Areas Related to Circles

NCERT solution Exercise 12.3

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Question (1) Find the area of the shaded region in figure, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.

Solution

[Strategy solve the question First find the circle of semi-circle. (b) Find the area of triangle PQR (c) Subtract the area of triangle from the semicircle, which will give the area of shaded region. ]

Given, PQ = 24 cm

PR = 7 cm

Then, area of the shaded region = ?

We know that Any angle inscribed in a semicircle is always a right angle.

Here, since angle RPQ is inscribed in a semi-circle, hence it is a right angle.

This means, ∠ RPQ = 90⁰

Now, in Δ RPQ

According to Pythagoras theorem, we know that

QR² = PQ² + RP²

= (24 cm)² + (7 cm)²

= 576 cm² + 49 cm²

= 625 cm²

Now, since QR = Diameter of the given circle = 25 cm

Thus, radius, r = 25/2 = 12.5 cm

Calculation of Area of semi-circle

Now, we know that Area of a circle = 1/2 ×π r²

Therefore, area of given circle = 1/2 × π (12.5 cm)²

Thus, area of semi circle = 245.53 cm²

Calculation of area of triangle

In triangle, PQR, base = PR = 7 cm

And, height PQ = 24 cm

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

Thus, area of Δ PQR = 1/2 × 24 cm × 7 cm

= 12 cm × 7 cm

= 84 cm²

Thus, area of Δ PQR = 84 cm²

Now, area of shaded portion as given in the figure,

Area of semi circle – Area of triangle PQR

= 245.53 cm² – 84 cm²

= 161.53 cm²

Thus, area of shaded portion as given in the figure = 161.53 cm² Answer

Question (2) Find the area of the shaded region in figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠ AOC = 40⁰.

Solution

[Strategy to solve the question (a) Find the area of sector, OAC of bigger circle (b) Find the area of sector OBD of smaller circle. (c) Subtract the area of sector of smaller circle from the area of sector of bigger circle, which will give the area of shaded portion as given in the figure.]

Given, OB = 7 cm

OA = 14 cm

And ∠AOC = ∠ BOD = 40⁰

Thus, area of shaded portion BACD = ?

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

Thus, area of sector, OBD of angle, 40⁰

= 17.11 cm²

Thus, area of sector OBD of smaller circle = 17.11 cm²

And area of sector, OAC of bigger sector, of anlge 40⁰

= 68.44 cm²

Thus, area of sector OAC of bigger circle = 68.44 cm²

Now, area of shaded portion = Area of sector OAC of bigger circle – Area of sector OBD of smaller circle

= 68.44 cm² – 17.11 cm²

= 51.33 cm²

Thus, area of shaded portion as given in the question = 51.33 cm² Answer

Alternate Method and Shortcut method

Given, Radius of smaller circle = OB = 7 cm

Radius of bigger circle = OA = 14 cm

Angle of sector= ∠ BOD = ∠ AOC = 40⁰

Since, both of the circle are concentric, hence

Thus, area of shaded portion = Area of sector OAC of bigger circle – Area of sector OBD of smaller circle

= 51.33 cm²

Thus, Area of shaded portion as given in the question = 51.33 cm² Answer

Question (3) Find the area of the shaded region in figure, if ABCD is a square of side 14 cm and APD and BPC are semicircle.

Solution

[Strategy to solve the question (a) Find the area of given Square (b) Find the sum of area of both of the semicircle (c) Subtract the sum of area of both of the semi circle from the area of square, which will give the area of shaded portion as given in the figure.]

Given, ABCD is a square

Side of square = 14 cm

APD and BPD are semicircle inscribed in the square,

Here diameter of both of the semi-circle are equal, because two of the side of the given square forms diameter of each of the circle.

Thus, radius of the semi-circles will also be equal.

This means both of the semi circles are equal.

Thus, Radius of both of the semi-circle APD and BPD

= side of square/2 = 14/2 = 7 cm

Area of shaded portion as given in the figure = ?

We know that, area of a square = side²

Thus, area of given square ABCD = (14 cm)²

⇒ Area of square ABCD = 196 cm²

Now we know that, area of a semi-circle = 1/2 π r²

Thus, area of given semi-circle

Thus, area of semi-circle = 77 cm²

Now, area of two same semi-circle = Area of one semi-circle × 2

Thus, area of two semi-circle in the given figure = 77 cm² × 2

⇒ Area of both of the semi circle as given in the question = 154 cm²

Thus, area of shaded portion = Area of square – total area of both of the semi-circle

= 196 cm² – 154 cm²

= 42 cm²

Thus, Area of given shaded portion = 42 cm² Answer

Question (4) Find the area of the shaded region in figure, where a semi circular arc of radius 6 cm has been drawn with vertex O of an equilateral triangle OAB of side 12 cm as centre.

