Question

From given figure, in ∆PQR, if ∠QPR = 90°, PM ⊥ QR, PM = 10, QM = 8, then for finding the value of QR, complete the following activity.

Activity: In ∆PQR, if ∠QPR = 90°, PM ⊥ QR,  ......[Given]

In ∆PMQ, by Pythagoras Theorem,

∴ PM² + `square` = PQ²     ......(I)

∴ PQ² = 10² + 8² 

∴ PQ² = `square` + 64

∴ PQ² = `square`

∴ PQ = `sqrt(164)`

Here, ∆QPR ~ ∆QMP ~ ∆PMR

∴ ∆QMP ~ ∆PMR

∴ `"PM"/"RM" = "QM"/"PM"`

∴ PM² = RM × QM

∴ 10² = RM × 8

RM = `100/8 = square`

And,

QR = QM + MR

QR = `square` + `25/2 = 41/2`

Answer 1

In ∆PQR, if ∠QPR = 90°, PM ⊥ QR,  ......[Given]

In ∆PMQ,

by Pythagoras Theorem,

∴ PM² + QM² = PQ²     ......(I)

∴ PQ² = 10² + 8² 

∴ PQ² = 100 + 64

∴ PQ² = 164

∴ PQ = `sqrt(164)`

Here, ∆QPR ~ ∆QMP ~ ∆PMR

∴ ∆QMP ~ ∆PMR

∴ `"PM"/"RM" = "QM"/"PM"`   ......[Corresponding sides of similar triangles]

∴ PM² = RM × QM

∴ 10² = RM × 8

RM = `100/8` = `25/2`

And,

QR = QM + MR

QR = 8 + `25/2 = 41/2`