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प्रश्न
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.
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उत्तर
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).
