CBSE Class 9 Answered

QUESTIONS

Asked by gpnkumar0 | 19 Sep, 2016, 08:31: PM

Expert Answer

Question 1:

Solution:

begin mathsize 12px style left parenthesis straight i right parenthesis space In space increment MNO comma MN space plus space straight N straight O space greater than space MO.... left parenthesis straight i right parenthesis In space increment MPO comma PM space plus space OP space greater than space MO.... left parenthesis ii right parenthesis Adding space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space get MN space plus space NO plus space OP plus PM space greater than space 2 MO Hence space proved. left parenthesis ii right parenthesis space space In space increment MNO comma MN space plus space NO space greater than space MO.... left parenthesis straight i right parenthesis In space increment MPO comma space MO plus space OP space greater than space PM.... left parenthesis ii right parenthesis Adding space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis comma space we space get MN space plus space NO plus MO plus space OP greater than MO space plus space PM rightwards double arrow MN space plus space NO plus space OP greater than space PM Hence space proved. end style

Question 2:

Solution:

Similarly, you can prove that AC > DC.

NOTE: Question 4 is same as the first part of this question.

Question 3:

Solution:

In a triangle, the sum of the lengths of either two sides is always greater than the third side.
Considering ΔOAB,
OA + OB > AB (i)
In ΔOBC,
OB + OC > BC (ii)
In ΔOCD,
OC + OD > CD (iii)
In ΔODA,
OD + OA > DA (iv)
Adding equations (i), (ii), (iii), and (iv), we obtain
OA + OB + OB + OC + OC + OD + OD + OA > AB + BC + CD + DA
2OA + 2OB + 2OC + 2OD > AB + BC + CD + DA
2OA + 2OC + 2OB + 2OD > AB + BC + CD + DA
2(OA + OC) + 2(OB + OD) > AB + BC + CD + DA
2(AC) + 2(BD) > AB + BC + CD + DA
2(AC + BD)> AB + BC + CD + DA
Yes, the given expression is true.

Question 5:

Solution: