What is the sum of all integers between: ✓11 and 39
1 answer:
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Answer:
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 brainly (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
