# Class 10 RD Sharma Solutions – Chapter 15 Areas Related to Circles – Exercise 15.4 | Set 2

### Question 18. In the given figure, Find the area of the shaded region. (Use π = 3.14).

**Solution:**

The side of the square = 14 cm

So, area = side

^{2}14

^{2}= 196 cm^{2}Let’s assume the radius of each semi-circle be r cm.

Then,

r + 2r + r = 14 – 3 – 3

4r = 8

r = 2

So, the radius of each semi-circle is 2 cm.

Area of 4 semi-circles = (4 x 1/2 x 3.14 x 2 x 2) = 25.12 cm

^{2}Now,

the length of side of the smaller square = 2r = 2 x 2 = 4 cm

So, the area of smaller square = 4×4 = 16 cm

^{2}

Now we find the area of unshaded region = Area of 4 semi-circles + Area of smaller square= (25.12 + 16) = 41.12 cm

^{2}

Now we find thearea of shaded region = Area of square ABCD – Area of unshaded region= (196 – 41.12) = 154.88 cm

^{2}

Hence, the area of the shaded region 154.88 cm^{2}

### Question 19. In the 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 that,

Radius of small quadrant, r = 2 cm

Radius of big quadrant, R = 3.5 cm

(i)Area of quadrant OACB = 1/4 πR^{2}= 1/4 (22/7)(3.5)

^{2}= 269.5/28 = 9.625 cm

^{2}

(ii)Area of shaded region = Area of big quadrant – Area of small quadrant= 1/4 π(R

^{2}– r^{2})= 1/4 (22/7)(3.52 – 22)

= 1/4 (22/7)(12.25 – 4)

= 1/4 (22/7)(8.25)

= 6.482 cm

^{2}

Hence, the area of the quadrant OACB is 9.625 cm^{2 }and shaded region is 6.482 cm^{2}

### Question 20. In the figure, a square OABC is inscribed in a quadrant OPBQ of a circle. If OA = 21 cm, find the area of the shaded region.

**Solution:**

Given,

Side of the square = 21 cm = OA

Area of the square = OA

^{2}= 21^{2}= 441 cm^{2}Diagonal of the square OB = √2 OA = 21√2 cm

From the figure its seen that,

The diagonal of the square is equal to the radius of the circle, r = 21√2 cm

So,

The area of the quadrant = 1/4 πr

^{2}= 1/4 (22/7)(21√2)^{2}= 693 cm^{2}

Now we find thearea of the shaded region = Area of the quadrant – Area of the square= 693 – 441

= 252 cm

^{2}

Hence, the area of the shaded region is 252 cm^{2}

### Question 21. In the figure, OABC is a square of side 7 cm. If OAPC is a quadrant of a cirice with centre O, then find the area of the shaded region. (Use π = 22/7)

**Solution:**

Given that,

Side of OABC square = 7 cm

So, OA = AB = BC = OC = 7 cm

Now, the area of square OABC = side

^{2}= 7^{2}= 49 cm^{2}It is given that OAPC is a quadrant of a circle with centre O.

So, the radius of the quadrant = OA = OC = 7 cm

Area of the OAPC quadrant = 90/360 x πr

^{2}= 1/4 x (22/7) x 7

^{2}= 77/2 = 38.5 cm

^{2}

Now we find the area of shaded portion = Area of square OABC – Area of quadrant OAPC= (49 – 38.5) = 10.5 cm

^{2}

Hence, the area of shaded portion is 10.5 cm^{2}

### Question 22. In the figure, OE = 20 cm. In sector OSFT, square OEFG is inscribed. Find the area of the shaded region.

**Solution:**

Given that,

Side of OEFG square = 20 cm.

