Chapter 7.2, Problem 5ST

To determine

To find: The dimensions of the rectangle.

Answer to Problem 5ST

The dimension of the rectangle is 4.303 cm and 0.697 cm.

Explanation of Solution

Given:

The area of the rectangle = 3 cm².

The perimeter of the rectangle = 10 cm

Calculation:

The area of the rectangle is given by

  $\begin{array}{l}A=l\times b\\ \text{Here, }l\text{is the length of the rectangle and }b\text{ is the bredth of the rectangle}\text{.}\end{array}$

The initial area of the rectangle

  $\begin{array}{l}{A}_1=3{\text{cm}}^2\\ 3=l\times b\end{array}$

The perimeter of the rectangle is given by

  $\begin{array}{l}P=2\left(l+b\right)\\ 10=2\times \left(l+b\right)\\ l+b=5\end{array}$

Substitute $b=\frac{3}{l}$ in the above equation.

  $\begin{array}{l}l+\frac{3}{l}=5\\ l^2-5l+3=0\end{array}$

For a quadratic equation in the form of $a{x}^2+bx+c=0$ , the solution of this equation is given by

  $x_{1,2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Comparing the given equation with the standard quadratic equation, the coefficients are:

  $\begin{array}{l}a=1\\ b=-5\\ c=3\end{array}$

Substituting these in the formula for the roots.

First take the positive sign.

  $\begin{array}{c}{l}_1=\frac{-\left(-5\right)+\sqrt{{\left(-5\right)}^2-4\times 1\times \left(3\right)}}{2\times 1}\\ =\frac{5+\sqrt{25-12}}{2}\\ =4.303\end{array}$

Take the negative sign.

  $\begin{array}{c}{l}_2=\frac{-\left(-5\right)-\sqrt{{\left(-5\right)}^2-4\times 1\times \left(3\right)}}{2\times 1}\\ =\frac{5-\sqrt{25-12}}{2}\\ =0.697\end{array}$

And

The breadth of the rectangle

  $\begin{array}{l}b=\frac{3}{4.303}=0.697\\ b=\frac{3}{0.697}=4.303\end{array}$

The dimension of the rectangle is 4.303 cm and 0.697 cm.

Therefore, the dimension of rectangle is 4.303 cm and 0.697 cm.

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