Question

From fig., seg PQ || side BC, AP = x + 3, PB = x – 3, AQ = x + 5, QC = x – 2, then complete the activity to find the value of x.

In ΔPQB, PQ || side BC

`(AP)/(PB) = (AQ)/(square)`   ...`[square]`

`(x + 3)/(x - 3) = (x + 5)/(square)`

(x + 3) `square` = (x + 5)(x – 3)

x² + x – `square` = x² + 2x – 15

x = `square`

Answer 1

In ΔPQB, PQ || side BC

`(AP)/(PB)` = \[\frac{{AQ}}{{\boxed{QC}}}\]    ...[\[\boxed{\text{Basic proportionality theorem}}\]]

`(x + 3)/(x - 3)` = \[\frac{{x + 5}}{{\boxed{x - 2}}}\]

(x + 3) \[\boxed{x - 2}\] = (x + 5)(x – 3)

x² + x – \[\boxed{6}\] = x² + 2x – 15

∴ x – 6 = 2x – 15

∴ 2x – x = 15 – 6

∴ x = \[\boxed{9}\]