NCERT Solutions involve detailed solutions of all questions listed under Exercise 12.3. Get free NCERT Solutions for Maths Chapter 12, Exercise 12.3 is solved by BYJUS experts using various tricks and with a proper understanding of concepts and logic. Class 10 Maths Chapter 12 Area Related to Circles Exercise 12.3 Questions and answers helps students to clear their concepts. NCERT Solutions for Class 10 Maths Chapter 12 Areas Related to Circles crafted to provide an accurate and logical solution for each question listed under Exercise 12.3. The sense of achievement on scoring well in first term exams will outshine a studentâ€™s performance in the schoolâ€™s internal exams. NCERT Solutions also prove to be of valuable help to students in their assignments and preparation of boards and competitive exams.

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Exercise 12.1 Solutions : 5 Solved Questions

Exercise 12.2 Solutions : 14 Solved Questions

### Access Answers to NCERT Class 10 Maths Chapter 12 Areas Related to Circles Exercise 12.3

**1. Find the area of the shaded region in Fig. 12.19, if PQ = 24 cm, PR = 7 cm and O is the centre of the circle.**

**Solution:**

Here, P is in the semi-circle and so,

P = 90Â°

So, it can be concluded that QR is hypotenuse of the circle and is equal to the diameter of the circle.

âˆ´ QR = D

Using Pythagorean theorem,

QR^{2 }= PR^{2}+PQ^{2}

Or, QR^{2 }= 7^{2}+24^{2}

QR= 25 cm = Diameter

Hence, the radius of the circle = 25/2 cm

Now, the area of the semicircle = (Ï€R^{2})/2

= (22/7)Ã—(25/2)Ã—(25/2)/2 cm^{2}

= 13750/56 cm^{2 }= 245.54 cm^{2}

Also, area of the Î”PQR = Â½Ã—PRÃ—PQ

=(Â½)Ã—7Ã—24 cm^{2}

= 84 cm^{2}

Hence, the area of the shaded region = 245.54 cm^{2}-84 cm^{2 }

= 161.54 cm^{2}

**2. Find the area of the shaded region in Fig. 12.20, if radii of the two concentric circles with centre O are 7 cm and 14 cm, respectively and AOC = 40Â°.**

**Solution:**

Given,

Angle made by sector = 40Â°,

Radius the inner circle = r = 7 cm, and

Radius of the outer circle = R = 14 cm

We know,

Area of the sector = (Î¸/360Â°)Ã—Ï€r^{2}

So, Area of OAC = (40Â°/360Â°)Ã—Ï€r^{2 }cm^{2}

= 68.44 cm^{2}

Area of the sector OBD = (40Â°/360Â°)Ã—Ï€r^{2 }cm^{2}

= (1/9)Ã—(22/7)Ã—7^{2 }= 17.11 cm^{2}

Now, area of the shaded region ABDC = Area of OAC â€“ Area of the OBD

= 68.44 cm^{2} â€“ 17.11 cm^{2 }= 51.33 cm^{2}

**3. Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles.**

**Solution:**

Side of the square ABCD (as given) = 14 cm

So, Area of ABCD = a^{2}

= 14Ã—14 cm^{2} = 196 cm^{2}

We know that the side of the square = diameter of the circle = 14 cm

So, side of the square = diameter of the semicircle = 14 cm

âˆ´ Radius of the semicircle = 7 cm

Now, area of the semicircle = (Ï€R^{2})/2

= (22/7Ã—7Ã—7)/2 cm^{2Â }

= 77 cm^{2 }

âˆ´ Area of two semicircles = 2Ã—77 cm^{2 }= 154 cm^{2}

Hence, area of the shaded region = Area of the Square â€“ Area of two semicircles

= 196 cm^{2 }-154 cm^{2}

= 42 cm^{2}

**4. Find the area of the shaded region in Fig. 12.22, 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:**

It is given that OAB is an equilateral triangle having each angle as 60Â°

Area of the sector is common in both.

Radius of the circle = 6 cm.

Side of the triangle = 12 cm.

