Question

Solve this Plz ..?!?!?!?!?!​

Answer 1

☆ANSWER:-

━☞GIVEN:-

$PM \perp QR$

$and \: SN \perp QR$

$PM = 3 \: and \: SN = 4.5$

━☞FIND:-

$\frac{A(PQR)}{A(SQR)}$

━☞SOLUTION:-

$in \: \triangle \: PQR, PM \: and \: QR$

$So, \: a \: of \: \triangle(PQR) = \frac{1}{2} \times QR \times PM \: \: - - (i)$

$IN \triangle SQR , SN \perp QR$

$So, \: a \: of \: \triangle(SQR) = \frac{1}{2} \times SN \times QR \: \: - - (i)$

$= > \frac{ a(PQR )}{a(SQR)} = \frac{ \frac{1}{2} \times QR \times PM }{ \frac{1}{2} \times SN \times QR } \: \: \binom{from \: eq. \: (i) \: and \: (ii)}{}$

$= \frac{PM}{ SN}$

$= \frac{3}{4.5 } \: \: ( \therefore \: PM = 3 \: and \: SN = 4.5)$

$= \frac{ \cancel {30}}{ \cancel{ 4.2}} = \frac{2}{3}$

$\therefore a \triangle( PQR ) \ratio a \triangle ( SQR) = 2 \ratio3$

Answer 2

Answer:

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