So its diagonal = √2 side = 20√2 cm

And, the radius of the quadrant(r) = diagonal of the square

r = 20√2 cm

Now we find the area of the shaded portion = Area of quadrant – Area of square= 1/4 πr

^{2}– side^{2}= 1/4 (22/7)(20√2)

^{2}– (20)^{2}= 1/4 (22/7)(800) – 400

= 400 x 4/7 = 1600/7 = 228.5 cm

^{2}

Hence, the area of the shaded region 228.5 cm^{2 }

### Question 23. Find the area of the shaded region in the figure, if AC = 24 cm, BC = 10 cm and O is the centre of the circle. (Use π = 3.14)

**Solution:**

Given,

AC = 24 cm and BC = 10 cm

From the figure AB is the diameter of the circle

So, ∠ACB = 90

^{o}By using Pythagoras theorem

AB

^{2}= AC^{2}+ BC^{2}= 24^{2}+ 10^{2}= 576 + 100 = 676So, AB = √676 = 26 cm

So, the radius of the circle(r) = 26/2 = 13 cm

Now we find thearea of shaded region = Area of semi-circle – Area of triangle ACB= 1/2 πr

^{2 }– 1/2 x b x h= 1/2 (22/7)13

^{2}– 1/2 x 10 x 24= 265.33 – 120

= 145.33 cm

^{2}

Hence, the area of shaded region is 145.33 cm^{2}

### Question 24. A circle is inscribed in an equilateral triangle ABC is side 12 cm, touching its sides (see figure). Find the radius of the inscribed circle and the area of the shaded part.

**Solution:**

Given that,

Side of ABC triangle = 12 cm

So, the area of the equilateral triangle = √3/4(side)

^{2}= √3/4(12)

^{2}= 36√3 cm^{2}Also the perimeter of triangle ABC = 3 x 12 = 36 cm

Now we find theradius of incircle = Area of triangle/ ½ (perimeter of triangle)= 36√3/ 1/2 x 36

= 2√3 cm

Also, we find the area of the shaded part = Area of equilateral triangle – Area of circle= 36√3 – πr

^{2}= 36(1.732) – (3.14)(2√3)

^{2}= 62.352 – 37.68

= 24.672 cm

^{2}

Hence, the area of the shaded part is 24.672 cm^{2}

### Question 25. In the figure, an equilateral triangle ABC of side 6 cm has been inscribed in a circle. Find the area of the shaded region. (Take π = 3.14)

**Solution:**

Given that,

The side of ABC triangle = 6 cm

So, the area of the equilateral triangle = √3/4(side)

^{2}= √3/4(6)

^{2}= √3/4(36)

= 9√3 cm

^{2}Let us mark the centre of the circle as O, OA and OB are the radii of the circle.

In triangle BOD,

sin 60

^{o}= BD/ OB√3/2 = 3/ OB

OB = 2√3 cm = r

Now we find thearea of shaded region = Area of the circle – area of the equilateral triangle= πr

^{2}– 9√3= 3.14 x (2√3)

^{2 }– 9√3= 3.14 x 12 – 9 x 1.732

= 37.68 – 15.588

= 22.092 cm

^{2}

Hence, the area of the shaded region 22.092 cm^{2}

### Question 26. A circular field has a perimeter of 650 m. A square plot having its vertices on the circumference of the field is marked in the field. Calculate the area of the square plot.

**Solution:**

Given that, the circular field has a perimeter = 650 m

As we know that the circumference of circle = 2πr

650 = 2×22/7xr

r = (650×7)/44

r = 103.409 m

As the diagonal of the square plot is the diameter of the circle.

Hence, r × 2 = d

So, diameter = 103.409 x 2

As we know that the Diameter of circle = 206.818 m = Diagonal of square plot

Now we find the area of the square plot = 1/2 × d²= 1/2 × (206.818)²

= 1/2 × 42773.68

Hence, the area of the square plot is 21386.84 m^{2}

### Question 27. Find the area of a shaded region in the figure, where a circular arc of radius 7 cm has been drawn with vertex A of an equilateral triangle ABC of side 14 cm as centre. (use π = 22/7 and √3 = 1.73)

**Solution:**

Given that,

The radius of circle (r) = 7 cm.

