Question

Q4 ABCD is a quadrilateral.

Is AB + BC + CD + DA > AC + BD?

Solution:

Yes, Given quadrilateral ABCD in which AC and BD

are its diagonals.

In ∆ABC, we have

AB + BC > AC …(i)

[Sum of any two sides is greater than the third

side]

In ∆BDC, we have

BC + CD > BD …(ii)

[Sum of any two sides is greater than the third

side]

In ∆ADC, we have

CD + DA > AC …(iii)

[Sum of any two sides is greater than the third

side]

In ∆DAB, we have

DA + AB > BD …(iv)

[Sum of any two sides is greater than the third

side]

Adding eq. (i), (ii), (iii) and (iv), we get

2AB + 2BC + 2CD + 2DA > 2AC + 2BD or AB +

BC + CD + DA > AC + BD [Dividing both sides by

2]

Hence, proved.


In this answer, I can't understand from - adding eq. Please help! I wanna score good marks in Maths... Please, please​

Answer 1

Answer:

ABCD is a quadrilateral and AC, and BD are the diagonals.

Sum of the two sides of a triangle is greater than the third side.

So, considering the triangle ABC, BCD, CAD and BAD, we get

AB + BC > AC

CD + AD > AC

AB + AD > BD

BC + CD > BD

Adding all the above equations,

2(AB + BC + CA + AD) > 2(AC + BD)

⇒ 2(AB + BC + CA + AD) > 2(AC + BD)

⇒ (AB + BC + CA + AD) > (AC + BD)

HENCE, PROVED