Gujarat Board GSEB Solutions Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 10 Maths Chapter 12 Areas Related to Circles Ex 12.3

Question 1.

Find the area of shaded region in the given figure. If PQ = 24 cm, PR = 7 cm, and O is the centre of the circle. (CBSE 2012)

Solution:

∠QPR is right angle (Angle subtended in semicircle is right angle)

Therefore

QR^{2} = PR^{2} + PQ^{2} = 24^{2} + 7^{2} = 576 + 49

QR^{2} = 625

QR = 25 cm

r = \(\frac {25}{2}\) cm

Hence, area of shaded region

= Area of semicircle – Area of ΔPQR

= \(\frac {1}{2}\) πr^{2} – \(\frac {1}{2}\) x PQ x PR

= \(\frac {1}{2}\) x \(\frac {22}{7}\) x \(\frac {25}{2}\) x \(\frac {25}{2}\) – \(\frac {1}{2}\) x 24 x 7

= \(\frac {6875}{28}\) – 84 = \(\frac {6875 – 84 × 28}{28}\) = \(\frac {4523}{28}\) cm^{2}

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

Question 2.

Find the area of the shaded region in given figure, if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and

∠AOC = 40°. (CBSE 2012)

Solution:

Here r_{1} = 7 cm, r_{1} = 14 cm and θ = 40°

Area of shaded region

= \(\frac{\pi r_{2}^{2} \theta}{360^{\circ}}\) – \(\frac{\pi r_{1}^{2} \theta}{360^{\circ}}\) = \(\frac{\pi \theta}{360^{\circ}}\) = [r_{2}^{2} – r_{1}^{2}

= \(\frac{\frac{22}{7} \times 40^{\circ}}{360^{\circ}}\) [14^{2} – 7^{2}]

= \(\frac{22 \times 40}{7 \times 360}\) [196 – 49]

= \(\frac{22}{7 \times 9}\) x 147 = \(\frac{22}{9}\) x 21 = \(\frac{154}{3}\) cm^{2}

Question 3.

Find the area of shaded region in the given figure, if ABCD is a square of side 14 cm and APD and BPC are semicircies.

Solution:

Side of square = 14 cm

Radius of semicircle = \(\frac{14}{2}\) = 7 cm

Area of shaded region

= Area of square – 2 x area of semicircle

= Side x Sue – 2 x πr^{2} = 14 x 14 – πr^{2}

= 196 – \(\frac{22}{7}\) x 7 x 7

= 196 – 154 = 42 cm^{2}

Question 4.

Find the area of the shaded region in the given figure, 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. (CBSE 2016)

Solution:

Here r = 6 cm

ΔAOB is an equilateral triangle.

Hence θ = 60°

Area of shaded region

= Area of an equilateral triangie + Area of circle – Area of sector

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:

Radius of circle = \(\frac{2}{1}\) = 1 cm

Radius of quadrant = 1 cm

Area of shaded region

= Area of square – (Area of 4 quadrants + Area of circle)

= 4 x 4 -[ 4 x \(\frac{1}{4}\)πr^{2} + πr^{2}]

= 16 – 2πr^{2}

= 16 – 2 x \(\frac{22}{7}\) x 1^{2} = 16 – \(\frac{44}{7}\)

= \(\frac{112 – 44}{7}\) = \(\frac{68}{7}\) cm^{2}

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 (Shaded region).

Solution:

ABC is an equilateral triangle.

Centre of circle coincide the centroid of the equilateral triangle.

Hence, = \(\frac{OD}{OA}\) = \(\frac{1}{2}\)

\(\frac{OD + OA }{OA}\) = \(\frac{3}{2}\)

\(\frac{h}{r}\) = \(\frac{3}{2}\)

h = 32 x \(\frac{3}{2}\) = 48 cm

Now in ft. ADB

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

⇒ a^{2} = 48^{2} + (\(\frac{a}{2}\))^{2}

⇒ a^{2} – \(\frac{a^{2}}{4}\) = 48 x 48

⇒ \(\frac{3a^{2}}{4}\) = 48 x 48

⇒ a^{2} = \(\frac{48 \times 48 \times 4}{3}\)

⇒ a^{2} = 64 x 48

Area of design or shaded region

= πr^{2} – \(\frac{\sqrt{3}}{4}\) x 64 x 48

= \(\frac{22}{7}\) x 32 x 32 – 16 x 48 \(\sqrt{3}\)

= (\(\frac{22528}{7}\) – 768\(\sqrt{3}\)) cm^{2}

Question 7.

