Areas Related to Circles

NCERT solution Exercise 12.3(part-2)

Question (8) Figure depicts a racing track whose left and right ends are semicircular.

The distance between the two inner parallel line segments is 60 m and they area each 106 m long. If the track is 10 m wide, find

(i) the distance around the track along its inner edge

(ii) the area of the track

Solution

[Strategy to solve the question (a) find the area of rectangular track, i.e. area of track excluding both of the semi circle (b) Find the circumference of inner semicircle (c) Add the circumference of both of the semi circle which will become circumference of one circle (d) sum of circumference of inner circle and length of track will give total length of track along its inner edge. (e) find the area of shaded region of semi-circle (f) Add the area of track excluding semicircle and areas of shaded region of semicircles, which will given the area of the track. ]

Given, inner length of the track = 106 m

Width of the track = 10 m

This means that total given circular track is divided into four regions

(a) 106 m long and 10 m wide two rectangular track

(b) And two semi circles having inner diameter = 60 m and outer diameter = 60+10+10 = 80 m

This means radius of inner semicircle = 60/2 = 30 m and radius of outer semi circle = 80/2 = 40 m

Calculation of circumference of inner semicircular section of the track

We know that, circumference of inner semicircle = 1/2 × 2 π r

= π r

= π 30m

Now, since there are two semicircular ends

Thus, total length of both of the semi circular ends = π 30m × 2

Thus, total inner length of circular section of track = 188.57 m

And total inner length of rectangular section of path = 106 m + 106 m = 212 m

Thus, total inner length of track

= total inner length of rectangular section of track + total inner length of circular section of track

= 212 m + 188.57

= 400.57 m

Thus, total inner length of track = 400.57 m

Calculation of area of rectangular track

We know that area of rectangle = length × breadth

Thus, area of one rectangular track = 106 m × 10 m

= 1060 m²

Thus, area of two rectangular track = 2 × Area of one rectangular track

= 2 × 1060 m²

= 2120 m²

Thus, total area of rectangular track = 2120 m

Calculation Area of semicircular track

Radius of inner semi-circle = 30 m

Radius of outer semi-circle = 40 m [Because track is 10 m wide]

Now, since there are two semicircular sections at two ends of track,

Thus, there is total one circle we get after combining these two semicircles.

Now, we know that area of circle = π r²

Thus, area of inner circle = π (30 m)²

= π 900 m²

And area of outer circle = π (40 m)²

= π 1600 m²

Now, area of circular section of track = area of outer circle – area of inner circle

= π 1600 m² – π 900 m²

= pi; (1600 – 900) m²

= 2200 m²

Thus, area of both of the circular region of track = 2200 m²

Now, total area of circular track = area of rectangular regions of track + areas of circular regions of track

= 2120 m + 2200 m

= 4320 m²

Thus, area of the track = 4320 m²

Thus, (a) the distance around the track along its inner edge = 400.57 m Answer

And (b) the area of track = 4320 m² Answer

Question (9) In figure AB and CD are two diameters of a circle (with center O) perpendicular to each other and OD is the diameter of the smaller circle. If OA = 7 cm, find the area of the shaded region.

Solution

[Strategy to solve the question (a) Find the area of semi circle ACB (b) Find the area of triangle ABC (c) Subtract the area of triangle ABC from the area of semi circle to find the area of shaded region of semicircle ACB (c) Find the area of smaller circle with center P. (d) Add the area of smaller circle and area of shaded region of semicircle, which will given the area of total shaded region.]

Given, OA and CD are two diameters and perpendicular to each other.

OA = 7 cm

Thus, OA = OB = OD = OC = 7 cm

(Because these all are radii of same circle)

Thus, area of shaded region = ?

