<p>From march 2005 Sat:
<a href="https://satonlinecourse.collegeboard.com/SR/digital_assets/assessment/images/cb_er_exam_15_question_06_SPR_VB340386_001.gif%5B/url%5D">https://satonlinecourse.collegeboard.com/SR/digital_assets/assessment/images/cb_er_exam_15_question_06_SPR_VB340386_001.gif</a>
In the figure above, PQRS is a rectangle. The area of triangle RST is 7 and PT = 2/5 PS What is the area of PQRS?</p>
<p>I didnt understand CB's explanation</p>
<p>This is how I would do it (though it may be complex):</p>
<p>Since the area of the triangle is 7, the area of a square made by doubling the triangle would be 14. Since PT = 2/5 PS, the area of the leftover space in the rectangle would be 2/3 that of the square. 14 + 2/3 * 14 = 70/3.</p>
<p>7 = (1/2) * rs * (3/5) ps
14 = rs * 3ps/5
70 = rs * 3ps
70/3 = rs * ps = A</p>
<p>Both solutions are very good.<br>
iin77's is more intuitive; it needs a minor correction: "a rectangle made by doubling the triangle".</p>
<p>I like assigning my own numbers when given a freedom:
PT, TS and RS could be anything, as long as PT=2/5 PS, and A(RST)=7.
Let's make PS=5, then PT=2, ST=3.
Since A(RST) = (1/2) x ST x RS = 7,
RS = 7 / ((1/2) x 3) = 14/3.
A(PQRS) = PS x RS = 5 x (14/3) = 70/3.</p>
<p>Wouldn’t the leftover space be 2/5 of 14? If so, wouldn’t it be (2/5 *14) + 14?
Which would give you 19.6.
I can see how to solve it if we assign 4 for PT and 6 for TS (4 is 2/5 of 10) and end up with 70/3 or 23.33 but do not see how to arrive at this solution this going the 2/5 of 14 method.
What are we missing?</p>
<p>No - 14 is 3/5 ( or .6) of the rectangle, so the leftover space is 2/5 of the unknown total area.</p>
<p>14 = .6
Looking for total area (1). I set up an equivalent fraction.</p>
<p>14 = .6 and x = 1. Cross-multiply and solve.</p>