In figure, 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 shaded region.

Solution:

Side of square = 14 cm

Radius of square = 7 cm

Area of shaded region

= Area of square – 4 x Area of quadrants

= Side x Side – 4 x \(\frac{1}{4}\) πr^{2}

= 14 x 14 – x \(\frac{22}{7}\) x 7^{2}

= 196 – 22 x 7 = 196 – 154 = 42 cm^{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 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:

(i) Distance between two inner parallel lines = 60 m

Radius of inner circular ends = \(\frac{60}{2}\) = 30m

The distance around the track along its inner track

= 106 + 106 + 2 x circumference of semicircle

= 212 + 2 x \(\frac{1}{2}\) x 2πr

= 212 + 2 x \(\frac{22}{7}\) x 30 = 212 + \(\frac{1320}{7}\)

= \(\frac{1484 + 1320}{7}\) = \(\frac{2804}{7}\) m

(ii) Area of the track

= Area of two rectangular track

+ 2 x [\(\frac{1}{2}\) πr^{2}_{2} – \(\frac{1}{2}\)r^{2}_{1}]

= 2 x 106 x 10 + 2 x \(\frac{1}{2}\)π[(30 + 10)^{2} – (30)^{2}]

= 2120 + \(\frac{22}{7}\) [40^{2} – 30^{2}]

= 2120 + \(\frac{22}{7}\) x (40 + 30) x (40 – 30)

= 2120 + \(\frac{22}{7}\) x 70 x 10

= 2120 + 2200

= 4320 m^{2}

Question 9.

In figure, 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. (CBSE 2000, 2010, 2013)

Solution:

Here, OA is the radius of larger circle, i.e.,

r = 7 cm

AB = 2r = 2 x 7 = 14cm

Radius of smaller circle,

r_{2} = \(\frac{OD}{2}\) = \(\frac{7}{2}\) cm

Area of shaded region = Area of smaller circle + Area of semicircle – Area of ΔABC

= πr^{2}_{2} + (\(\frac{1}{2}\)πr^{2} – \(\frac{1}{2}\) x AB x OC)

= π x (\(\frac{7}{2}\))^{2} + \(\frac{1}{2}\)π x 7^{2} – \(\frac{1}{2}\) x 14 x 7

= \(\frac{49π}{4}\) + \(\frac{49π}{2}\) – 49 = \(\frac{147 π}{4}\) – 49

= \(\frac{147 – 22}{4 x 7}\) – 49 = \(\frac{21 x 11}{2}\) – 49

= \(\frac{231 – 98}{2}\) = \(\frac{133}{4 x 7}\) = 66.5 cm^{2}

Question 10.

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

Solution:

Hence, area of an equilateral triangle

= \(\frac{\sqrt{3}}{4}\) a^{2}

\(\frac{\sqrt{3}}{4}\)a^{2} = 17320.5

⇒ a^{2} = \(\frac{17320.5 \times 4}{\sqrt{3}}\)

⇒ a^{2} = \(\frac{17320.5 \times 4}{\sqrt{3}}\)

a^{2} = \(\frac{17320.5 \times 4}{1.73205}\)

a^{2} = 40,000

a^{2} = 200 cm

Radius of circle = \(\frac{200}{2}\) = 100 cm

Area of shaded region = Area of an equilateral triangle – 3 x Area of sectors

= 17320.5 – 3 x \(\frac{\pi r^{2} \theta}{360^{\circ}}\)

= 17320.5 – 3 x \(\frac{3.14 \times 100^{2} \times 60^{\circ}}{360^{\circ}}\)

= 17320.5 – 3 x \(\frac{3.14 \times 10000}{6}\)

= 17320.5 – 15700 = 1620.5 cm^{2}

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:

Here, side of square ABCD

= 3 x diameter of a circular design

= 3 x d = 3 x 2r

= 6 x 7 = 42cm

Area of remaining handkerchief = Area of square ABCD – Area of 9 circular designs

= Side x Side – 9πr^{2}

= 42 x 42 – 9 x \(\frac{22}{7}\) x 7 x 7

= 1764 – 9 x 22 x 7 = 1764 – 1386

= 378 cm^{2}

Question 12.

