Areas Of Regular Polygon Circles And Plane Figures

EXERCISE 1. A rectangular lawn, 75m x 60 m, has two roads, each 40 m wide, running through the middle of the lawn, one parallel to the length and other parallel to the breadth, as shown in figure 9-21. The area of the these roads is

a. 540m 2 b. 504m 2 c. 524m 2 d. 542m 2 Sol:

Correct option is (c)

The shaded portion in figure 9-21 is the area of the cross-roads. Now, area of cross roads in a rectangular field = (length of the field + breadth of the field – width of the roads) x width of the roads

= (75 + 60 − 4) × 4 = (135 − 4) × 4 = 131× 4 = 524m 2

2. In figure 9-22, the area of the shaded portion between two concentric circles is 770cm2 . If the radius of the outer circle is 21 cm, then the radius of the inner circle is

a. 14 cm b. 7 cm c. 28 cm d.

35 cm 2

Sol:

Correct option is (a)

The shaded portion of figure 9-22 is called a ring The area of a ring = π ( R 2 − r 2 ) , where R is the radius of the outer circle and r is the radius of the inner circle. Let the radius of the inner circle = x cm Then, π (212 − x 2 ) = 770 or

22 (441 − x 2 ) = 770 7

or (441 − x 2 ) = 770 ×

7 22

or (441 − x 2 ) = 245 or − x 2 = 245 − 441 or − x 2 = −196 or x 2 = 196 or x = 196 = 14cm Therefore, Radius of the inner circle is 14 cm 3. The area of the shaded portion in figure 9-23 is

a. 441m2 b.

1155 2 m 8

c. 144m 2 d.

1225 2 m 8

Sol:

Correct option is (b)

The area of the shaded portion is = Area of semicircle AE – area of semicircle AB + area of semicircle BC – area of semicircle CD + area of semicircle CD.

2 2 2 2 1 ⎡ ⎛ 21 ⎞ ⎛7⎞ ⎛7⎞ ⎛7⎞ ⎤ = ⎢π ⎜ ⎟ − π ⎜ ⎟ + π − π ⎜ ⎟ + π ⎜ ⎟ ⎥ 2 ⎣⎢ ⎝ 4 ⎠ ⎝2⎠ ⎝2⎠ ⎝ 2 ⎠ ⎦⎥

=

2 2 ⎤ 1 ⎡ ⎛ 21 ⎞ ⎛7⎞ ⎢π ⎜ ⎟ − π ⎜ ⎟ + π ⎥ 2 ⎣⎢ ⎝ 4 ⎠ ⎝2⎠ ⎥⎦

=

1 ⎡ 441 49 ⎤ π −π +π ⎥ ⎢ 2 ⎣ 16 4 ⎦

=

1 ⎡ 441π − 196π + 1225π ⎤ ⎥⎦ 2 ⎢⎣ 16

=

1 ⎡1666π − 196π ⎤ ⎥⎦ 2 ⎢⎣ 16

=

1 ⎡1470π ⎤ 2 ⎢⎣ 16 ⎥⎦

1 1470 22 = × × 2 16 7 =

1155 2 m 8

4. The area of an equilateral triangle is 49 3cm 2 . Taking each angular point as centre, a circle is described with radius equal to half the length of the side of the triangle as shown in figure 9-24 below. Then the area of the shaded portion is a. 50 3 − 70cm 2 b. 49 3 − 70cm 2 c. 40 2 − 77cm 2 d. 49 3 − 77cm 2 Sol:

Correct option is (d)

Area of the shaded portion = Area of the equilateral Triangle ABC -Area of the three quadrant Let the length of the each side of the equilateral Triangle = x cm Therefore, Area of equilateral =

