Chapter 9.4, Problem 23PSB

a.

To determine

To calculate:Thearea of the plot.

Answer to Problem 23PSB

The area of the plot is $1500m^2$ .

Explanation of Solution

Given information:

Length of rectangle $(AB)=40m,$

Width of rectangle $(AC)=30m,$

Side $(CE)=60m,$

Base of a triangle $(DE)=20m,$

Height oftriangle $(BD)=30m.$

Formula used:

Area of rectangle: $A=w\ast l$

w = width of rectangle,

l = length of rectangle.

Area of triangle: $A=\frac{1}{2}\ast b\ast h$

b = base of triangle,

h = height of triangle.

Calculation:

  

Draw altitude BD.

  $A=A_{ABDC}+A_{BDE}$

  $A_{ABDC}$ = Area of rectangle ABDC .

  $A_{BDE}$ = Area of triangle BDE .

  $\begin{array}{l}A=(AC\ast AB)+(\frac{1}{2}\ast DE\ast BD)\hfill \\ \begin{array}{l}A=\left(30\ast 40\right)+(\frac{1}{2}\ast 20\ast 30)\\ A=1200+300\end{array}\hfill \end{array}$

  $A=1500m^2$

b.

To determine

To calculate: The length of a side BE .

Answer to Problem 23PSB

The length of a side BEis $\text{36 }m$ .

Explanation of Solution

Given information:

Length of rectangle $(AB)=40m,$

Width of rectangle $(AC)=30m,$

Side $(CE)=60m,$

Base of triangle $(DE)=20m,$

Height of triangle $(BD)=30m.$

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 BE can be calculated by using Pythagoras theorem.

In right angle triangle BDE , we get

  ${\left(BE\right)}^2={\left(BD\right)}^2+{\left(DE\right)}^2$

  $\begin{array}{l}{\left(BE\right)}^2={\left(30\right)}^2+{\left(20\right)}^2\\ =900+400\\ =1300\\ \therefore BE\approx 36m\end{array}$