Question

PQR is a right angled triangle, where angle P = 90 degree
PD is perpendicular to QR.
PQ : √PR =​

Answer 1

Answer:

hope it helps you friend.

Answer 2

$\frac{PQ}{\sqrt{PR}} = \sqrt{\sqrt{QR} \times \frac{QD}{\sqrt{DR}}}$

Step-by-step explanation:

First we take here $\triangle PQR$  

so in this triangle

h² = p² + b²   ..............1

(QR)² = (QP)² + (PR)²   ..........2

and

in $\triangle PDQ$

(PQ)² = (PD)² + (QD)²   ...........3

and

In $\triangle PDR$  

(PR)² = (PD)² + (DR)²    ...........4

so from equation 3 and 4,  we equate (PD)²

(PQ)² - (QD)² = (PR)² - (DR)²

(PQ)² - (PR)² = (QD)² - (DR)²

(PQ)² - (PR)² = (QD + DR ) × (QD -DR)

(PQ)² - (PR)² = QR  ×  ( QD -DR ) .......5

and

(PQ)² - (PR)² = (QR)²    .............6

so from above 5 and 6 equation

2 × (PQ)² = QR × ( QR + QD - DR )

2 × (PQ)² = QR × ( QD + DR  + QD - DR )

2 × (PQ)² = 2 × QD ×  QR

(PQ)² =  QD ×  QR   .........7

and

2 (PR)² = (QR)²  -  QR × ( QD - DC )

2 (PR)² = QR × ( QD - DR - QD + DR )

(PR)²  = DR + QR  ........... 8

and

so now for PQ : √PR

$\frac{PQ^2}{PR} = \frac{QD \times QR}{\sqrt{QR \times DR}}$    ..........9

$(\frac{PQ}{\sqrt{PR}})^2 = \sqrt{QR} \times \frac{QD}{\sqrt{DR}}$  

$\frac{PQ}{\sqrt{PR}} = \sqrt{\sqrt{QR} \times \frac{QD}{\sqrt{DR}}}$  

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