Perimeter and Area of Irregular Figures

Here we will get the ideas how to solve the problems on finding the perimeter and area of irregular figures.

1. The figure PQRSTU is a hexagon.

PS is a diagonal and QY, RO, TX and UZ are the respective distances of the points Q, R, T and U from PS. If PS = 600 cm, QY = 140 cm, RO = 120 cm, TX = 100 cm, UZ = 160 cm, PZ = 200 cm, PY = 250 cm, PX = 360 cm and PO = 400 cm. Find the area of the hexagon PQRSTU.

Solution:

Area of the hexagon PQRSTU = area of ∆PZU + area of trapezium TUZX + area of ∆TXS + area of ∆PYQ + area of trapezium QROY + area of ∆ROS

= {$$\frac{1}{2}$$ × 200 × 160 + $$\frac{1}{2}$$ (100 + 160)(360 – 200) + $$\frac{1}{2}$$ (600 – 360) × 100 + $$\frac{1}{2}$$ × 250 × 140 + $$\frac{1}{2}$$ (120 + 140) (400 – 250) + $$\frac{1}{2}$$ (600 – 400) × 120} cm$$^{2}$$

= (16000 + 130 × 160 + 120 × 100 + 125 × 140 + 130 × 150 + 100 × 120) cm$$^{2}$$

= (16000 + 20800 + 12000 + 17500 + 19500 + 12000) cm$$^{2}$$

= 97800 cm$$^{2}$$

= 9.78 m$$^{2}$$

2. In a square lawn of side 8 m, an N-shaped path is made, as shown in the figure. Find the area of the path.

Solution:

Required area = area of the rectangle PQRS + area of the parallelogram XRYJ + area of the rectangle JKLM

= (2 × 8 + PC × BE + 2 × 8) m$$^{2}$$

= (16 + 2 × 4 + 16) cm$$^{2}$$

= 40 m$$^{2}$$

We can solve this problem using another method:

Required area = Area of the square PSLK – Area of the ∆RYM – Area of the ∆XQJ

= [8 × 8 - $$\frac{1}{2}$${8 – (2 + 2)} × 6 - $$\frac{1}{2}$${8 – (2 + 2)} × 6] m$$^{2}$$

= (64 – 12 – 12) m$$^{2}$$

= 40 m$$^{2}$$

You might like these

• Area and Perimeter of Combined Figures | Circle | Triangle |Square

Here we will solve different types of problems on finding the area and perimeter of combined figures. 1. Find the area of the shaded region in which PQR is an equilateral triangle of side 7√3 cm. O is the centre of the circle. (Use π = $$\frac{22}{7}$$ and √3 = 1.732.)

• Area and Perimeter of a Semicircle | Solved Example Problems | Diagram

Here we will discuss about the area and perimeter of a semicircle with some example problems. Area of a semicircle = $$\frac{1}{2}$$ πr$$^{2}$$ Perimeter of a semicircle = (π + 2)r. Solved example problems on finding the area and perimeter of a semicircle

• Area of a Circular Ring | Radius of the Outer Circle and Inner Circle

Here we will discuss about the area of a circular ring along with some example problems. The area of a circular ring bounded by two concentric circle of radii R and r (R > r) = area of the bigger circle – area of the smaller circle = πR^2 - πr^2 = π(R^2 - r^2)

• Area and Circumference of a Circle |Area of a Circular Region |Diagram

Here we will discuss about the area and circumference (Perimeter) of a circle and some solved example problems. The area (A) of a circle or circular region is given by A = πr^2, where r is the radius and, by definition, π = circumference/diameter = 22/7 (approximately).

• Perimeter and Area of Regular Hexagon | Solved Example Problems

Here we will discuss about the perimeter and area of a Regular hexagon and some example problems. Perimeter (P) = 6 × side = 6a Area (A) = 6 × (area of the equilateral ∆OPQ)