Hence

3 2 x 4

3 2 x = 49 3 4

or x 2 = 49 3 ×

4 3

or x 2 = 49 × 4 or x 2 = 196 or x = 196 = 14cm Now radius of the circle =

14 = 7cm 2

Therefore, Area of three sector = 3 ×

60 22 × × 7 × 7 = 77cm 2 360 7

Thus, area of the shaded portion = 49 3 − 77cm 2 5. ABCD is a square lawn with side AB = 42 m. Two circular flower beds are there on the sides AD and BC with centre at the intersection of its diagonals. Then the area of both the flower beds (shaded portion in figure 9-25) is

a. 405m 2 b. 500m 2 c. 504m 2 d. 450m 2 Sol:

Correct option is (c)

We know that the diagonals of a square bisect each other at right angles. Therefore, ∠AOD = 90o and OA=OD

Using Pythagorean Theorem, in right Triangle AOD , we have

OA2 + OD 2 = AD 2 or OA2 + OA2 = AD 2

( Since OA = OD and AD=42m)

or 2OA2 = AD 2 or 2OA2 = 422 or OA2 =

422 2

or OA2 = 882 or OA = 882 = 21 2m Now, area of sector OAED =

90 × π × (21 2) 2 360

2 22 = × × 441× 2 = 693m 2 4 7

1 Area of OAD = × OA × OD 2 1 = × 21 2 × 21 2 = 441m2 2 Therefore, Area of flower bed = Area of sector OAED – Area of OAD = 693-441 = 252m2 Therefore, Area of both the flower beds = 2 × 252 = 504m2 6. The area of the shaded portion in figure 9-26 in which ABCD is a rectangle 28 cm long, and the three circles have the same radii is

a.

21 2 cm 2

b.

21 2 cm 4

c. 21cm 2 d. 10cm2 Sol:

Correct option is (a)

From the figure, it is clear that the diameter of the two circles, touching each other is the length of the rectangle and the breadth of the rectangle equal to the diameter of the circle. Let the radius of each circle = rcm its diameter = 2r Hence, length of the rectangle = 2 × 2r = 4rcm and the breadth = 2r

Therefore, 4r = 28 or r =

28 = 7cm 4

Thus, breadth of the rectangle = 2 × 7 = 14cm

Now if we draw a line EF, such that it touches the two circles, then AFED will be a square. Now, the area enclosed between the corner of a square and the 1 circumference of the circle is th of the difference in areas of the square and 4 the circle. Since, there are two such circles and two squares, therefore, the 1 shaded area = [Area of the rectangle – area of two circles] 8

1⎡ 22 ⎤ = ⎢ 28 ×14 − 2 × × 7 × 7 ⎥ 8⎣ 7 ⎦ =

1 [392 − 308] 8

=

1 [84] 8

=

21 2 cm 2

7. In figure 9-27, a circle is inscribed in a semi-circle. If the radius of the semicircle is 6 cm, then the area of the un-shaded portion is

a.

376 2 cm 7

b.

198 2 cm 7

c.

99 2 cm 7

d.

297 2 cm 7

Sol:

Correct option is (b)

Since, the circle has been inscribed in a semicircle; therefore, the radius of the inscribed circle is half the radius of the semicircle. Hence radius of the circle =

6 = 3cm 2

Now, the area of the un-shaded portion is the difference of the areas of the semicircle and the circle. Therefore, Area of the un-shaded portion = Area of the semicircle – Area of the circle Therefore, required area =

=

397 198 − 7 7

=

397 − 198 7

=

198 2 cm 7

1 22 22 × × 6 × 6 − × 3× 3 2 7 7

8. In figure 9-28, the radius of quarter circular plot taken is 2m and the radius of the flower bed is also 2m. Then the area of the un-shaded portion is

a.

175 2 cm 7

b.

165 2 cm 7

c.

170 2 cm 7

d.

