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In gives figure ST !! RQ,PS- …

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In gives figure ST !! RQ,PS- 3cm and SR-4 cm .Find the ratio of the area of PST to the to area of PRQ

Posted by Golu Yadav 6 years, 6 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 6 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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