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PQRS is a square. T and U are the mid-points of sides PS and QR respectively. Find the area of ?OTS, if PQ= 8 cm, where O is the point of intersection of TU and OS.

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PQRS IS SQUARE AND T AND U are mid pts of PS &QR, TU//PS AND TU=PS(DINCRVPQRS IS A SQUARE), JOIN QS WHICH INTERSECTS TU AT O, IN ∆PQS ST/TP=1(T IS MID PT OF SP) AND TO//PQ, THEREFORE O IS THE MID POINT OF SQ, by theorem, TO//PQ AND TO=1/2 OF PQ =4CM, AR. OF ∆OTS= 1/2*TS*OT=1/2*4*4=8 SQ.cm
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I think the above question needs a correction and `O` is the point of intersection of `TU` and `QS`. Assuming the above condition, The required area OTS is a triangle having two sides are equal and angle `STO` is right angle. So that the triangle OTS is a regular right angled triangle. Here PQ = PS...
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I think the above question needs a correction and `O` is the point of intersection of `TU` and `QS`. Assuming the above condition, The required area OTS is a triangle having two sides are equal and angle `STO` is right angle. So that the triangle OTS is a regular right angled triangle. Here PQ=PS = 8cm=> ST=4cm => area OTS = 1/2 * 4 * 4 sqcms = 8 sqcms read less
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16 sq.cm
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8 cm^2
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Maths Tutor

8
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See solution
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