Chapter 11.4, Problem 2PSA

a

To determine

To calculate: The area of kite with diagonal $8$ and $14$

Answer to Problem 2PSA

Area of kite is $56$ square units.

Explanation of Solution

Given information : Figure; length of diagonals

Formula used : Area of kite = $\frac{1}{2}\times d_1\times d_2$ (where $d_1$ and $d_2$ are diagonals of the rhombus)

Calculation: According to the given figure,

  $d_1=4+4=8$ and $d_2=5+9=14$

  $\therefore$ Diagonals of rhombus are $8$ and $14$ .

Area of rhombus = $\frac{1}{2}\times d_1\times d_2$

  $\begin{array}{l}\Rightarrow \frac{1}{2}\times 8\times 14\\ \Rightarrow \frac{112}{2}\\ \Rightarrow 56\end{array}$

b

To determine

To calculate: The area of kite with one side $13$

Answer to Problem 2PSA

Area of kite is $85$ square units.

Explanation of Solution

Given information : One side of kite is $13$ .

Formula used : Area of kite = $\frac{1}{2}\times d_1\times d_2$ (where $d_1$ and $d_2$ are diagonals of the rhombus)

Calculation: Let ABCD be the kite with diagonals AC and BD intersecting at O.

Now, in $\Delta BOC$ , $OB=OD$ (Given)

Now, by using Pythagoras theorem,

  $\begin{array}{l}$ {(OB)}^2+{(OC)}^2={(BC)}^2$$ \Rightarrow {(5)}^2+O{C}^2={(13)}^2$\Rightarrow 25+O{C}^2=169\\ \Rightarrow O{C}^2=169-25\\ \Rightarrow O{C}^2=144\\ \Rightarrow OC=\sqrt{144}\\ \Rightarrow OC=12\end{array}$

Now, According to the figure, $OD=OA=5$

  $AC=OA+OC=5+12=17$ and $BD=OB+OD=5+5=10$

  $\therefore$ Diagonals are $17$ and $10$ .

Area = $\frac{1}{2}\times d_1\times d_2$

  $\begin{array}{l}\Rightarrow \frac{1}{2}\times AC\times BD\\ \Rightarrow \frac{1}{2}\times 17\times 10\\ \Rightarrow \frac{170}{2}\\ \Rightarrow 85\end{array}$

c

To determine

To calculate: The area of kite with side $16$ and diagonal $20$ .

Answer to Problem 2PSA

Area of kite is $160$ square units.

Explanation of Solution

Given information : One side of kite is $16$ and one diagonal is $20$ .

Formula used : Area of kite = $\frac{1}{2}\times d_1\times d_2$ (where $d_1$ and $d_2$ are diagonals of the rhombus)

Calculation: Let ABCD be the kite with diagonals AC and BD intersecting at O.

Now, in $\Delta BOA$ , as angles are $90°,60°,30°$ ,

  $\Rightarrow$ Hypotenuse will be $2a$ and perpendicular will be $a$ .

Now, as hypotenuse = $a$ ,

  $\begin{array}{l}\Rightarrow 16=2a\\ \Rightarrow a=\frac{16}{2}\\ \Rightarrow a=8\end{array}$

  $\therefore$ Perpendicular (OB) = $a=8$

Now, $BD=2(OB)$ (Longer diagonal of kite bisects the shorter diagonal into two equal halves)

  $\begin{array}{l}\Rightarrow BD=2(8)\\ \Rightarrow BD=16\end{array}$

  $\therefore$ Diagonals are $20$ and $16$ .

Area = $\frac{1}{2}\times d_1\times d_2$

  $\begin{array}{l}\Rightarrow \frac{1}{2}\times AC\times BD\\ \Rightarrow \frac{1}{2}\times 20\times 16\\ \Rightarrow \frac{320}{2}\\ \Rightarrow 160\end{array}$