# Properties of Triangles

## Exercise 6.4

Question 1: Is it possible to have a triangle with the following sides?

1. 2 cm , 3 cm, 5 cm
2. 3 cm, 6 cm, 7 cm
3. 6 cm, 3 cm, 2 cm

Answer: Making a triangle is possible only with sides given in option ‘b’. In other options, sum of two sides is either equal to or less than the third side.

Question 2: Take any point O in the interior of a triangle PQR. Is 1. OP + OQ > PQ?
2. OQ + OR > QR?
3. OR + OP > RP?

Answer: The answer is yes in each case, because sum of any two sides of a triangle is always greater than the third side.

Question 3: AM is a median of a triangle ABC. Is AB + BC + CA > 2AM? (Consider the sides of ΔABM and ΔAMC) Answer: In ΔABM; AB + BM > AM
Similarly, in ΔAMC; AC + CM > AM
AB + BM + CM + AC > 2AM
Or, AB + BC + CA > 2AM

Question 4: ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD? Answer: In ΔABC; AB + BC > AC
In ΔDAC; DA + CD > AC
In ΔDAB; DA + AB > DB
In ΔDCB; CD + CB > DB

2AB + 2BC + 2 CD + 2AD > 2AC + 2BD
Or, 2(AB + BC + CD + AD) > 2(AC + BD)
Or, AB + BC + CD + AD > AC + BD

Question 5: ABCD is a quadrilateral. Is AB + BC + CD + DA < 2(AC + BD)?

Answer: Let us assume a point O at the point of intersection of diagonals AC and BD.

In ΔOAB; OA + OB > AB
In Δ OBC; OB + OC > BC
In ΔODC; OD + OC > CD

AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OD + OA
Or, AB + BC + CD + DA < OA + OA + OC + OC + OD + OD + OB + OB
Or, AB + BC + CD + DA < 2(OA + OC + OD + OB)
Or, AB + BC + CD + DA < 2(AC + BD)

Question 6: The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?

Answer: Sum of given two sides = 12 cm + 15 cm = 27 cm
Hence, the third side should always be less than 27 cm.
The difference between given sides = 15 cm – 12 cm = 3 cm
If the third side will be = 3 cm then 12 + 3 = 15 cm shall be equal to one of the given sides.
Hence, the third side should be more than 3 cm
So, range of measure of third side = 4 cm to 26 cm.