Q35. - Geometry

Question

1. Suppose you have40m of fencing with which with to make a rectangle pen for a dog. If one side of the rectangle is x long, explain why the other side is$(20-x)m$ long.

Step-by-Step Solution

Answer

$(20-x)m$

Step 1. Given information.

Suppose you have40m of fencing with which to make a rectangular pen for a dog. If one side of the rectangle is$xm$ long.

. Step 2. Concept Used.

The perimeter of the rectangle is $P=2(length+breadth)$

The area of the square is$A={\left(side\right)}^2$

Step 3. Let’s find the other side length.

Perimeter of the pen is40m

$\begin{array}{c}P=2(l+b)\\ 40=2(l+b)\\ 20=l+b\\ l+b=20\\ x+b=20\\ b=(20-x)m\end{array}$

Therefore, the other side is$(20-x)m$.

1. Express the area of pen in terms of x .

Step 1. Given information.

Suppose you have40m of fencing with which to make a rectangular pen for a dog. If one side of the rectangle is$xm$ long.

. Step 2. Concept Used.

The area of the rectangle is 
$A=length\times breadth$

Step 3. Let’s find the area of pen.

Length=$(20-x)m$

Width=$xm$

Thus 

Area of the pen is, 

$\begin{array}{l}A=(20-x)\left(x\right)\\ A=20x-x^2\end{array}$

Therefore, the area of the pen is$20x-x^2$ .

1. Find the area of the pen for each value of :$x:\text{ 0,2,4,6,8,10,12,14,16,18,20}$ . Record your answers on a set of axes like the one shown.

Step 1. Given information.

Suppose you have 40mof fencing with which to make a rectangular pen for a dog. If one side of the rectangle is$xm$ long.

Step 2. Concept Used.

The area of the rectangle is 
$A=length\times breadth$

Step 3. Let’s find the areas.

Area of the pen for$x=0,2,4,6,8,10,12,16,18,20$

$A=20x-x^2$

Substitute the values of x in the area.

The table shows the area for given x values.

The area on the axes is shown below.

1. Give the dimensions of the pen with the greatest area.

Step 1. Given information.

Suppose you have40m of fencing with which to make a rectangular pen for a dog. If one side of the rectangle is$xm$ long.

Step 2. Concept Used.

The area of the rectangle is 
$A=length\times breadth$

Step 3. Let’s find the dimensions of the pen.

When$x=10$

The pen has the greatest area$A=100sq.units$

Therefore, the maximum area at $x=10$