Question

In fig., PS = 2, SQ = 6, QR = 5, PT = x and TR = y. Then find the pair of value of x and y such that ST || side QR.

Answer 1

In ΔPQR,

line ST || side QR   ...[Given]

∴ `(PS)/(SQ) = (PT)/(TR)`   ...[Basic proportionality theorem]

∴ `2/6 = (PT)/(TR)`   ...[Given]

∴ `1/3 = x/y`

∴ y = 3x   ...(i)

PR = PT + TR   ...[P-T-R]

∴ PR = x + y   ...(ii) [Given]

In ΔPQR, PQ + QR > PR   ...`[("The sum of the lengths of"),("any two sides of a triangle is"),("greater than the third side")]`

∴ (PS + SQ) + QR > PR   ...[P-S-Q]

∴ 2 + 6 + 5 > PR   ...[Given]

∴ 13 > PR

∴ x + y < 13   ...[From (ii)]

∴ x + 3x < 13   ...[From (i)]

∴ 4x < 13   ...(ii)

∴ Integer values of x satisfying equation (ii) are 1, 2, 3.

∴ `{:("If"  x = 1",", "we get"  y = 3x = 3),("If"  x = 2",", "we get"  y = 3x = 6),("If"  x = 3",", "we get"  y = 3x = 9):}}`   ...[From (i)]

∴ Some of the pairs of values of x and y are (1, 3), (2, 6), (3, 9).