160 2 cm 7

Sol:

Correct option is (d)

The required area = Area of the rectangular field – (Area of the four quadrants + Area of the central circle)

22 ⎛ 1 22 ⎞ = 8× 6 − ⎜ 4× × × 2× 2 + × 2× 2 ⎟ 4 7 7 ⎝ ⎠ ⎛ 88 88 ⎞ = 48 − ⎜ + ⎟ ⎝ 7 7 ⎠ ⎛ 88 + 88 ⎞ = 48 − ⎜ ⎟ ⎝ 7 ⎠ ⎛ 176 ⎞ = 48 − ⎜ ⎟ ⎝ 7 ⎠ =

236 − 176 7

=

160 2 m 7

9. The area of the square ABCD is 16cm 2 . The area of the square formed by joining the mid points of the sides of the square ABCD is a. 8cm 2 b. 4cm 2 c. 16cm2 d. 12cm 2 Sol:

Correct option is (a)

The square formed by joining the mid points of the sides of the square is 16 = 2cm EFGH. Now , AE=AH = 2

But area of square EFGH = ( HE ) 2

… (1)

Now, let us find HE In right triangle HAE, using Pythagorean Theorem, we have AE 2 + AH 2 = HE 2

or 22 + 22 = HE 2 or 4 + 4 = HE 2 or 8 = HE 2 From (1), we have area of the square EFGH = HE 2 = 8cm 2 10. In figure 9-30, all the lengths are in cm. The area of the figure is a. (800 + 10 3)cm2 b. (80 + 10 3)cm2 c. (800 + 100 3)cm2 d. (800 + 1000 3)cm2

Sol:

Correct option is (d)

The area of the figure is Area of the rectangle + Area of the two equilateral triangles with sides 40 cm + Area of the two equilateral triangles with sides 20 cm Therefore, required area = 40 × 20 + 2 ×

3 3 × 40 × 40 + 2 × × 20 × 20 40 4

= 800 + 800 3 + 200 3 = 800 + 1000 3cm 2

11. In figure 9-31, ABCD is a square. If the area of the right triangle DEC is 16cm2 , then the area of the shaded portion is a. 80cm2 b. 64cm 2 c. 48cm 2 d. 96cm2

Sol:

Correct option is (b)

Since the area of the right triangle is 16cm2

Therefore,

1 × 2 x × x = 16 2

or x 2 = 16 or x = 16 = 4cm Hence DE = 2 x 4 = 8 cm and EC = 4 cm Using Pythagorean Theorem, we have

DE 2 + EC 2 = DC 2 or 82 + 42 = DC 2 or 64 + 16 = DC 2 or 80 = DC 2

… (1)

Now, area of the square = ( side) 2 = ( DC ) 2 = DC 2 From, (1), we have

Area of the square ABCD = DC 2 = 80cm 2 Now, area of the shaded portion = 80 − 16 = 64cm2 12. The length and breadth of a rectangular piece of paper are 21 cm and 14 cm respectively. A semi-circle portion is cut off from the breadth side and semicircle paper is added to the length side, as shown in figure 9-32. Then the area of the shaded region is

a. 1128cm2 b. 1064cm2 c. 2064cm2 d. 1084cm2 Sol:

Correct option is (b)

The area of the shaded region = Area of the rectangle + Area of the semi-circle on length – Area of semi circle on breadth

= 21×14 + π × 212 − π × 142 = 294 +

22 22 × 21× 21 − × 14 × 14 7 7

= 294 + 1386 − 616 = 1064cm 2 13. In figure 9-33, a semi circle is drawn on AB as diameter and O is the centre. Semi circles are formed on AO and OB as diameter. If AB = 28 cm then the area of the shaded region is

a. 308cm2 b. 231cm2 c. 154cm 2 d. 208cm2 Sol:

Correct option is (c)

Clearly, area of the shaded region = Area of the big semi circle – Area of the two small semi circles

1 1 = × π × 142 − 2 × × π × 7 2 2 2 1 22 1 22 = × × 14 ×14 − 2 × × × 7 × 7 2 7 2 7

= 308 − 154 = 154cm 2 14. In figure 9-34, the area of the shaded region is

a. 117cm 2 b. 119cm2 c. 60cm 2 d. 118cm 2 Sol:

Correct option is (b)

Obviously, the area of the shaded region = Area of the square – Area of the four semicircles

1 22 7 7 = 14 ×14 − 4 × × × × 2 7 2 2 = 196 − 77 = 119cm 2 15. The area of the shaded portion of figure 9-35 is

a. 77cm 2 b.

