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geometry problem from SAT prep

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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
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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