1.

ll y-WHich is nearest to the point (-2,5),5. In given figure, ST I I RQ, PS = 3 cm and SR = 4 cm. Find the ratio of the area of AST to the area ofIf cos A , find the value of 4 + 4 tan2 A.

Answer»

Given:

ST || RQ

PS= 3 cm

SR = 4cm

We know that the ratio of areas of two similar triangles is equal to the ratio of squares of their corresponding sides.

ar(∆PST) /ar(∆PRQ)= (PS)²/(PR)²

ar(∆PST) /ar(∆PRQ)= 3²/(PS+SR)²

ar(∆PST) /ar(∆PRQ)= 9/(3+4)²= 9/7²=9/49

Hence, the required ratio ar(∆PST) :ar(∆PRQ)= 9:49



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