77 2 cm 4

c.

231 2 cm 4

d.

321 2 cm 4

Sol:

Correct option is (c)

The area of the shaded portion = Area of the sector AOB – Area of sector COD

=

45 45 × π × 142 − × π × 72 360 360

=

45 22 45 22 × × 14 × 14 − × × 7× 7 360 7 360 7

= 77 −

77 4

=

308 − 77 4

=

231 2 cm 4

16. In figure 9-36, if all the steps are region is

a. 4m 2 b. 8m 2 c. 16m 2 d. 32m 2

1 m high, then the area of the shaded 2

Sol:

Correct option is (a)

The area of the shaded region = Area of square A + Area of rectangle B + Area of rectangle C + Area of rectangle D

1 1 ⎛1 3⎞ 1 5 1 7 = × +⎜ × ⎟+ × + × 2 2 ⎝2 2⎠ 2 2 2 2

=

1 3 5 7 + + + 4 4 4 4

=

1+ 3 + 5 + 7 4

=

16 = 4m 2 4

17. In figure 9-37, if ABCD is a square of side 28 cm, then the area of the shaded portion is

a. 784cm2 b. 168cm2 c. 616cm2

d. none of the above Sol:

Correct option is (b)

The area of the shaded region =Area of the square ABCD-area of two semi circle 1 22 = 28 × 28 − 2 × × × 14 × 14 2 7 = 784 − 616 = 168cm 2

18. In figure 9-38, the area of the shaded region is

a.

349 + 62 3 2 cm 28

b.

351 + 63 2 2 cm 28

c.

63 + 351 3 2 cm 28

d.

351 + 63 3 2 cm 28

Sol:

Correct option is (d)

Area of the shaded region = Area of the semi circle A + Area of square B + Area of equilateral triangle C

1 22 3 3 3 = × × × + 3× 3 + × 3× 3 2 7 2 2 4

=

99 9 +9+ 3 28 4

=

99 + 252 + 63 3 28

=

351 + 63 3 28

19. In figure 9-39, two circle, O and P intersect each other at A and B. Then the area of their shaded portion is

a.

4279 7 95 + 49 3 − 21 4

b.

4279 7 95 + 49 3 + 21 4

c.

4279 7 95 + 49 2 − 21 4

d.

4279 7 95 + 49 3 − 4 21

Sol:

Correct option is (b)

It is clear from the figure the shaded area = Area of circle O + Area of circle P – (area of segment of circle O + area of segment of circle P) In Triangle OAB, OA = OB

(Radii of the same circle are equal)

Therefore ∠OBA = ∠OAB

(Base angles of isosceles triangles are equal)

Now, in Triangle OAB,

∠AOB + ∠OBA + ∠OAB = 180o

(Angle sum property of triangles)

or 60o + ∠OAB + ∠OAB = 180o

( Since ∠OBA = ∠OAB )

or 60o + 2∠OAB = 180o or 2∠OAB = 180o − 60o or 2∠OAB = 120o or ∠OAB =

120o 2

or ∠OAB = 60o Therefore, In Triangle OAB,

∠AOB = ∠OBA = ∠OAB = 60o Therefore, Triangle OAB is an equilateral triangle Thus, OA=OB=AB=7 cm

Therefore, Area of Triangle OAB =

3 2 3 ×7 = × 49 4 4

…

(1) Since AR is perpendicular to OP, Therefore, Triangle ARP is a right triangle in which AR =

7 cm 2

(Since OR is altitude of equilateral Triangle OAB, and altitude of an equilateral triangle bisects the base) Now, in right Triangle ARP, we have

