**Exercise: 12.3 (Page No: 234)**

**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 = 90°

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

∴ 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 = (πR^{2})/2

= (22/7)×(25/2)×(25/2)/2 cm^{2}

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

Also, area of the ΔPQR = ½×PR×PQ

=(½)×7×24 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 = 40°.**

**Solution:**

Given,

Angle made by sector = 40°,

Radius the inner circle = r = 7 cm, and

Radius of the outer circle = R = 14 cm

We know,

Area of the sector = (θ/360°)×πr^{2}

So, Area of OAC = (40°/360°)×πr^{2 }cm^{2}

= 68.44 cm^{2}

Area of the sector OBD = (40°/360°)×πr^{2 }cm^{2}

= (1/9)×(22/7)×7^{2 }= 17.11 cm^{2}

Now, area of the shaded region ABDC = Area of OAC – Area of the OBD

= 68.44 cm^{2} – 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}

= 14×14 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

∴ Radius of the semicircle = 7 cm

Now, area of the semicircle = (πR^{2})/2

= (22/7×7×7)/2 cm^{2 }=

= 77 cm^{2}

∴ Area of two semicircles = 2×77 cm^{2 }= 154 cm^{2}

Hence, area of the shaded region = Area of the Square – 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 60°

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 = (√3/4) (OA)^{2}= (√3/40×12^{2 }= 36√3 cm^{2}

Area of the circle = πR^{2} = (22/7)×6^{2 }= 792/7 cm^{2}

Area of the sector making angle 60° = (60°/360°) ×πr^{2 }cm^{2}

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

Area of the shaded region = Area of the equilateral triangle + Area of the circle – Area of the sector

= 36√3 cm^{2} +792/7 cm^{2}-132/7 cm^{2}

= (36√3+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 = (πR^{2})/4 cm^{2} = (22/7)×(1^{2})/4 = 11/14 cm^{2}

∴ Total area of the 4 quadrants = 4 ×(11/14) cm^{2} = 22/7 cm^{2}

Area of the circle = πR^{2 }cm^{2} = (22/7×1^{2}) = 22/7 cm^{2}

Area of the shaded region = Area of square – (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.

⇒ BD = AB/2

Since, AD is the median of the triangle

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

⇒ (2/3)AD = 32 cm

⇒ AD = 48 cm

In ΔADB,

By Pythagoras theorem,

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

⇒ AB^{2 }= 48^{2}+(AB/2)^{2}

⇒ AB^{2 }= 2304+AB^{2}/4

⇒ 3/4 (AB^{2})= 2304

⇒ AB^{2 }= 3072

⇒ AB= 32√3 cm

Area of ΔADB = √3/4 ×(32√3)^{2 }cm^{2 }= 768√3 cm^{2}

Area of circle = πR^{2} = (22/7)×32×32 = 22528/7 cm^{2}

Area of the design = Area of circle – Area of ΔADB

= (22528/7 – 768√3) 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.

∴ Radius of the circles = 14/2 cm = 7 cm

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

Area of the quadrant = (πR^{2})/4 cm^{2} = (22/7) ×7^{2}/4 cm^{2}

= 77/2 cm^{2}

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

Area of the shaded region = Area of the square ABCD – Area of the quadrant

= 196 cm^{2 }– 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 = O’C

= 60/2 m = 30 m

Radius of outer semicircle, R = OA = O’B

= 30+10 m = 40 m

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

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

= 106+106+(2×πr) m = 212+(2×22/7×30) m

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

Area of the track = Area of ABCD + Area EFGH + 2 × (area of outer semicircle) – 2 × (area of inner semicircle)

= (AB×CD)+(EF×GH)+2×(πr^{2}/2) -2×(πR^{2}/2) m^{2}

= (106×10)+(106×10)+2×π/2(r^{2}-R^{2}) m^{2}

= 2120+22/7×70×10 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 ΔBCA = OC = 7 cm

Base of ΔBCA = AB = 14 cm

Area of ΔBCA = 1/2 × AB × OC = (½)×7×14 = 49 cm^{2}

Area of larger circle = πR^{2 }= (22/7)×7^{2} = 154 cm^{2}

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

Area of smaller circle = πr^{2} = (22/7)×(7/2)×(7/2) = 77/2 cm^{2}

Area of the shaded region = Area of larger circle – Area of triangle – 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 π = 3.14 and √3 = 1.73205)**

**Solution:**

ABC is an equilateral triangle.

∴ ∠ A = ∠ B = ∠ C = 60°

There are three sectors each making 60°.

Area of ΔABC = 17320.5 cm^{2}

⇒ √3/4 ×(side)^{2} = 17320.5

⇒ (side)^{2} =17320.5×4/1.73205

⇒ (side)^{2} = 4×10^{4}

⇒ side = 200 cm

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

Area of the sector = (60°/360°)×π r^{2 }cm^{2}

= 1/6×3.14×(100)^{2 }cm^{2}

= 15700/3cm^{2}

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

Thus, area of the shaded region = Area of equilateral triangle ABC – 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.

∴ Side of the square = 3×diameter of circle = 3×14 = 42 cm

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

Area of the circle = π r^{2 }= (22/7)×7×7 = 154 cm^{2}

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

Area of the remaining portion of the handkerchief = Area of the square – Total area of the design = 1764 – 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 = (πR^{2})/4 cm^{2}

= (22/7)×(7/2)×(7/2)/4 cm^{2}

= 77/8 cm^{2}

**(ii)** Area of triangle BOD = (½)×(7/2)×2 cm^{2}

= 7/2 cm^{2}

Area of shaded region = Area of quadrant – 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 π = 3.14)**

**Solution:**

Side of square = OA = AB = 20 cm

Radius of the quadrant = OB

OAB is right angled triangle

By Pythagoras theorem in ΔOAB,

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

⇒ OB^{2 }= 20^{2 }+20^{2}

⇒ OB^{2 }= 400+400

⇒ OB^{2 }= 800

⇒ OB= 20√2 cm

Area of the quadrant = (πR^{2})/4 cm^{2 }= (3.14/4)×(20√2)^{2 }cm^{2 }= 628cm^{2}

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

Area of the shaded region = Area of the quadrant – 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 ∠AOB = 30°, 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 = 30°

Area of the larger sector = (30°/360°)×πR^{2 }cm^{2}

= (1/12)×(22/7)×21^{2 }cm^{2}

= 231/2cm^{2}

Area of the smaller circle = (30°/360°)×πr^{2 }cm^{2}

= 1/12×22/7×7^{2 }cm^{2}

=77/6 cm^{2}

Area of the shaded region = (231/2) – (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 ΔABC,

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

⇒ BC^{2 }= 14^{2 }+14^{2}

⇒ BC = 14√2 cm

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

Area of ΔABC =( ½)×14×14 = 98 cm^{2}

Area of quadrant = (¼)×(22/7)×(14×14) = 154 cm^{2}

Area of the semicircle = (½)×(22/7)×7√2×7√2 = 154 cm^{2}

Area of the shaded region =Area of the semicircle + Area of ΔABC – 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 ΔABC = Area of ΔADC = (½)×8×8 = 32 cm^{2}

Area of quadrant AECB = Area of quadrant AFCD = (¼)×22/7×8^{2}

= 352/7 cm^{2}

Area of shaded region = (Area of quadrant AECB – Area of ΔABC) = (Area of quadrant AFCD – Area of ΔADC)

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

= 2×(352/7-32) cm^{2}

= 256/7 cm^{2}