Solution

[Strategy to solve the question (a) Find the area of given equilateral triangle. (b) Find the area of major sector (c) Add the area of major sector and area of triangle, which will give the area of shaded region.]

Given, Side of equilateral triangle AOB = 12 cm

Radius of circle = 6 cm

Thus, area of shaded region = ?

We know that, Area of equilateral triangle

Thus, area of given triangle OAB

= 62.352 cm²

Now, since angle of equilateral triangle = 60⁰

Thus, angle of major sector DEC in given figure = Angle of circle – 60⁰

= 360⁰ – 60⁰ = 300⁰

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

Thus, area of major sector (DEC), of angle 300⁰

= 94.285 cm²

Thus, area of shaded region = Area of major sector, DEC + Area of triangle ABC

= 94.285 cm² + 62.352²

= 156.367 cm²

Thus, area of given shaded region = 156.367 cm² Answer

Question (5) From each corner of a square of side 4 cm a quadrant of a circle of radius 1 cm is cut and also a circle of diameter 2 cm is cut as shown in figure. Find the area of the remaining portion of the square.

Solution

[Strategy to solve the question (a) Find the area of square (b) Find the sum of areas of circles (c) Subtract the sum of the areas of circle from the area of square, which will give the area of shaded region.]

Given, ABCD is a square.

Side of square = 4 cm

Radius of quadrant = 1 cm

Diameter of circle = 2 cm

Thus, radius of circle = 1 cm

Thus, area of shaded region = ?

Here, there are four quadrant of circle, this means 1/4 th of circle

And one full circle.

Thus, total number of circles

= 1/4 × 4 + 1 = 1 + 1 = 2

Thus, total number of circle = 2

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

Thus, areas of given circle

= 3.142 cm²

Thus, total areas of given 2 circles = Area of 1 circle × 2

= 3.142 × 2 = 6.284 cm²

Thus, total areas of all circles and quadrant cut from the given square = 6.284 cm²

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

Thus, area of given square = (4 cm)²

= 16 cm²

Thus, area of shaded region as given in the figure = area of square – total area of circles and quadrant

= 16 cm² – 6.284 cm²

= 9.716 cm²

Thus, area of shaded region = 9.716 cm² Answer

Question (6) In a circular table cover of radius 32 cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in figure. Find the area of the design.

Solution

[Strategy to solve the question (a) find the area of triangle after calculating the side of triangle (b) Find the area of circle (c) Subtract the area of triangle from the area of circle, which will give the area of design. ]

Let, O is the centre of the circle given in question.

Given, Radius of circle, OA= 32 cm

Construction

A perpendicular AE is drawn from vertex A from centre on side BC

Since Δ ABC is an equilateral triangle

Therefore, ∠ A = ∠ B = ∠ C = 60⁰

And, OE or AOE is the median

Thus, ∠ OEC = 90⁰

And ∠ OCE = 30⁰

Now, in triangle OEC

Now, BC = BE + EC

⇒ BC = EC + EC

[Because BE = EC]

Now, we know that area of an equilateral triangle

Therefore, Area of given Δ ABC =

= 1.732 ×32× 24 cm²

= 1330.176 cm²

We know that area of a circle = π r²

Therefore, Area of given circle

= 3218.285 cm²

Now, area of design

= Area of circle – Area of triangle

= 3218.285 – 1330.176 cm²

= 1888.109 cm² Answer

Question (7) In figure, ABCD is a square of side 14 cm. With centers A, B, C and D, four circles are drawn such that each circle touch externally two of the remaining three circles. Find the area of the shaded region.

Solution

[Strategy to solve the question (a) Find the area of square (b) Find the area of all four quadrant, i.e. area of one circle (c) Subtract the area of sum of areas of four quadrant, i.e. one circle from the area of square, which will give the area of shaded region.]

Given, ABCE is a square and Side of square = 14 cm

Since, each circle touch externally two of the remaining three circles,

Thus, radius of circle = 14/2 = 7 cm

Thus, Area of shaded region = ?

We know that area of square = side²

Therefore, area of given square = (14 cm)²

= 196 cm²

As per given figure, there is one quadrant of each circle falls in the region of square.

And there are total four quadrants of circles.

Thus, total number of circles = 1/4 circle × 4 = 1 circle

Now, we know that, are of circle = π r²

Thus, area of 4 quadrants, i.e. one circle

Thus, area of circles = 154 cm²

Now, Area of shaded region = Area of square – Total Areas of 4 quadrant, i.e. area of 1 cirlce

= 196 cm² – 154 cm²

= 42 cm²

Thus, area of shaded region = 42 cm²

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