Chapter 6, Problem 13STP

To determine

To Find :

The value of m .

Answer to Problem 13STP

  $\text{m=4 and m= -4}\text{.875}$

Explanation of Solution

Given Statement:

The perimeter of the given quadrilateral is 160.

  

Concept Used :

• Perimeter of any polygon is the sum of the lengths of its sides.

Calculation:

  $\begin{array}{l}Given\\ \text{Sides of the quadrilateral:}\\ {\text{a= 4m}}^2+1\\ b=2m+6\\ c=3m-4\\ d=4m^2+2m+1\\ Perimeter=a+b+c+d=160\\ \text{⇒}{\text{4m}}^2+1+2m+6+3m-4+4m^2+2m+1=160\\ \text{⇒}(4+4)m^2+(2+3+2)m+(6+1+1-4)=\mathrm{160.........}[\text{add like terms]}\\ \text{⇒}{\text{8m}}^2+7m+4=160\\ \text{⇒}8m^2+7m+4-160=\mathrm{0............................................}[\text{subtract 160 from each side]}\\ \text{⇒}8m^2+7m-156=0\\ {\text{Roots of a quadratic equation ax}}^2+bx+c=0\\ are\\ x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ Comparing,a=8,b=7,c=-156\\ m=\frac{-7\pm \sqrt{7^2-4(8)(-156)}}{2(8)}\\ m=\frac{-7\pm \sqrt{49+4992}}{16}\\ m=\frac{-7\pm \sqrt{5041}}{16}\\ m=\frac{-7\pm 71}{16}\\ m=\frac{-7+71}{16}\text{ and }m=\frac{-7-71}{16}\\ m=4\text{ and }m=-4.875\\ \text{Since measurement of the side cannot be negative}\text{.}\\ \text{and both the values of m here does not make any side's}\\ \text{length negative}\text{.}\\ \text{So, m=4 and m= -4}\text{.875}\\ \text{both satisfy the conditions}\text{.}\end{array}$