Advertisements
Advertisements
Question
Find the area of the irregular polygon shaped fields given below.
Advertisements
Solution
Area of the field = Area of trapezium FBCH + Area of ∆DHC + Area of ∆EGD + Area of ∆EGA + Area of ∆BFA
Area of the triangle = `1/2 "bh sq. units"`
Area of the trapezium = `1/2 xx "h" xx ("a" + "b") "sq. units"`
Area of the trapezium FBCH = `1/2 xx (10 + 8) xx (8 + 3) "m"^2`
= 9 × 11 = 99 m2 ...(1)
Area of the ∆DHC = `1/2 xx 8 xx 5 "m"^2`
= 20 m2 ...(2)
Area of ∆EGD = `1/2 xx 8 xx 15 "m"^2`
= 60 m2 ....(3)
Area of ∆EGA = `1/2 xx 8 xx (8 + 6) "m"^2`
= 4 × 14 m2
= 56 m2
Area of ∆BFA = `1/2 xx 3 xx 6 "m"^2`
= 9 m2
∴ Area of the field = 99 + 20 + 60 + 56 + 9 m2
= 244 m2
Area of the field = 244 m2
APPEARS IN
RELATED QUESTIONS
Find the area of the shaded part in the following figure. (π = 3.14)
Find the area of the shaded part in the following figure. (π = 3.14)
Find the area of the combined figure given which is got by joining of two parallelograms
The doormat which is in a hexagonal shape has the following measures as given in the figure. Find its area.

Find the perimeter and area of the following shape
The following figure are of equal area. Which figure has the least perimeter?
If two identical rectangles of perimeter 30 cm are joined together, then the perimeter of the new shape will be
A rectangle has length 40 cm and breadth 20 cm. How many squares with side 10 cm can be formed from it
Two plots have the same perimeter. One is a square of side 10 m and another is a rectangle of breadth 8 m. Which plot has the greater area and by how much?
Look at the picture of the house given and find the total area of the shaded portion.