AP 2 − AR 2 = RP 2

(Using Pythagorean Theorem)

2

⎛7⎞ or 62 − ⎜ ⎟ = RP 2 ⎝2⎠ or 36 −

49 = RP 2 4

or

144 − 49 = RP 2 4

or

95 = RP 2 4

or RP =

95 2

Now area of the sector PAB =

… (2)

120 22 264 2 × × 6× 6 = cm 360 7 7

Therefore, Area of segment of circle P =

=

264 1 95 − ×7× 7 2 2

=

264 7 95 − 7 4

264 - Area of Triangle ABP 7

… (3)

Area of segment of circle O = Area of sector BOA – Area of Triangle OAB

=

60 22 3 × ×7×7 − × 49 360 7 4

=

154 49 3 − 6 4

=

77 49 3 − 3 4

Now, area of the two segments =

(using 1)

… (4)

264 7 95 77 49 3 − + − 7 4 3 4

Area of the shaded portion

=

⎛ 264 7 95 77 49 3 ⎞ 22 22 × 7 × 7 + × 6 × 6 − ⎜⎜ − + − ⎟ 7 7 4 3 4 ⎟⎠ ⎝ 7

= 154 +

792 264 7 95 77 49 3 − + − + 7 7 4 3 4

=

3234 + 2376 − 792 − 539 7 95 49 3 + + 21 4 4

=

5610 − 1331 7 95 + 49 3 + 21 4

=

4279 7 95 + 49 3 2 cm + 21 4

20. The area of the shaded region of figure 9-40 is

… (5)

a. 462cm2 b. 308cm 2 c. 468cm 2 d. 434cm 2 Sol:

Correct option is (c)

Area of the bigger semi-circle =

1 22 × × 14 × 14 = 308cm 2 2 7

⎛ 1 22 ⎞ Area of the smaller semi-circle = 2 ⎜ × × 7 × 7 ⎟ = 154cm 2 ⎝2 7 ⎠ Area of the triangle =

1 × 2 × 6 = 6cm 2 2

Therefore, Total shaded area = 308 + 154 + 6 = 468cm2

21. An athletic track 14 m wide consists of two straight sections 120 m long joining semicircular ends. If the inner radius of the circular ends is 35 m and their outer radius is 49 m, then the area of the shaded region is

a. 7056m 2 b. 8400m 2 c. 1680m 2 d. 4200m 2 Sol:

Correct option is (a)

From figure 9-41, we find that the area of the shaded region = Area of rectangle LBCM + Area of rectangle AKND + 2 ( area of semi circle with radius 49 m – area of semi circle with radius 35 m) 1 22 ⎡ 1 22 ⎤ = (14 × 120) + (14 × 120) + 2 ⎢ × × 492 − × × 352 ⎥ 2 7 ⎣2 7 ⎦ = 1680 + 1680 + 2 [3773 − 1925] = 3660 + 3696 = 7056m 2

22. In figure 9-42, AB parallel to CD parallel to EF . The area of the shaded region is

a. 397cm2 b. 280cm 2 c. 400cm 2 d. 329cm2 Sol:

Correct option is (d)

Since AB parallel to CD parallel to EF . Therefore, quadrilaterals ABCD and DCEF are trapezoids. So, area of the shaded portion = Area of trapezoid ABCD + area of trapezoids DCEF + area of semi circle AGF

=

1 1 1 22 (20 + 16) × 7 + (20 + 16) × 7 + × × 7 × 7 2 2 2 7

1 1 = × 36 × 7 + × 36 × 7 + 77 2 2 = 126 + 126 + 77 = 329cm 2

23. A boy is cycling such that the wheels of the cycle are making 150 revolutions per minute. If the diameter of the wheel is 70 cm, then the speed of the boy in km/hr is a. 20 km/hr b.