Side of the triangle = 14 cm

Now we find the area of shaded region = Area of the circle + Area of the triangle – 2 area of the sector EAF

= πr

^{2 }+ √3a^{2}/4 – 2 × πr^{2 }× 60/360= 22/7 × (7)

^{2}+ (1.73)(14)^{2}/4 – 2 × 22/7 × (7)^{2 }× 60/360= 283.77 – 51.33

= 187.43cm

^{2}

Hence, the area of the shaded region is 187.43cm^{2}

### Question 28. A regular hexagon is inscribed in a circle. If the area of hexagon is 24√3 cm^{2}, find the area of the circle. (Use π = 3.14)

**Solution:**

Given that,

A regular hexagon ABCDEF is inscribed in a circle

Area of hexagon = 24 √3 cm

^{2}Let us considered r be the radius of circle

So, the side of regular hexagon = r

Area of equilateral ΔOAB = √3/3 r

^{2}cm^{2}But the area of triangle OAB = 1/6 x area of hexagon

24√3/6 = 4√3 r

^{2}So, r

^{2 }= 4√3 x 4/√3 = 16r = 4 cm

Now we find the area of circle = πr

^{2}= 3.14 x (4)2 cm^{2}= 3.14 x 16 cm

^{2}= 50.24 cm

^{2}

Hence, the area of circle = 50.24 cm^{2}

### Question 29. ABCDEF is a regular hexagon with centre O (see figure). If the area of triangle OAB is 9 cm^{2}, find the area of :

### (i) the hexagon and

### (ii) the circle in which the hexagon is inscribed.

**Solution:**

The area of hexagon = 54 cm

^{2}and area of circle = 65.324 cm

^{2}Given that,

Area of ∆OAB = 9 cm

^{2}Let us considered ‘r’ be the radius of a circle and ‘a’ be the side of a equilateral triangle.

(i)As we know that the area of hexagon = 6 × Area of equilateral triangle= 6 × 9 = 54 cm²

= 54 cm²

(ii)Area of equilateral ∆OAB = √3/4 × a^{2}9 = √3/4 × a

^{2}9 × 4 = √3a

^{2}36 = √3a

^{2}a

^{2}= 36/√3Side

^{2}= 36/√3 cmAs we know that in regular hexagon inscribed in a circle,

its side is equal to the radius of a Circle

So, the radius of circle(r) = Side of a hexagon

r

^{2}= 36/√3 cmNow we find the area of circle = πr

^{2}= 22/7 × 36/√3

= 22/7 × 36/1.732

= (22 × 36) /(7 ×1.732)

= 792/12.124

= 65.324 cm

^{2}

Hence, the area of hexagon is 54 cm^{2}and area of circle is 65.324 cm^{2}

### Question 30. Four equal circles, each of radius 5 cm, touch each other as shown in the figure. Find the area included between them. (Take π = 3.14).

**Solution:**

Given that,

Radius of a circle = 5 cm

Side of a square = 2 × Radius of a circle

= 2 × 5

Side of a square = 10 cm

Area of a square = Side

^{2}Area of a square = 10

^{2}= 100 cm^{2}Area of a square = 100 cm

^{2}Area of the quadrant of one circle = 1/4πr

^{2}Area of the quadrant of four circles = 4 × 1/4πr

^{2}= πr^{2}= 3.14 × 5 × 5

= 3.14 × 25

= 78.5cm²

Area of the quadrant of four circles = 78.5cm

^{2}Now we find the area of the shaded portion = Area of the square – Area of the quadrant of four circles

= 100 – 78.5

= 21.5 cm

^{2}

Hence, the area of the shaded portion is 21.5 cm^{2}.

