Question

Please solve this fast...​

Answer 1

Given:-

• 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

Answer 2

Answer:

Sorry...but aapne mere question me jo answer kiya tha na kisi ne delete kar diya... dekhne bhi nhi diya...