Question

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) plse and mee its urgent

Answer 1

Answer:

Total area = 237.09 $cm^2$

Step-by-step explanation:

Area of equilateral triangle is given as:

$A = \dfrac{\sqrt3}{4} \times side^2$

$\triangle ABC$ is equilateral triangle with side = 13 cm

$A_{II} = \dfrac{\sqrt3}{4} \times 13^2 = 73.09 cm^2$

Side EA = ED + DA

CDEF is a rectangle, so CF = ED

EA = CF+DA

19=7+DA

DA = 12 cm

Looking in the region I, i.e. right angle $\triangle CDA$.

According to pythagoras theorem:

$\text{Hypotenuse}^{2} = \text{Base}^{2} + \text{Perpendicular}^{2}$

$13^{2} = 12^{2} + \text{CD}^{2}\\\text{CD}^{2} = 25\\CD = 5 cm$

Area of a right angled Triangle:

$A = \dfrac{1}{2}\times Base \times Perpendicular$

$A_I = \dfrac{1}{2}\times 12 \times 5\\A_I = 30 cm^2$

Area of rectangle is given as: Length $\times$ Width

$A_{III} = 7 \times 5 = 35cm^2$

For finding area of trapezium i.e. region IV, let us draw a line parallel to side FG at E that will cut GH at a point P and 'h' is the height of triangle or distance between parallel sides of trapezium.

Now, we have a triangle EPH whose 3 sides are given as a=12, b=9 and c=15.

$s=\dfrac{a+b+c}{2}\\\Rightarrow s=\dfrac{12+9+15}{2} = 18$

Using hero's formula for area of a triangle:

$A =\sqrt{s(s-a)(s-b)(s-c)}\\A =\sqrt{18(6)(9)(3)} = 54cm^2$

Comparing with:

$A = \dfrac{1}{2}\times Base \times Perpendicular$

$54 = \dfrac{1}{2}\times 12 \times h\\\Rightarrow h = 9cm$

h is distance between parallel sides of trapezium.

Area of trapezium:

$A_{IV} = \dfrac{1}{2} (FE+GH)\times h\\\Rightarrow A_{IV} = \dfrac{1}{2} (5+17)\times 9 =99cm^2$

Total area = $A = A_{I}+A_{II}+A_{III}+A_{IV} = 30 +73.09+35+99 = 237.09cm^2$