In figure OACB is a quadrants of circle with centre O and radius 3.5 cm. If OD = 2 cm, find the area of the

(i) quadrant OACB

(ii) shaded region. (CBSE 2017)

Solution:

(i) Area of quadrant AOBC

\(\frac{1}{4}\)πr^{2} = \(\frac{1}{4}\) x \(\frac{22}{7}\) x 3.5^{2}

= \(\frac{1}{4}\) x \(\frac{22}{7}\) = \(\frac{35}{10}\) x \(\frac{35}{10}\) = \(\frac{77}{8}\)cm^{2}

(ii) Area of shaded region = Area of the quadrant OACB – area of the ΔOBD

= \(\frac{77}{8}\) – \(\frac{1}{2}\) OB x OD

= \(\frac{77}{8}\) – \(\frac{1}{2}\) x 3.5 x 2 = \(\frac{77}{8}\) – \(\frac{35}{10}\)

= \(\frac{77}{8}\) – \(\frac{7}{2}\) = \(\frac{77 – 28}{8}\) = \(\frac{49}{8}\) cm^{2}

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) (CBSE 2012, 2014)

Solution:

In rt. ΔOAB

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

OB^{2} = 20^{2} + 20^{2}

OB^{2} = 400 + 400

OB = 20\(\sqrt{2}\) cm

r = OB = 20\(\sqrt{2}\) cm

Area of shaded region = Area of quadrant – Area of square OABC

= \(\frac{1}{4}\)πr^{2} – OA x OA

= \(\frac{1}{4}\) x 3.14 x (20\(\sqrt{2}\))^{2} – 20 x 20

= \(\frac{1}{4}\) x 3.14 x 20 x 20 x 2 – 400

= 628 – 400 = 228 cm^{2}

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 = 30°, find the area of shaded region.

Solution:

Here r_{1} = 7 cm

r_{2} = 21 cm

and θ = 30°

Area of shaded region = Area of shaded region OAB – Area of shaded region OCD

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 shaded region. (CBSE 2008,2012)

Solution:

Here, radius of quadrant, r = 14 cm

In rt. ΔBAC

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

BC^{2} = 14^{2} + 14^{2}

= 196 + 196

BC^{2} = 2 x 196

BC = \(\sqrt{196 \times 2}\)

BC = 14\(\sqrt{2}\) cm

r_{1} = \(\frac{14 \sqrt{2}}{2}\) = 7 \(\sqrt{2}\) cm

Area of segment BCPB = Area of quadrant ACPB – Area of ΔABC

= \(\frac{1}{4}\)πr^{2} – \(\frac{1}{2}\)AC x AB

= \(\frac{1}{4}\) x \(\frac{22}{7}\) x 14^{2} – \(\frac{22}{7}\) – 14 x 14

= \(\frac{22}{4 x 7}\) x 14 x 14 – 98

= 154 – 98 = 56 cm^{2}

Area of shaded region = Area of semicircle – Area of segment BCPB

= \(\frac{1}{2}\) πr^{2}_{2} – 56

= \(\frac{1}{2}\) x \(\frac{22}{7}\) x (7\(\sqrt{2}\))^{2} -56

= \(\frac{11}{7}\) x 49 x 2 – 56

= 154 – 56 = 98 cm^{2}

Question 16.

Calculate the area of the diagonal region in figure common between the two quadrants of circles of radius 8 cm each.

Solution:

Area of segment

= Area of sector – Area of ΔABC

Area of shaded region = 2 x \(\frac{128}{7}\) = \(\frac{256}{7}\) cm^{2}