Question

Field book of an 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)

Answer 1

Answer:

Total area = 237.09 cm²

Step-by-step explanation:

Given question is incomplete; here is the complete 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)

From the figure attached,

Area of the right triangle I = $\frac{1}{2}(\text{Base})\times (\text{Height})$

Area of ΔADC = $\frac{1}{2}(\text{CD})(\text{AD})$

                        = $\frac{1}{2}(\sqrt{(AC)^2-(AD)^2})(\text{AD})$

                        = $\frac{1}{2}(\sqrt{(13)^2-(19-7)^2} )(19-7)$

                        = $\frac{1}{2}(\sqrt{169-144})(12)$

                        = $\frac{1}{2}(5)(12)$

                        = 30 cm²

Area of equilateral triangle II = $\frac{\sqrt{3} }{4}(\text{Side})^2$

Area of equilateral triangle II = $\frac{\sqrt{3}}{4}(13)^2$

                                                = $\frac{(1.73)(169)}{4}$

                                                = 73.0925

                                                ≈ 73.09 cm²

Area of rectangle III = Length × width

                                 = CF × CD

                                 = 7 × 5

                                 = 35 cm²

Area of trapezium EFGH = $\frac{1}{2}(\text{EF}+\text{GH})(\text{FJ})$

Since, GH = GJ + JK + KH

17 = $\sqrt{9^{2}-x^{2}}+5+\sqrt{(15)^2-x^{2}}$

12 = $\sqrt{81-x^2}+\sqrt{225-x^2}$

144 = (81 - x²) + (225 - x²) + 2$\sqrt{(81-x^2)(225-x^2)}$

144 - 306 = -2x² + $2\sqrt{(81-x^2)(225-x^2)}$

-81 = -x² + $\sqrt{(81-x^2)(225-x^2)}$

(x² - 81)² = (81 - x²)(225 - x²)

x⁴ + 6561 - 162x² = 18225 - 306x² + x⁴

144x² - 11664 = 0

x² = 81

x = 9 cm

Now area of plot IV = $\frac{1}{2}(5+17)(9)$

                                = 99 cm²

Total Area of the land = 30 + 73.09 + 35 + 99

                                    = 237.09 cm²