Calculation of area of semi-circle OACB

We know that, area of semi circle = 1/2 × π r²

Thus, area of given semi circle OACB

= 77 cm²

In triangle ABC

Base= AB = OA + OB

⇒ Base = 7 + 7 = 14 cm

Height = OC = 7 cm

Now, we know that, area of a triangle = 1/2 × height × base

= 1/2 × 7 cm × 14 cm

= 49 cm²

Now, area of smaller circle i.e. circle with center (P)

Here, since OD = 7 cm

Thus, radius of smaller circle = 7/2 = 3.5 cm

Now, we know that, area of a circle = π r²

= 38.5 cm²

Now, area of shaded region

= Area of semi circle – Area of triangle + area of smaller circle

= 77 cm² – 49 cm² + 38.5 cm²

= 66.5 cm²

Thus, area of shaded region = 66.5 cm² Answer

Question (10) The area of an equilateral triangle ABC is 17320.5 cm². With each vertex of the triangle as centre, a circle is drawn with radius equal to half the length of the side of the triangle (see figure). Find the area of the shaded region. (Use π = 3.14 and √3 = 1.73205)

Solution

[Strategy to solve the question (a) Find the side of given equilateral triangle using area given (b) Find the radius of circle by dividing side of triangle by 2 (c) Find the area of minor sector of circle (d) subtract the sum of area of sectors from the area of triangle which will give the area of shaded region. ]

Given, Area of equilateral triangle, ABC = 17320.5 cm²

Area of shaded region= ?

We know that area of an equilateral triangle

⇒ side = 100 × 2 cm = 200 cm

Thus, side of the given equilateral triangle = 200 cm

And, hence radius of the circle = 100 cm

Now, since triangle is an equilateral triangle, thus angle of the triangle = 60⁰

It is clear from the given figure, that there three sectors are formed with three circles.

Now, we know that area of sector of angle, θ

Thus, area of three (3) sectors of angle, 60⁰

Now, area of shaded region

= Area of triangle – Area of sum of three sectors

= 17320.5 cm² – 15700 cm²

= 1620.5 cm²

Thus, area of shaded region = 1620.5 cm² Answer

Question (11) On a square handkerchief, nine circular designs each of radius 7 cm are made (see figure). Find the area of the remaining portion of the handkerchief.

Solution

[Strategy to solve the question (a) Find the side of square handkerchief using radius of circle. (b) Find the area of handkerchief (c) Find the area of circle (d) multiply the area of circle to find the area of nine circles (e) Find the area of remaining part of handkerchief by subtracting sum of areas of all nine circles from the area of square. ]

Given, radius of circle = 7 cm

Total number of circles = 9

Number of circles in each row = 3

Then area of remaining part of handkerchief after the design, i.e. after the circle = ?

We know that, area of circle = Π r²

Therefore, area of given circle

= 154 cm²

Thus total areas of 9 circles = 9 × Area of 1 circle

= 9 × 154 cm²

= 1386 cm²

Calculation of area of square handkercheif

Radius of circle = 7 cm

Thus, diameter of circle = 7 × 2 = 14 cm

Since in one row the number of circle = 3

Thus, side of square = 3 circles = 3 × 14 cm

= 42 cm

Thus, side of square = 42 cm

Now, we know that, area of square = (side)²

Thus, area of given square = (42 cm)²

= 1764 cm²

Thus, area of given square handkerchief = 1764 cm²

Now, Area of remaining portion of handkerchief

= Area of square – Sum of areas of all 9 square

= 1764 cm² – 1386 cm²

= 378 cm²

Thus, area of remaining portion of square handkerchief = 378 cm² Answer

Question (12) In figure OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the

(i) quadrant OACB

(ii) shaded region

Solution

Given, Radius of the circle of which quadrant is given = 3.5 cm

OD = 2 cm

(i) Area of quadrant OACB

We know that, area of a quadrant of a circle

Thus, area of given quadrant

= 9.625 cm²

Thus, area of quadrant = 9.625 cm²

(ii) Area of shaded region

In triangle OBD,

Base (OB) = 3.5 cm and height (OD) = 2 cm

Now, we know that area of triangle = 1/2 × base × height

Thus area of triangle OBC = 3.5 cm²

Now area of shaded region = Area of quadrant – Area of triangle

= 9.625 cm² – 3.5 cm²

= 6.125 cm²

Thus, (i) area of quadrant OACB = 9.625 cm² and (ii) Area of shaded region = 6.125 cm² Answer