### Question 31. Four equal circles, each of radius ‘a’ touch each other. Show that the area between them is 6a^{2}/7. (Take π = 22/7)

**Solution:**

Given that,

The radius of a circle = a

Side of a square = 2 × Radius of a circle

= 2 × a

Side of a square = 2a cm

Area of a square = Side

^{2}= (2a)

^{2}= 4a

^{2}Area of the quadrant of one circle = 1/4πr

^{2}Area of the quadrant of four circles = 4 × 1/4πr

^{2}= πr^{2}Now we find the area of the shaded portion = Area of the square – Area of the quadrant of four circles

= 4a

^{2}– 22/7 × a^{2}= 4a

^{2}– 22a^{2}/7= (28a

^{2}-22a^{2})/7= 6a

^{2}/7

Hence, proved the area between them is 6a^{2}/7

### Question 32. A child makes a poster on a chart paper drawing a square ABCD of side 14 cm. She draws four circles with centre A, B, C, and D in which she suggests different ways to save energy. The circles are drawn in such a way that each circle touches externally two of the three remaining circles in the given figure. In the shaded region she write a message ‘Save Energy’. Find the perimeter and area of the shaded region. (Use π = 22/7)

**Solution:**

Given that,

Side of a square = 14 cm

Radius of a Circle(r) = Side of a square/2 = 14/2 = 7 cm

Central angle, θ = 90°

So, the perimeter of the shaded portion = 4 × length of the arc having Central angle 90°

= 4 × θ/360° × 2πr

= 4 × 90°/360° × 2 × 22/7 × 7

= 4 × 1/4 × 44

= 44 cm

As we know that area of a square = Side

^{2}= (14)

^{2}= 196 cm^{2}Also, the area of a square = 196 cm

^{2}Area of the quadrant of one circle = 1/4πr

^{2}Area of the quadrant of four circles = 4 × 1/4πr

^{2}= πr^{2}= 22/7 × 7

^{2}= 22 × 7

= 154 cm

^{2}Now we find the area of the shaded portion = Area of the square ABCD – Area of the quadrant of four circles

= 196 – 154

= 42 cm

^{2}

Hence, the Perimeter of the shaded portion is 44 cm and area of the shaded portion is 42 cm^{2}

### Question 33. The diameter of a coin is 1 cm (see figure). If four such coins be placed on a table so that the rim of each touches that of the other two, find the area of the shaded region (Take π = 3.1416).

**Solution:**

Given that,

The diameter of a coin (circle) = 1 cm

Find: the area of the shaded region

As we know that the radius(r) of a coin (circle) = Diameter of a coin (circle)/2

r = 1/2 = 0.5 cm

Side of a square = 2 × Radius of a coin (circle)

= 2 × 0.5

= 1 cm

Area of the quadrant of one circle = 1/4πr

^{2}Area of the quadrant of four circles = 4 × 1/4πr

^{2}= πr^{2}Now we find the area of the shaded portion = Area of the square – Area of the quadrant of four circles

A = Side

^{2}– πr^{2}= 1

^{2}– 3.1416 × (0.5)^{2}= 1 – 3.1416 × 0.25

= 1 – 0.7854

= 0.2146 cm

^{2}

Hence, the area of the shaded portion is 0.2146 cm^{2}

### Question 34. Two circular pieces of equal radii and maximum area, touching each other are cut out from a rectangular card board of dimensions 14 cm x 7 cm. Find the area of the remaining card board. (Use π = 22/7)

**Solution:**

Given that,

Length of a rectangle(l) = 14 cm

Breadth of a rectangle(b) = 7 cm

So, the diameter of 1 circle = 7 cm

Radius of 1 circle(r) = 7/2 cm

Now we find the area of the remaining cardboard = Area of rectangular cardboard – 2 × Area of circle

= l × b – 2 (πr

^{2})= (14× 7) – 2 × (22/7) × (7/2)

^{2}= 98 – (44/7)(49/4)

= 98 – 77

= 21 cm

^{2}

Hence, the area of the remaining cardboard is 21 cm^{2}.