Question

The width of a rectangle is 6 less than twice its length. If the area of the rectangle is 170 cm2 , what is the length of the diagonal?The length of the diagonal is cm.Give your answer to 2 decimal places.Submit QuestionQuestion 25

Answer 1

The formula to find the area of a rectangle is:

`\begin{gathered} A=l\cdot w \\ \text{ Where} \\ \text{ A is the area} \\ l\text{ is the length} \\ w\text{ is the width} \end{gathered}`

Since the rectangle area is 170cm², we can write the following equation.

`170=l\cdot w\Rightarrow\text{ Equation 1}`

On the other hand, we know that the width of the rectangle is 6 less than twice its length. Then, we can write another equation.

`\begin{gathered} w=2l-6\Rightarrow\text{ Equation 2} \\ \text{ Because} \\ 2l\Rightarrow\text{ Twice length} \\ 2l-6\Rightarrow\text{ 6 less than twice length} \end{gathered}`

Now, we solve the found system of equations.

`\begin{cases}170=l\cdot w\Rightarrow\text{ Equation 1} \\ w=2l-6\Rightarrow\text{ Equation 2}\end{cases}`

For this, we can use the substitution method.

Step 1: we replace the value of w from Equation 2 into Equation 1. Then, we solve for l.

`\begin{gathered} 170=l(2l-6) \\ \text{Apply the distributive property} \\ 170=l\cdot2l-l\cdot6 \\ 170=2l^2-6l \\ \text{ Subtract 170 from both sides} \\ 0=2l^2-6l-170 \end{gathered}`

We can use the quadratic formula to solve the above equation.

`\begin{gathered} x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}\Rightarrow\text{ Quadratic formula} \\ \text{ For }ax^2+bx+c=0 \end{gathered}`

Then, we have:

`\begin{gathered} a=2 \\ b=-6 \\ c=-170 \\ l=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ l=\frac{-(-6)\pm\sqrt[]{(-6)^2-4(2)(-170)}}{2(2)} \\ l=\frac{6\pm\sqrt[]{1396}}{4} \\ \end{gathered}`

There are two solutions for l.

`\begin{gathered} l_1=\frac{6+\sqrt[]{1396}}{4}\approx10.84 \\ l_2=\frac{6-\sqrt[]{1396}}{4}\approx-7.84 \\ \text{ The symbol }\approx\text{ is read 'approximately'.} \end{gathered}`

Since the value of l can not be negative, the value of l is 10.84.

Step 2: We replace the value of l into any of the equations of the system to find the value of w. For example, in Equation 1.

`\begin{gathered} 170=l\cdot w\Rightarrow\text{ Equation 1} \\ 170=10.84\cdot w \\ \text{ Divide by 10.84 from both sides} \\ (170)/(10.84)=(10.84\cdot w)/(10.84) \\ 15.68\approx w \end{gathered}`

Now, the long side, the wide side and the diagonal of the rectangle form a right triangle.

Then, we can use the Pythagorean theorem formula to find the length of the diagonal.

`\begin{gathered} a^2+b^2=c^2 \\ \text{ Where} \\ a\text{ and }b\text{ are the legs} \\ c\text{ is the hypotenuse} \end{gathered}`

In this case, we have:

`\begin{gathered} a=10.84 \\ b=15.68 \\ a^2+b^2=c^2 \\ (10.84)^2+(15.68)^2=c^2 \\ 117.51+245.86=c^2 \\ 363.37=c^2 \\ \text{ Apply square root to both sides of the equation} \\ \sqrt[]{363.37}=\sqrt[]{c^2} \\ 19.06=c \end{gathered}`

Therefore, the length of the diagonal of the given rectangle is 19.06 cm rounded to 2 decimal places.