Question (13) In figure, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use π = 3.14)

Solution

[Strategy to solve the question (a) Find the radius of quadrant using side of square (b) Find the area of quadrant (c) Find the area of square (b) Find the area of shaded region after subtracting area of square from area of quadrant]

Given, OABC is a square and inscribed in quadrant OPBQ

OA = 20 cm

Area of shaded region = ?

construction

Join the point O and B of square OABC

Since, OABC is a square thus, ∠ A = 90⁰

And, OA = AB = 20 cm

Now, in triangle OAB

According to Pythagoras Theorem

OB² = OA² + AB²

= (20 cm)² + (20 cm)²

= 400 cm² + 400 cm²

= 800 cm²

= 400 × 2 cm²

Now, OB = 20√2 cm = Radius of circle

Now we know that, area of quadrant of a circle

= 628 cm²

Now, we know that area of an square = (side)²

Thus, area of given square OABD = (20 cm)²

= 400 cm²

Now, Area of shaded region = Area of quadrant – Area of square

= 628 cm² – 400 cm²

= 228 cm²

Thus, area of shaded region = 228 cm² Answer

Question (14) AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see figure). If ∠ AOB = 30⁰, find the area of the shaded region.

Solution

Given, radius of bigger circle = 21 cm

Radius of smaller circle = 7 cm

∠ AOB = 30⁰

This means Area of sector = 30⁰

Thus, Area of shaded region = ?

We know that area of sector of angle θ

Thus, area of sector of smaller circle of angle 30⁰

= 12.833 cm²

Thus, area of sector of bigger circle of angle 30⁰

= 115.5 cm²

Now, area of shaded region = Area of sector of bigger circle – Area of sector of smaller circle

= 115.5 cm² – 12.833 cm²

= 102.667 cm²

Thus, area of shaded region = 102.667 cm² Answer

Question (15) In figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region.

Solution

[Strategy to solve the question (a) Find the BC using Pythagoras theorem in triangle ABC (b) Find the radius of semi circle (c) Find the area of semi-circle (d) Find the area of triangle quadrant (e) Difference between area of semicircle and area of area of will give the area of shaded region.]

Given, AC = AB = 14 cm

And ∠ A = 90⁰

Thus, Area of shaded region = ?

In triangle, ABC

According to Pythagoras theorem,

BC² = AB² + AC²

= (14 cm)² + (14 cm)²

= 196 cm² + 196 cm²

= 392 cm²

We know that, Area of quadrant

Thus, area of given quadrant, ABFC

= 154 cm²

We know that area of triangle = 1/2 × base × height

= 98 cm²

Now, area of segment BFC = Area of sector i.e. quadrant ABFC – Area of triangle ABC

= 154 cm² – 98 cm²

= 56 cm²

We know that area of semicircle = 1/2 × π × r²

= 154 cm²

Now, area of shaded region = Area of semicircle BEC – Area of segment BFC

= 154 cm² – 56 cm²

= 98 cm²

Thus, area of shaded region = 98 cm² Answer

Question (16) Calculate the area of the designed region in figure common between the two quadrants of circles of radius 8 cm each.

Solution

[Strategy to solve the question Since, given designed region is common between two quadrants, thus (a) Find the area of segment of one quadrant (b) Multiply the area of segment by two which will give the area of common region.]

Given, radius of quadrant of circle = 8 cm.

And ∠ C = ∠ D = 90⁰

We know that, Area of quadrant

Thus, area of given sector of quadrant

Now, area of triangle ABC

We know that area of triangle = 1/2 × base × height

= 1/2 × 8 cm × 8 cm

= 32 cm²

Now, area of segment = Area of sector of quadrant – Area of triangle

= 50.28 cm² – 32 cm²

= 18.28 cm²

Now, since there are two segments in designed region, thus area of designed region

= Area of one segment × 2

= 18.28 cm² × 2

= 36.56 cm²

Thus, area of designed region = 36.56 cm² Answer

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