Areas Of Combinations Of Plane Figures

In the previous section, we have learnt to find areas of sectors and segments of a circle. In this section, we will learn to calculate the areas of some plane figures which are combinations of more than one plane figure.

Solution: Area of the square lawn ABCD = (side⁾² = 56 × 56 m² ... (i)

Let OA = OB = x metres

In ΔAOB, ∠AOB = 90° [Diagonals of square bisect at 90°]

So, [By Pythagoras theorem]

or, .. (ii)

Now, area of sector OAB =

= [From (ii)] ... (iii)

Also, area of ΔOAB =

 1/2 x 28 x 56 [∠AOB = 90º] ... (iv)

So area of flower bed AB = area of sector OAB − area of ΔOAB

Similarly, area of the other flower bed

= . ... (vi)

Therefore, total area

Solution: Length of inner curved portion = (340 – 2 × 60) m = 220 m.

∴ The length of each inner curved part = 110 m.

Let the radius of each inner curved part be r.

Then πr = 110 m

∴ Inner radius = 35 m, outer radius = (35 + 7) m = 42 m.

∴ Area of the track = (area of 2 rectangles) + (area of the circular ring)

Hence, the outer perimeter of the track is 384 m.

Solution: Area of the shaded region =

Area of the major sector of the circle + Area of the equilateral triangle OAB.

Solution: .

Then for the quadrant OPBQ, r = 20√2 cm.

The area of the shaded region

= The area of the quadrant OPBQ – The area of the square OABC

So, the total area of the handkerchief = (42)cm²  = 1764 cm².

Area of the nine circular designs = 9 x πr²

= 9 x 22/7 (7)² cm² = 9 x 22 x 7 cm² = 1386cm².

Then the required area of the remaining portion

= 1764cm² - 1386cm² = 378cm²

So, area of one circle = πr² = 22/7 x 7/2 x 7/2 cm²

= 77/2 cm².

Therefore, area of the four circles = 4 x 77/2 cm² = 154cm2.

Hence, area of the shaded region = (196 – 154) cm² = 42 cm².