Area of the equilateral triangle = (âˆš3/4) (OA)^{2}= (âˆš3/4)Ã—12^{2 }= 36âˆš3 cm^{2}

Area of the circle = Ï€R^{2} = (22/7)Ã—6^{2 }= 792/7 cm^{2}

Area of the sector making angle 60Â° = (60Â°/360Â°) Ã—Ï€r^{2 }cm^{2}

= (1/6)Ã—(22/7)Ã— 6^{2 }cm^{2 }= 132/7 cm^{2}

Area of the shaded region = Area of the equilateral triangle + Area of the circle â€“ Area of the sector

= 36âˆš3 cm^{2} +792/7 cm^{2}-132/7 cm^{2}

= (36âˆš3+660/7) cm^{2}

**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 Fig. 12.23. Find the area of the remaining portion of the square.**

**Solution:**

Side of the square = 4 cm

Radius of the circle = 1 cm

Four quadrant of a circle are cut from corner and one circle of radius are cut from middle.

Area of square = (side)^{2}= 4^{2 }= 16 cm^{2}

Area of the quadrant = (Ï€R^{2})/4 cm^{2} = (22/7)Ã—(1^{2})/4 = 11/14 cm^{2}

âˆ´ Total area of the 4 quadrants = 4 Ã—(11/14) cm^{2} = 22/7 cm^{2}

Area of the circle = Ï€R^{2 }cm^{2} = (22/7Ã—1^{2}) = 22/7 cm^{2}

Area of the shaded region = Area of square â€“ (Area of the 4 quadrants + Area of the circle)

= 16 cm^{2}-(22/7) cm^{2}-(22/7) cm^{2}

= 68/7 cm^{2}

**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 Fig. 12.24. Find the area of the design.**

**Solution:**

Radius of the circle = 32 cm

Draw a median AD of the triangle passing through the centre of the circle.

â‡’ BD = AB/2

Since, AD is the median of the triangle

âˆ´ AO = Radius of the circle = (2/3) AD

â‡’ (2/3)AD = 32 cm

â‡’ AD = 48 cm

In Î”ADB,

By Pythagoras theorem,

AB^{2 }= AD^{2 }+BD^{2}

â‡’ AB^{2 }= 48^{2}+(AB/2)^{2}

â‡’ AB^{2 }= 2304+AB^{2}/4

â‡’ 3/4 (AB^{2})= 2304

â‡’ AB^{2 }= 3072

â‡’ AB= 32âˆš3 cm

Area of Î”ADB = âˆš3/4 Ã—(32âˆš3)^{2 }cm^{2 }= 768âˆš3 cm^{2}

Area of circle = Ï€R^{2} = (22/7)Ã—32Ã—32 = 22528/7 cm^{2}

Area of the design = Area of circle â€“ Area of Î”ADB

= (22528/7 â€“ 768âˆš3) cm^{2}

**7. In Fig. 12.25, ABCD is a square of side 14 cm. With centres 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:**

Side of square = 14 cm

Four quadrants are included in the four sides of the square.

âˆ´ Radius of the circles = 14/2 cm = 7 cm

Area of the square ABCD = 14^{2 }= 196 cm^{2}

Area of the quadrant = (Ï€R^{2})/4 cm^{2} = (22/7) Ã—7^{2}/4 cm^{2}

= 77/2 cm^{2}

Total area of the quadrant = 4Ã—77/2 cm^{2 }= 154cm^{2 }

Area of the shaded region = Area of the square ABCD â€“ Area of the quadrant

= 196 cm^{2 }â€“ 154 cm^{2}

= 42 cm^{2 }

**8. Fig. 12.26 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 are 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:**

Width of the track = 10 m

Distance between two parallel lines = 60 m

Length of parallel tracks = 106 m

DE = CF = 60 m

Radius of inner semicircle, r = OD = Oâ€™C

= 60/2 m = 30 m

Radius of outer semicircle, R = OA = Oâ€™B

= 30+10 m = 40 m

Also, AB = CD = EF = GH = 106 m

Distance around the track along its inner edge = CD+EF+2Ã—(Circumference of inner semicircle)

