Field book of an agricultural land is given in the figure. It is divided into 4 plots . Plot I is a right triangle , plot II is an equilateral triangle, plot III is a rectangle and plot IV is a trapezium, Find the area of each plot and the total area of the field. ( use √3 =1.73)

Plot I area= 32.5cm²

Plot II area = 73.09cm²

Plot III area = 35cm²

Plot IV area = 54cm²

Total the area of Field = 194.59cm²

Step-by-step explanation:

The field is made up of four plots with different shapes. So we would find the area of the 4 shapes to get the area of the plots.

A question related to this can be found at (question ID: 18861101)

Find attached the diagram

Given:

AC = 13cm

AE = 19cm

CF = DE = 7cm

AD = AE - DE

AD =  19-7 = 12cm

GF = 9cm, EH = 15cm

GH = 17cm

Plot I: A right angle triangle

Area = ½ × base × height

Base = CD, height = AD

Using Pythagoras theorem

CD = √(AC² - AD)²

CD = √(13² - 12²) = √(169-144)

CD = √25 = 5

Area = ½ × 5 ×13= 32.5

Area plot I = 32.5cm²

Plot II: An equilateral triangle

Area of the equilateral triangle = a²/4 ×(√3)

√3=1.73

a = side = AC

Area = (13)²/4 ×(√3) = 42.25 × 1.73 = 73.0925

Area of Plot II = 73.09cm²

Plot III: A rectangle

Area of a rectangle = length × width

length = 7cm

width = 5cm

Area of plot III = 7×5 = 35cm²

Plot IV: A trapezium

Area of trapezium = ½(base 1 + base2) × height

Base 1= FE = CD

Base 2= GH

To get height using diagram 2. When you draw the lines from the two points on base1, you would have 1 rectangle in the middle with the two triangles by the side.

We would apply Pythagoras theorem to find h in the two right angled triangles:

Hypotenuse ² = opposite ²+adjacent ²

1st ∆: 9² = h²+a²

h² = 81-a²

2nd ∆: 15² = h² + (12-a)²

225 = h² +144 - 24a+ a²

225-144 = h²-24a+ a²

Insert value for h² in the 2nd

81 = 81-a² - 24a + a²

24a = 81-81

a= 0

h² = 81-0²

h = √81 = 9

Area of trapezium = ½(5+7) × 9

= 6 × 9

Area of plot IV= 54cm²

Total the area of field = area of plot I + area of plot II + area of plot III +area of plot IV

= 32.5 + 73.09 + 35+ 54

Total the area of Field = 194.59cm²

-6 is the correct option from the above given options