Chapter 11.2, Problem 31PSC

To determine

To find the area of kite ABCD.

Answer to Problem 31PSC

The area of kite ABCDis $120$ .

Explanation of Solution

Given:

  $\begin{array}{l}BD=10,\\ AC=24\end{array}$

Formula Used:

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ b=Base,\\ h=Height\end{array}$

Calculation:

  

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

If $BD=10,$

  $\overline{AC}$ is a perpendicular bisector of $\overline{BD}$ , then

  $BX=5,XD=5.$

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

  $\begin{array}{l}\overline{AB}\cong \overline{AD},\\ \overline{BC}\cong \overline{DC}\end{array}$

Two triangles are congruent by SSS congruence rule.

  $△ABC\cong △ADC$

  $\begin{array}{l}Areaofkite=2*Areaoftriangle\\ A(ABCD)=2*\frac{1}{2}\ast b\ast h\\ =2(\frac{1}{2})(AC)(BX)\\ =2\frac{1}{2}(24*5)=120\\ A(ABCD)=120\end{array}$