= 106+106+(2Ã—Ï€r) m = 212+(2Ã—22/7Ã—30) m

= 212+1320/7 m = 2804/7 m

Area of the track = Area of ABCD + Area EFGH + 2 Ã— (area of outer semicircle) â€“ 2 Ã— (area of inner semicircle)

= (ABÃ—CD)+(EFÃ—GH)+2Ã—(Ï€r^{2}/2) -2Ã—(Ï€R^{2}/2) m^{2}

= (106Ã—10)+(106Ã—10)+2Ã—Ï€/2(r^{2}-R^{2}) m^{2}

= 2120+22/7Ã—70Ã—10 m^{2}

= 4320 m^{2 }

**9. In Fig. 12.27, AB and CD are two diameters of a circle (with centre 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:**

Radius of larger circle, R = 7 cm

Radius of smaller circle, r = 7/2 cm

Height of Î”BCA = OC = 7 cm

Base of Î”BCA = AB = 14 cm

Area of Î”BCA = 1/2 Ã— AB Ã— OC = (Â½)Ã—7Ã—14 = 49 cm^{2 }

Area of larger circle = Ï€R^{2 }= (22/7)Ã—7^{2} = 154 cm^{2 }

Area of larger semicircle = 154/2 cm^{2 }= 77 cm^{2 }

Area of smaller circle = Ï€r^{2} = (22/7)Ã—(7/2)Ã—(7/2) = 77/2 cm^{2}

Area of the shaded region = Area of larger circle â€“ Area of triangle â€“ Area of larger semicircle + Area of smaller circle

Area of the shaded region = (154-49-77+77/2) cm^{2}

= 133/2 cm^{2} = 66.5 cm^{2}

**10. The area of an equilateral triangle ABC is 17320.5 cm ^{2}. 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 Fig. 12.28). Find the area of the shaded region. (Use Ï€ = 3.14 and âˆš3 = 1.73205)**

**Solution:**

ABC is an equilateral triangle.

âˆ´ âˆ A = âˆ B = âˆ C = 60Â°

There are three sectors each making 60Â°.

Area of Î”ABC = 17320.5 cm^{2}

â‡’ âˆš3/4 Ã—(side)^{2} = 17320.5

â‡’ (side)^{2} =17320.5Ã—4/1.73205

â‡’ (side)^{2} = 4Ã—10^{4 }

â‡’ side = 200 cm

Radius of the circles = 200/2 cm = 100 cm

Area of the sector = (60Â°/360Â°)Ã—Ï€ r^{2 }cm^{2}

= 1/6Ã—3.14Ã—(100)^{2 }cm^{2}

= 15700/3cm^{2}

Area of 3 sectors = 3Ã—15700/3 = 15700 cm^{2 }

Thus, area of the shaded region = Area of equilateral triangle ABC â€“ Area of 3 sectors

= 17320.5-15700 cm^{2 }= 1620.5 cm^{2}

**11. On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig. 12.29). Find the area of the remaining portion of the handkerchief.**

**Solution:**

Number of circular designs = 9

Radius of the circular design = 7 cm

There are three circles in one side of square handkerchief.

âˆ´ Side of the square = 3Ã—diameter of circle = 3Ã—14 = 42 cm

Area of the square = 42Ã—42 cm^{2} = 1764 cm^{2}

Area of the circle = Ï€ r^{2 }= (22/7)Ã—7Ã—7 = 154 cm^{2}

Total area of the design = 9Ã—154 = 1386 cm^{2}

Area of the remaining portion of the handkerchief = Area of the square â€“ Total area of the design = 1764 â€“ 1386 = 378 cm^{2}

**12. In Fig. 12.30, 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:**

Radius of the quadrant = 3.5 cm = 7/2 cm

**(i)** Area of quadrant OACB = (Ï€R^{2})/4 cm^{2}

= (22/7)Ã—(7/2)Ã—(7/2)/4 cm^{2}

= 77/8 cm^{2}

**(ii)** Area of triangle BOD = (Â½)Ã—(7/2)Ã—2 cm^{2}

= 7/2 cm^{2}

Area of shaded region = Area of quadrant â€“ Area of triangle BOD

= (77/8)-(7/2) cm^{2 }= 49/8 cm^{2 }

= 6.125 cm^{2}

**13. In Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. If OA = 20 cm, find the area of the shaded region. (Use Ï€ = 3.14)**

