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[tex]\Large{\underbrace{\sf{\purple{Required\:Answer:}}}}[/tex]
[tex]\underline{\underline{\huge{\gray{\tt{\textbf Given :-}}}}}[/tex]
[tex]\underline{\underline{\huge{\red{\tt{\textbf Find :-}}}}}[/tex]
[tex]\underline{\underline{\huge{\blue{\tt{\textbf construction :-}}}}}[/tex]
[tex]\underline{\underline{\huge{\gray{\tt{\textbf Proof :-}}}}}[/tex]
AP║ BQ║DS║CR [ given]
∴ PQ : QR: RS = AB:BC:CD
PQ : QR: RS = 6:9:12
Let PQ= 6x , QR= 9x RS= 12x
PS = PQ+QR+RS
∴ PQ+QR+RS=36 [PS=36cm]
6x + 9x + 12x = 36
27x = 36
x = 36/27 = 4/3
x = 4/3
∴ PQ = 6x = 6× 4/3
PQ = 8cm
QR = 9x = 9 × 4/3
QR= 12cm
RS = 12x = 12× 4/3
RS = 16 cm
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Answer:
RequiredAnswer:
[tex]\underline{\underline{\huge{\gray{\tt{\textbf Given :-}}}}}
[/tex]
PA, QB, RC,& SD are perpendicular on line l.
[tex]
\underline{\underline{\huge{\red{\tt{\textbf Find :-}}}}} [/tex]
PQ, QR & RS
[tex]\underline{\underline{\huge{\blue{\tt{\textbf construction :-}}}}} [/tex]
produce SP & l to meet each other at E.
[tex]\underline{\underline{\huge{\gray{\tt{\textbf Proof :-}}}}}
[/tex]
In ΔEDS,
AP║ BQ║DS║CR [ given]
∴ PQ : QR: RS = AB:BC:CD
PQ : QR: RS = 6:9:12
Let PQ= 6x , QR= 9x RS= 12x
PS = PQ+QR+RS
∴ PQ+QR+RS=36 [PS=36cm]
6x + 9x + 12x = 36
27x = 36
x = 36/27 = 4/3
x = 4/3
∴ PQ = 6x = 6× 4/3
PQ = 8cm
QR = 9x = 9 × 4/3
QR= 12cm
RS = 12x = 12× 4/3
RS = 16 cm
Hence PQ = 8cm , QR = 12cm, RS = 16 cm
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