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From fig., seg PQ || side BC, AP = x + 3, PB = x – 3, AQ = x + 5, QC = x – 2, then complete the acitivity to find the vlaue of x.In ΔPQB, PQ || side BC APPB=AQ[______] .......[______] x+3x-3=x+5[___ - Geometry Mathematics 2

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प्रश्न

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"/(["______"])`    ...[______]

`(x + 3)/(x - 3) = (x + 5)/(["______"])`

(x + 3) [______] = (x + 5)(x – 3)

x2 + x – [______] = x2 + 2x – 15

x = [______]

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योग

उत्तर

In ΔPQB,

PQ || side BC

`"AP"/"PB" = "AQ"/bb"QC"`    ...[Base proportionality theorem]

`(x + 3)/(x - 3)` = `(x + 5)/bb(x - 2)`

(x + 3) (x – 2) = (x + 5)(x – 3)

x2 + x – 6 = x2 + 2x – 15

∴ x – 6 = 2x – 15

∴ 2x – x = 15 – 6

∴ x = 9

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अध्याय 1: Similarity - Q.3 (A)

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