**Solution:**

Side of square = OA = AB = 20 cm

Radius of the quadrant = OB

OAB is right angled triangle

By Pythagoras theorem in Î”OAB,

OB^{2 }= AB^{2}+OA^{2}

â‡’ OB^{2 }= 20^{2 }+20^{2}

â‡’ OB^{2 }= 400+400

â‡’ OB^{2 }= 800

â‡’ OB= 20âˆš2 cm

Area of the quadrant = (Ï€R^{2})/4 cm^{2 }= (3.14/4)Ã—(20âˆš2)^{2 }cm^{2 }= 628cm^{2}

Area of the square = 20Ã—20 = 400 cm^{2}

Area of the shaded region = Area of the quadrant â€“ Area of the square

= 628-400 cm^{2 }= 228cm^{2}

**14. AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig. 12.32). If âˆ AOB = 30Â°, find the area of the shaded region.**

**Solution:**

Radius of the larger circle, R = 21 cm

Radius of the smaller circle, r = 7 cm

Angle made by sectors of both concentric circles = 30Â°

Area of the larger sector = (30Â°/360Â°)Ã—Ï€R^{2 }cm^{2}

= (1/12)Ã—(22/7)Ã—21^{2 }cm^{2}

= 231/2cm^{2}

Area of the smaller circle = (30Â°/360Â°)Ã—Ï€r^{2 }cm^{2}

= 1/12Ã—22/7Ã—7^{2 }cm^{2}

=77/6 cm^{2}

Area of the shaded region = (231/2) â€“ (77/6) cm^{2}

= 616/6 cm^{2} = 308/3cm^{2}

**15. In Fig. 12.33, 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:**

Radius of the quadrant ABC of circle = 14 cm

AB = AC = 14 cm

BC is diameter of semicircle.

ABC is right angled triangle.

By Pythagoras theorem in Î”ABC,

BC^{2 }= AB^{2 }+AC^{2}

â‡’ BC^{2 }= 14^{2 }+14^{2}

â‡’ BC = 14âˆš2 cm

Radius of semicircle = 14âˆš2/2 cm = 7âˆš2 cm

Area of Î”ABC =( Â½)Ã—14Ã—14 = 98 cm^{2}

Area of quadrant = (Â¼)Ã—(22/7)Ã—(14Ã—14) = 154 cm^{2}

Area of the semicircle = (Â½)Ã—(22/7)Ã—7âˆš2Ã—7âˆš2 = 154 cm^{2}

Area of the shaded region =Area of the semicircle + Area of Î”ABC â€“ Area of quadrant

= 154 +98-154 cm^{2 }= 98cm^{2}

**16. Calculate the area of the designed region in Fig. 12.34 common between the two quadrants of circles of radius 8 cm each.**

**Solution:**

AB = BC = CD = AD = 8 cm

Area of Î”ABC = Area of Î”ADC = (Â½)Ã—8Ã—8 = 32 cm^{2}

Area of quadrant AECB = Area of quadrant AFCD = (Â¼)Ã—22/7Ã—8^{2}

= 352/7 cm^{2}

Area of shaded region = (Area of quadrant AECB â€“ Area of Î”ABC) = (Area of quadrant AFCD â€“ Area of Î”ADC)

= (352/7 -32)+(352/7- 32) cm^{2}

= 2Ã—(352/7-32) cm^{2}

= 256/7 cm^{2}

This exercise explains Areas of Combinations of Plane Figures. In this section, students calculate the areas of some combinations of plane figures which we might see in our daily life like window designs, designs on bedsheets, carpet designs, etc. through some examples. To solve such NCERT exemplar problems, it is necessary to have good knowledge in finding areas of different plane figures.