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. Find the perimeter of the given figure.

Solution:

Perimeter = AB + BC + CD + DE + EF + FG + GA

          = 3.2 cm + 1.5 cm + 5 cm + 5 cm + 1.5 cm + 3.2 cm + 2 cm

          = 21.4 cm

2. Find the perimeter of each of the following figures:

(i) Perimeter of the region = (2 + 19 + 2 + 9 + 10 + 3 + 10 + 7) cm

                                       = 62 cm.

(ii) Perimeter = AB + BC + CD + DE + EF + AF

                    = (100 + 120 + 90 + 45 + 60 + 80) m

                    = 495 m .

3. 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\(^{2}\)

4. 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\(^{2}\)

9th Grade Math

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