Chapter 9, Problem 29RP

a.

To determine

Tocalculate: Thelength of a kite.

Answer to Problem 29RP

The length of a kite is 12.

Explanation of Solution

Given information:

In kite, KITE

  $\begin{array}{l}\angle KIT=\angle KET={90}^{\circ }\\ KP=9,\\ TP=4.\end{array}$

Formula used:

The below property is used:

Altitude to hypotenuse is mean proportional between segment of hypotenuse.

Calculation:

  

In a kite, one diagonal is perpendicular bisector of other.

  $\overline{KT}\perp bisects\overline{IE}$ which makes $\overline{IP}$ the altitude of $\Delta KIT$ .

  $\overline{IP}\cong \overline{PE}$

  $\Delta KIT$ is a right angle triangle because $\angle KIT={90}^{\circ }$ .

Altitude to hypotenuse is mean proportional between segment of hypotenuse.

  $\begin{array}{l}\frac{PT}{IP}=\frac{IP}{PK}\\ \frac{4}{IP}=\frac{IP}{9}\\ I{P}^2=36\\ IP=6\\ IE=2\ast IP=2\ast 6=12\end{array}$

b.

To determine

To find: The perimeter of a kite.

Answer to Problem 29RP

The perimeter of a kite KITE is $10\sqrt{13}$ .

Explanation of Solution

Given information:

In kite, KITE

  $\begin{array}{l}\angle KIT=\angle KET={90}^{\circ }\\ KP=9,\\ TP=4.\end{array}$

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Calculation:

  

Side IK and IT can be calculated by applying Pythagoras Theorem.

In a right $\Delta IPK$ ,we get

  $\begin{array}{l}{\left(IK\right)}^2={\left(KP\right)}^2+{\left(IP\right)}^2\\ {\left(IK\right)}^2=9^2+6^2\\ {\left(IK\right)}^2=81+36\\ {\left(IK\right)}^2=117\\ IK=\sqrt{117}\\ IK=3\sqrt{13}\end{array}$

In a right $\Delta IPT$ ,we get

  $\begin{array}{l}{\left(IT\right)}^2={\left(PT\right)}^2+{\left(IP\right)}^2\\ {\left(IT\right)}^2=4^2+6^2\\ {\left(IT\right)}^2=16+36\\ {\left(IT\right)}^2=52\\ IT=\sqrt{52}\\ IT=2\sqrt{13}\end{array}$

In kite, two distinct pairs of consecutive sides are congruent.

  $\begin{array}{l}\overline{KI}\cong \overline{KE},\\ \overline{IT}\cong \overline{ET}.\end{array}$

  $\begin{array}{l}Perimeter=2(IK)+2(IT)\\ Perimeter=2(3\sqrt{13})+2(2\sqrt{13})\\ Perimeter=10\sqrt{13}\end{array}$