# ABCD is quadrilateral. Is AB + BC + CD + DA < 2(AC + BD)?

**Solution:**

In this question, it is asked to check Is AB + BC + CD + DA < 2(AC + BD) or not.

This question is based on the property 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 center of the quadrilateral.

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)

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

**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

ABCD is quadrilateral. Is AB + BC + CD + DA < 2(AC + BD)?

Yes, it is true