1.

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Answer»

en:-

  • PT = 2cm
  • TR = 4CM

To find:-

  • \sf ratio \: of \: triangle \: PST \:and\:PQR

Theorem:-

  • Ratio of area of two similar triangles is equal to square of the ratio of their corresponding sides

Solution:-

→ ∠S = ∠Q {correspomding angles}

→ ∠ T= ∠R {correspomding angles}

∴ ΔPST ~ Δ PQR {AA}

PT = 2 CM and PR = PT + TR = 4+2 = 6 cm

→\sf \dfrac{PT}{PR} =\dfrac{\cancel{2}}{\cancel{6}} = \dfrac{1}{3} \\\\→ \dfrac{ar(ΔPST)}{ar(ΔPQR)} = \bigg( \dfrac{PT}{PR} \bigg)^{2} = \bigg( \dfrac{1}{3} \bigg)^{2} =\dfrac{1}{9}

hence the REQ ratio is 1:9


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