Question

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PQR is a right angled at P and M is a point on QR such that PM Perpendicular to QR. Show that PM^2 = QM.MR .

Class - 10th
Chapter - Triangles.

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Answer 1

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• PQR is a right angled at P and M is a point on QR such that PM Perpendicular to QR. Show that PM² = QM.MR .

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• $\sf{In\: \triangle PQR \:and\: \triangle MPR,}$

• $\textsf{angle R = angle R (common angle)}$

• $\textsf{angle QPR = angle PMR = 90°.}$

• $\sf{\triangle PQR\: \sim\: \triangle MPR\: by \:AA\: similarly. \: \longrightarrow\: 1}$

• $\sf{In\: \triangle PQR \:and\: \triangle MQP,}$

• $\textsf{angle Q = angle Q (common angle)}$

• $\textsf{angle QPR = angle QMP = 90°}$

• $\sf{\triangle PQR \:\sim\: \triangle MQP\: by\: AA\: similarly. \: \longrightarrow \:2}$

• $\textsf{From 1 and 2,}$

• $\sf{\triangle MPR\: \sim \:\triangle MQP}$

• $\sf{In\: \triangle MPR\: \sim \:\triangle MQP,}$

• $\sf \dfrac{PM}{QM} \: = \: \dfrac{RM}{PM} \: = \: \dfrac{PR}{QR} \: (Ratio \: of \: corresponding \: sides)$

• $\textsf{Taking,}$

• $\sf \dfrac{PM}{QM} \: = \: \dfrac{RM}{PM}$

• $\textsf{Cross - multiplying,}$

• $\sf PM \: * \: PM \: = \: QM \: * \: MR$

• $\boxed{\sf PM^{2} \: = \: QM \: * \: MR}$

• $\textsf{Hence, proved.}$

Answer 2

★ Given :

PQR is a right angled at P and M is a point on QR such that PM Perpendicular to QR.

★ To prove :

PM = QM.MR

★ Solution :

⚘In ∆PQR

➞∠RPQ = 90°

➞So ∆PQR is a right triangle

⚘Using pythagoras theorem in PQR

➞Hypotenuse² = (Height)² + (Base)²

➞ RQ² = PQ² +PR² ____(i)

⚘In ∆ PMR

➞ PM ⊥QR

➞∠ PMR = 90°

∆PMR is a right triangle.

⚘Using pythagoras theorem in PMR

➞Hypotenuse = (Height)² + (Base)²

➞PR²= PM² + MR² _____ (ii)

⚘Similarly

➞In ∆ PQR

➞∠PMQ = 90°

➞∆PMQ is a right triangle

⚘Using pythagoras theorem in ∆PMQ

➞ Hypotenuse² = (Height)² + (Base)²

➞ PQ² = PM² + MQ²_____(iii)

⚘Putting (2) & (3) and (1)

➞ RQ² = RQ² + PR²

➞ RQ² = (PM²+ MQ²) +(PM² + MR²)

➞ RQ² = (PM²+ PM²) +(MQ²+ MR²)

➞ (MQ+ MR)² = 2PM² + (MQ² + MR²)

➞ 0 + 2MQ × MR = 2PM²

➞ 2MQ × MR = 2PM²

➞ MQ × MR = PM²

➞ PM² = MQ × MR

Hence Proved

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