# Areas Of Combinations Of Plane Figures

## Area Related To Circles of Class 10

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.

 question 1. In the figure, two circular flowerbeds have been shown on two sides of a square lawn ABCD of side 56 m. If the centre of each circular flowerbed is the point of intersection O of the diagonals of the square lawn, find the sum of the areas of the lawn and the flowerbeds.

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

 question 2. The inside perimeter of a running track, (figure) is 340 m. The length of each straight portion is 60 m, and the curved portions are semicircles. If the track is 7 m wide, find the area of the track. Also, find the outer perimeter of the track.

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.

 3. Find the area of the shaded region in the figure, where a circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

Solution: Area of the shaded region =

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

 question 4. In the figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (π = 3.14).

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

 question 5. On a square handkerchief, nine circular designs each of radius 7 cm are made (figure). Find the area of the remaining portion of the handkerchief. Solution: Radius of the circle = 7 cm then diameter of the circle = 14 cm. then length of the side of the square ABCD = 3 × 14 cm = 42 cm.

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

 question 6. Find the area of the shaded region in the figure, where ABCD is a square of side 14 cm. Solution: Area of square ABCD = 14 × 14 cm2 = 196 cm2. Diameter of each circle = 14/2 = 7cm. So, radius of each circle = 7/2 cm.

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.