This question is from CAT Mensuration in CAT Geometry.In this question, the sides of a square is given and with given condition we need to find the area of the new triangle formed. Mensuration Questions in the CAT exam can be solved with practical approach and thought process rather than just knowing theorems from CAT Geometry. Even if one is not completely prepared for CAT Geometry exam, he or she should be able to do well in Mensuration problems tested by the CAT Exam. Knowing Mensuration formulas is important. Make sure you master Mensuration questions for CAT exam. Its's another beautiful question that tests the understanding in properties of triangle.

Question 5: PQRS is a square of sides 2 cm & ST = 2 cm. Also, PT=RT. What is the area of ∆PST?

- 2 cm
^{2} - √3 cm
^{2} - √2 cm
^{2} - \\frac{1}{√2}\\)cm
^{2}

√2 cm

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Notice, ∆PST ~ ∆SRT (all three sides equal). So, TS extended will meet PR at cross points of diagonal as shown:-

Now, A ∆ PST = A ∆ POT – A ∆ POS ---------- Equation(1)

A ∆POT = \\frac{1}{2}\\) * PO * OT = \\frac{1}{2}\\) * \\frac{2√2}{2}\\) (PO = \\frac{2√2}{2}\\) as diagonal of square = 2√a) * (OS + ST)

=\\frac{√2}{2}\\) * (\\frac{2√2}{2}\\) + 2)

= 1 * \\frac{2 + √2}{√2}\\)

= √2 + 1

A ∆ POS = \\frac{1}{4}\\) * Area of Square

= \\frac{1}{4}\\) * 2^{2} = 1cm^{2}

=> A ∆ POT = √2 +1 - 1 = √2 cm^{2}

The question is **" What is the area of ∆PST? "**

Choice C is the correct answer.

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