Chapter 5, Problem 30RP

To determine

Find the three possible value of perimeter of kite.

Answer to Problem 30RP

  $4,60,102$

Explanation of Solution

Given information:

  

Calculation:

Consider the given figure,

  

Disjoint pairs of consecutive sides of a kite are congruent.

  $\begin{array}{l}firstcase:-\\ \overline{ET}\cong \overline{EK}\\ \overline{KI}\cong \overline{TI}\\ \sec ondcase\\ \overline{ET}\cong \overline{TI}\\ \overline{EK}\cong \overline{KI}\\ ET=EK=5y+6=y^2\\ TI=KI=2y+3\\ y^2=5y+6\\ y^2-5y-6=0\\ y^2-6y+y-6=0\\ y\left(y-6\right)+1\left(y-6\right)\\ \left(y-6\right)\left(y+1\right)=0\\ y_1=6,y_2=-1\end{array}$

There are two possibilities,

  $\begin{array}{l}{y}_1=6\\ ET=EK=6^2=36\\ TI=KI=2\left(6\right)+3=15\\ Perimeter=2ET+2TI\\ =2\left(36\right)+2\left(15\right)\\ =72+30=102\\ y_2=-1\\ ET=EK={\left(-1\right)}^2=1\\ TI=KI=2\left(-1\right)+3=1\\ Perimeter=2ET+2TI\\ =2\left(1\right)+2\left(1\right)\\ =2+2=4\end{array}$

  $\begin{array}{l}\sec ondcase:\\ ET=TI\Rightarrow y^2=2y+3\\ EK=KI=5y+6\\ y^2=2y+3\\ y^2-2y-3=0\\ y^2+y-3y-3=0\\ y\left(y+1\right)-3\left(y+1\right)=0\\ \left(y+1\right)\left(y-3\right)=0\\ y_1=-1,y_2=3\end{array}$

There are two possibilities,

  $\begin{array}{l}{y}_1=-1\\ ET=EK=KI=TI=1\\ Perimeter=4\left(1\right)=4\\ y_2=3\\ ET=TI=3^2=9\\ EK=KI=5\left(3\right)+6=21\\ Perimeter=2ET+2EK=2\left(9\right)+2\left(21\right)=60\\ Perimeter=4,60,102\end{array}$

Hence, three possible value of perimeter of kite are $4,60,102$