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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. |
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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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