OA + OB.........(1)
In triangle COD,
CD < OC + OD........(2)
In triangle AOD,
DA < OD + OA.........(3)
In triangle COB,
BC < OC + OB........(4)
Adding equation (1), (2), (3) and (4) we get,
AB + BC + CD + DA < OA + OB + OC + OD + OD + OA + OC + OB
= AB + BC + CD + DA < 2OA + 2OB + 2OC + 2OD
= AB + BC + CD + DA < 2(OA + OB) + 2(OC + OD)
= AB + BC + CD + DA < 2(AC + BD) [From diagram OA + OB = AC and OC + OD = BD]
Yes, AB + BC + CD + DA < 2(AC + BD) is true.
Useful Tip : Whenever you encounter problems of this kind, it is best to think of the property based on the sum of lengths of any two sides of a triangle is always greater than the third side
☛ Check: NCERT Solutions for Class 7 Maths Chapter 6
Video Solution:
ABCD is quadrilateral. Is AB + BC + CD + DA < 2(AC + BD)?
NCERT Solutions for Class 7 Maths Chapter 6 Exercise 6.4 Question 5
Summary:
ABCD is quadrilateral. Yes, it is true AB + BC + CD + DA < 2(AC + BD)?
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