99 km/hr 5

c. 40 km/hr d. 21 km/hr Sol:

Correct option is (b)

Radius of the wheel =

70 = 35cm 2

Since Circumference of the wheel = 2 ×

22 × 35 = 220cm 7

Thus distance covered by the cycle in 1 revolution = 220 cm So, distance covered by the cycle in 150 revolution = 220 x 150 = 33000 cm Hence, distance covered in one minute = 33000 cm Therefore, Distance covered in one hour (60 minutes) = 33000 x 60 = 1980000 cm So, the speed of the boy in km/hr =

1980000 198 99 = = km / hr 100 × 100 10 5

24. If the side of a rhombus is 5 cm and its area is 24cm2 then the sum of its diagonals is

a. 10 cm b. 12 cm c. 14 cm d. 16 cm Sol:

Correct option is (c)

We know that the diagonals of a rhombus bisect each other at right angles. We have to find AC + BD. Let AC = x cm and BD = y cm, Then OA =

y x and OB = 2 2

Area of a rhombus = 1 or 24 = × x × y 2

or 48 = xy

1 × product of diagonals 2

or xy = 48

… (1)

Now in right OAB, OA2 + OB 2 = AB 2

(Pythagorean Theorem)

2

2

⎛ x⎞ ⎛ y⎞ or ⎜ ⎟ + ⎜ ⎟ = 52 ⎝2⎠ ⎝ 2⎠

or

x2 y 2 + = 25 4 4

or

x2 + y 2 = 25 4

or x 2 + y 2 = 100

… (2)

We know that ( x + y ) 2 = x 2 + y 2 + 2 xy

Therefore ( x + y ) 2 = 100 + 2 × 48

(Using 1 and 2)

or ( x + y ) 2 = 196 or x + y = 196 or x + y = 14

… (3)

Again ( x − y ) 2 = x 2 + y 2 − 2 xy

Therefore ( x − y ) 2 = 100 − 2 × 48

(Using 1 and 2)

or ( x − y )2 = 4 or x − y = 4 or x − y = 2 Adding (3) and (4), we have

… (4)

x + y + x − y = 14 + 2 2 x = 16 x=

16 2

x= 8 cm

… (5)

Substituting the value of x in (3), we have

8 + y = 14 y = 14 − 8

y=6

… (6)

From (5) and (6), we find that AC = 8 cm and BD = 6 cm So AC + BD = 8 = 6 = 14 cm

2860 2 cm . If the 7 distance between their centers is 14 cm, then the difference of the two radii is

25. Two circles touch externally. The sum of their areas is

a. 8 cm b. 11 cm c. 7 cm d. 2 cm Sol:

Correct option is (a)

We know that when two circles touch externally, the distance between their centers is equal to the sum of their radii. Let the radii of the two circles be R and r respectively. Let C1 and C2 be the centers of the two circles

Now, C1C2 = R + r or 14 = R + r

… (1)

(Since Distance between their centers = 14 cm given)

It is given that the sum of the areas of the two circles =

π R2 + π r 2 =

2860 7

π (R2 + r 2 ) =

2860 7

22 2 2 2860 (R + r ) = 7 7 R2 + r 2 =

2860 7 × 7 22

R2 + r 2 =

2860 22

R 2 + r 2 = 130

… (2)

But ( R + r ) 2 = R 2 + r 2 + 2 Rr So (14) 2 = 130 + 2Rr

(Using 1 and 2)

2860 2 cm 7

196 = 130 + 2 Rr 196 − 130 = 2 Rr 66 = 2 Rr 66 = Rr 2 33 = Rr Rr = 33

… (3)

Now

( R − r ) 2 = R 2 + r 2 − 2 Rr So ( R − r ) 2 = 130 − 2 × 33

(Using 1 and 3)

( R − r ) 2 = 130 − 66 ( R − r ) 2 = 64 R − r = 64 R−r =8

… (4)

Adding (1) and (4), we have

R + r + R − r = 14 + 8 2 R = 22 R=

22 = 11cm 2

Substituting the value of R in (1), we have

8 + r = 11 r = 11 − 8 r = 3cm

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