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mingshum
Senior Member

USA
29 Posts

 Posted - 01/12/2009 :  11:25:06 The drawing shows a rectangle PQRS. There is a point T on segment PS such that PT = 2/5 PS. The area of triangle RST is 7.What is the area of PQRS?I know that triangle RST is a right triangle and that TS = 3/5 PS but that's it! Please enlighten me as to how the solution of 70/3 or 23.3 was obtained.

Mrspi

USA
998 Posts

 Posted - 01/12/2009 :  11:53:33 quote:Originally posted by mingshumThe drawing shows a rectangle PQRS. There is a point T on segment PS such that PT = 2/5 PS. The area of triangle RST is 7.What is the area of PQRS?I know that triangle RST is a right triangle and that TS = 3/5 PS but that's it! Please enlighten me as to how the solution of 70/3 or 23.3 was obtained.Suppose we let x = PS, and y = SR. Since PQRS is a rectangle, area(PQRS) = xyYou know that TS = (3/5)(PS), so TS = (3/5)xSR = y.The area of triangle RST = (1/2)*TS*SRArea(triangle RST) = (1/2)*(3/5)x*y7 = (3/10)xyMultiply both sides of the equation by (10/3):(10/3)*7 = (10/3)*(3/10)*xy70/3 = xyBut, xy is the area of rectangle PQRS!So, area of rectangle PQRS = 70/3

mingshum
Senior Member

USA
29 Posts

 Posted - 01/12/2009 :  12:19:55 Ahhhh... thank you for the clera explanation Mrspi!

sahsjing

USA
2399 Posts

 Posted - 02/17/2009 :  20:50:01 mingshum, You can also use proportion to get the answer.Let x be the area of PQRS,x/(2*7) = 5/3x = 70/3Can you explain why this way works?
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