Grid in from 10 reals (dont read if you havent taken all the tests yet)

Test Preparation
1

<p>Anyway, this a grid in that perplexes me.
Here it is <a href="http://www.geocities.com/aynul2788/Scan0001.jpg%5B/url%5D"&gt;http://www.geocities.com/aynul2788/Scan0001.jpg&lt;/a&gt;&lt;/p>

2

<p>Triangle Area = BH1/2</p>

<p>a=B
a=H</p>

<p>A=4
(4)(4)1/2=8</p>

3

<p>Rahul's answer for A is correct but I think the way he solved it is wrong.</p>

<p>Try to imagine a right triangle. Look at the point S. The x value for that point is 9; use the point (9,0) to make the right traingle (ignore the line ST). The right triangle has the coordinates (0,0), (9,0), and (0,A). The given area of the portion of that right triangle is 8. So,
The area of the right triangle subtracted by the unused portion(with corrdinates S,T, and (9,0))= 8.
(9)(A)(1/2) - (9-A)(A)(1/2)= 8
=> [(9A)-(9A-(A^2))]=16
=> A^2= 16
=> A=+/- 4 (in this case, +4)</p>

<p>Good luck!</p>

4

<p>What rahul did is 100% correct,</p>

<p>area of the triangle = 1/2 X base X hieght
Base = a
height = also equal to a, (it does not matter if the perpendicular lies inside the triangle or outside)</p>

<p>-> a^2 = 2 X 8
a=+-4</p>