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
\(OA + OB + OB + OC + OC + OD + OD + OA > AB + BC + CD + DA\)
\(\Rightarrow\) \(2\;OA + 2\;OB + 2\;OC + 2\;OD > AB + BC + CD + DA\)
\(\Rightarrow\) \(2\;OA + 2\;OC + 2\;OB + 2\;OD > AB + BC + CD + DA\)
\(\Rightarrow\) \(2(OA + OC) + 2(OB + OD) > AB + BC + CD + DA\)
\(\Rightarrow\) \(2(AC) + 2(BD) > AB + BC + CD + DA\)
\(\Rightarrow\) \(2(AC + BD) > AB + BC + CD + DA\)
Yes, the given expression is true.
Write equations for the following statements:
(i) The sum of numbers x and 4 is 9.
(ii) 2 subtracted from y is 8.
(iii) Ten times a is 70.
(iv) The number b divided by 5 gives 6.
(v) Three-fourth of t is 15.
(vi) Seven times m plus 7 gets you 77.
(vii) One-fourth of a number x minus 4 gives 4.
(viii) If you take away 6 from 6 times y, you get 60.
(ix) If you add 3 to one-third of z, you get 30