# Thread: rectangle inscribed in a triangle

1. ## rectangle inscribed in a triangle

domain question> http://i207.photobucket.com/albums/b...EvErSe_/16.jpg
triangle question>
http://i207.photobucket.com/albums/b..._/untitled.jpg

NOTE : PLEASE IGNORE THE NUMBER 1 MARKED ON TOP OF THE RECTANGLE IN THE TRIANGLE QUESTION (I DREW THAT).
i have the domain one but i'm not sure if I did it right, i got domain is (negative infinity, positive infinity) and my graph is going up with a top and then curves down like an upside down u and then for the secod question part a i got P(x,1/2) and part b i cant figure out the lengths of the rectangle ? if anyone could help it would be much appreciated thank you in advance. im just really worried ><

2. Originally Posted by rmn
you are correct here. this is a polynomial and so the domain is everywhere, that is, $\displaystyle (- \infty, \infty)$

the graph is an upside down parabola, with vertex (-1,2) and x-intercepts $\displaystyle \left( \frac {-2 \pm \sqrt {8}}{2},0 \right)$

the triangle is right-isosceles, so the line AB has the same length as the line from (-1,0) to B. thus, by Pythagoras, if we call the length of AB, $\displaystyle s$, then we have that $\displaystyle 2^2 = s^2 + s^2 \implies s = \sqrt {2}$

Now we can find another right triangle, ABO, again by Pythagoras, we can find the length of OB. We find that this is 1. therefore, B is the point (0,1).

From this we can deduce that AB is a segment of the line $\displaystyle y = 1 - x$

Therefore, the point P is given by $\displaystyle (x,f(x)) = (x, 1 - x)$, so the y-coordinate in terms of x is 1 - x. and that takes care of part (a)

for (b), the answer is now obvious. the area of a rectangle is length times width. the width, which i take to be the height if given by (1 - x), the length is the base which is 2x, thus we have:

$\displaystyle A = 2x \cdot (1 - x) = 2x - x^2$

3. thank you so much