Can’t see the model.......................
4/5 and 40/50 are both equivalent
In the process of laying out the question, you have given us important facts about the shaded area and the unshaded area. Now step back for just a second. If they're that deeply involved in the question, do you honestly expect that anybody can answer the question without seeing the picture of them ? ! ?
first subtract the last equation from the first. this gives:-
-x + y = -8 .....................(1)
Then multiply the first equation by 2 and add it to the 2nd equations This gives
9x = 18
so x = 2
and from equation (1) y = -8 + 2 = -6
Substituting for x and y in the second equation
z = (24 -5(2) -12) / 2 = 1
Answer is choice b.
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 = 
Area of ΔADC = 
= 
= 
= 
= 
= 30 cm²
Area of equilateral triangle II = 
Area of equilateral triangle II = 
= 
= 73.0925
≈ 73.09 cm²
Area of rectangle III = Length × width
= CF × CD
= 7 × 5
= 35 cm²
Area of trapezium EFGH = 
Since, GH = GJ + JK + KH
17 = 
12 = 
144 = (81 - x²) + (225 - x²) + 2
144 - 306 = -2x² + 
-81 = -x² + 
(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 = 
= 99 cm²
Total Area of the land = 30 + 73.09 + 35 + 99
= 237.09 cm²