# Thread: help me find dimensions of rectangle and triangles

1. ## help me find dimensions of rectangle and triangles

a rectangular pen is subdivided by two dividers that meet at the midpoint P of the base of the rectangle. ( the picture is a rectangle, in which there is a line segment coming from each upper corner of the rectangle, coming together at midpoint P) each divider is 10 m long, and the perimeter of the pen is 30 min. what are the dimensions of the rectangle?

2. Originally Posted by polakio92
a rectangular pen is subdivided by two dividers that meet at the midpoint P of the base of the rectangle. ( the picture is a rectangle, in which there is a line segment coming from each upper corner of the rectangle, coming together at midpoint P) each divider is 10 m long, and the perimeter of the pen is 30 min. what are the dimensions of the rectangle?
Let a and b denote the sides of the rectangle. Then you get 2 equations:

$2a+2b=30~\implies~ b=15-a$ ....... [1]

$\left(\frac12 a\right)^2+b^2=100$ ....... [2]

Plug in the term of b from [1] into [2]:

$\left(\frac12 a\right)^2+(15-a)^2=100$ and solve for a:

$a=12\pm2\sqrt{11}~\implies~b=3\mp 2\sqrt{11}$

The negative value of b isn't very plausible so there is only one solution left.

3. Originally Posted by earboth
Let a and b denote the sides of the rectangle. Then you get 2 equations:

$2a+2b=30~\implies~ b=15-a$ ....... [1]

$\left(\frac12 a\right)^2+b^2=100$ ....... [2]

Plug in the term of b from [1] into [2]:

$\left(\frac12 a\right)^2+(15-a)^2=100$ and solve for a:

$a=12\pm2\sqrt{11}~\implies~b=3\mp 2\sqrt{11}$

The negative value of b isn't very plausible so there is only one solution left.

for some reason i can't get a to end in the square root of 11. am i doing something wrong?

4. yes, it is not working for some reason. is there any further explanation?

5. Originally Posted by polakio92
for some reason i can't get a to end in the square root of 11. am i doing something wrong?
Do you mean that you can't get the result? If so:

$\left(\frac12 a\right)^2+(15-a)^2=100 ~\iff~ \frac14 a^2 + 225 -30a + a^2 = 100 ~\iff~ \frac54 a^2 - 30a + 125 = 0$

Now apply the formula to solve this quadratic equation:

$a = \frac{30 \pm \sqrt{900 - 4 \cdot \frac54 \cdot 125}}{2 \cdot \frac54} = \frac{30 \pm \sqrt{900 - 625}}{ \frac52} = \frac{ 2 \cdot 30 \pm 2 \cdot 5 \cdot \sqrt{11}}{5}$ which will yield the result I posted in my previous reply.