ST is parallel to RQ,PS=3cm and …
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Sia ? 6 years, 5 months ago
PS = 3 cm, SR = 4 cm and ST || RQ.
PR = PS + SR
= 3 + 4 = 7 cm
In {tex}\triangle{/tex}PST and {tex}\triangle{/tex}PRQ
{tex}\angle{/tex}SPT {tex}\cong{/tex} {tex}\angle{/tex}RPQ (common angle)
{tex}\angle{/tex}PST {tex}\cong{/tex} {tex}\angle{/tex}PRQ (Alternate angle)
{tex}\triangle{/tex}PST {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.
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