Author 
Topic 

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
Advanced Member
USA
998 Posts 
Posted  01/12/2009 : 11:53:33

quote: Originally posted by mingshum
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.
Suppose we let x = PS, and y = SR. Since PQRS is a rectangle,
area(PQRS) = xy
You know that TS = (3/5)(PS), so TS = (3/5)x
SR = y.
The area of triangle RST = (1/2)*TS*SR
Area(triangle RST) = (1/2)*(3/5)x*y
7 = (3/10)xy
Multiply both sides of the equation by (10/3):
(10/3)*7 = (10/3)*(3/10)*xy
70/3 = xy
But, 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
Advanced Member
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) = 5/3 x = 70/3
Can you explain why this way works? 



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