ST PARALLEL TO RQ,PS=3,SR=4.FIND THE RATIO OF THE AREA OF TRIANGLE PST TO THE AREA OF TRIANGLE PRQ

ST PARALLEL TO RQ,PS=3,SR=4.FIND THE RATIO OF THE AREA OF TRIANGLE PST

1.	ST PARALLEL TO RQ,PS=3,SR=4.FIND THE RATIO OF THE AREA OF TRIANGLE PST TO THE AREA OF TRIANGLE PRQ
Answer» PS = 3 cm, SR = 4 cm and ST \|\| RQ.PR = PS + SR= 3 + 4 = 7 cmIn {tex}\\triangle{/tex}PST and {tex}\\triangle{/tex}PRQ{tex}\\angle{/tex}SPT\xa0{tex}\\cong{/tex}{tex}\\angle{/tex}RPQ (common angle){tex}\\angle{/tex}PST\xa0{tex}\\cong{/tex}{tex}\\angle{/tex}PRQ (Alternate angle){tex}\\triangle{/tex}PST\xa0{tex}\\sim{/tex}{tex}\\triangle{/tex}PRQ (AA configuration){tex}\\frac { \\text { ar } \\Delta P S T } { \\text { ar } \\Delta P Q R } = \\frac { P S ^ { 2 } } { P R ^ { 2 } } {/tex}{tex}= \\frac { 3 ^ { 2 } } { 7 ^ { 2 } } = \\frac { 9 } { 49 }{/tex}Hence required ratio = 9 : 49.