TOPICS

Exercise - 12.3

Areas Related To Circle

**Question-1 :-** Find the area of the shaded region in Fig., if PQ = 24 cm, PR = 7 cm and O is the centre of the circle. (Take π = 22/7).

Here, RQ is the diameter of the circle. Therefore, ∠RPQ will be 90°. PQ = 24 cm, PR = 7 cm By applying Pythagoras theorem in ΔPQR, RP² + PQ² = RQ² (7)² + (24)² = RQ² 49 + 576 = RQ² RQ² = 625 RQ = ±25 cm Radius of circle, OR = RQ/2 = 25/2 cm Since RQ is the diameter of the circle, it divides the circle in two equal parts. Area of semi-circle RPQOR = 1/2 x πr² = 1/2 x 22/7 x (25/2)² = 11/7 x 625/4 = 6875/28 cm² Area of ΔPQR = 1/2 x PQ x PR = 1/2 x 24 x 7 = 84 cm² Area of shaded region = Area of semi-circle RPQOR − Area of ΔPQR = 6875/28 - 84 = (6875 - 2352)/28 = 4523/28 cm²

**Question-2 :-** Find the area of the shaded region in Fig. , if radii of the two concentric circles with centre O are 7 cm and 14 cm respectively and ∠ AOC = 40°. (Take π = 22/7).

Radius of inner circle = 7 cm Radius of outer circle = 14 cm ∠ AOC = 40° Area of shaded region = Area of sector OAFC − Area of sector OBED = 40/360 x 22/7 x 14 x 14 - 40/360 x 22/7 x 7 x 7 = 1/9 x 22/7 x 196 - 1/9 x 22/7 x 49 = 1/9 x 22/7(196 - 49) = 1/9 x 22/7 x 147 = 1/9 x 22 x 21 = 154/3 cm²

**Question-3 :-** Find the area of the shaded region in Fig., if ABCD is a square of side 14 cm and APD and BPC are semicircles. (Take π = 22/7).

It can be observed from the figure that the radius of each semi-circle is 7 cm. Area of each semi-circle = 1/2 x πr² = 1/2 x 22/7 x 7 x 7 = 11 x 7 = 77 cm² Area of square ABCD = (Side)² = (14)² = 196 cm² Area of the shaded region = Area of square ABCD − Area of semi-circle APD − Area of semi-circle BPC = 196 − 77 − 77 = 196 − 154 = 42 cm²

**Question-4 :-** Find the area of the shaded region in Fig., 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. (Take π = 22/7).

We know that each interior angle of an equilateral triangle is of measure 60°. Radius of circle = 6 cm Area of sector OCDE = θ/360 x πr² = 60/360 x 22/7 x 6 x 6 = 1/6 x 22/7 x 6 x 6 = 22/7 x 6 = 132/7 cm² Area of ΔOAB = √3/4 x (side)² = √3/4 x 12 x 12 = √3 x 36 = 36√3 cm² Area of circle = πr² = 22/7 x 6 x 6 = 792/7 cm² Area of shaded region = Area of ΔOAB + Area of circle − Area of sector OCDE = 36√3 + 792/7 - 132/7 = (36√3 + 660/7) cm²

**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 Fig. Find the area of the remaining portion of the square. (Take π = 22/7).

Each quadrant is a sector of 90° in a circle of 1 cm radius. Area of each quadrant = θ/360 x πr² = 90/360 x 22/7 x 1 x 1 = 1/4 x 22/7 = 11/14 cm² Area of square = (Side)² = (4)² = 16 cm² Area of circle = πr² = π (1)² = 22/7 cm² Area of the shaded region = Area of square − Area of circle − 4 × Area of quadrant = 16 - 22/7 - 4 x 11/14 = 16 - 22/7 - 22/7 = 16 - 44/7 = (112 - 44)/7 = 68/7 cm²

**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 Fig. Find the area of the design. (Take π = 22/7).

Radius (r) of circle = 32 cm AD is the median of ΔABC. AO = 2/3 x AD 32 = 2/3 x AD AD = 48 cm In ΔABD, AB² = AD² + BD² AB² = (48)² + (AB/2)² AB² = 2304 + AB²/4 AB² - AB²/4 = 2304 3AB²/4 = 2304 AB² = (2304 x 4)/3 AB² = 9216/3 AB = 96/√3 By rationalization AB = 96/√3 x √3/√3 AB = 96√3/3 AB = 32√3 cm Area of equilateral triangle, ΔABC = √3/4 x (side)² = √3/4 x 32√3 x 32√3 = 96 x 8 x √3 = 768√3 cm² Area of circle = πr² = 22/7 x 32 x 32 = 22/7 x 1024 = 22528/7 cm² Area of design = Area of circle − Area of ΔABC = (22528/7 - 768√3) cm²

**Question-7 :-** In Fig., 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. (Take π = 22/7).

Area of each of the 4 sectors is equal to each other and is a sector of 90° in a circle of 7 cm radius. Area of each sector = θ/360 x πr² = 90/360 x 22/7 x 7 x 7 = 1/4 x 22 x 7 = 77/2 cm² Area of square ABCD = (Side)² = (14)² = 196 cm² Area of shaded portion = Area of square ABCD − 4 × Area of each sector = 196 - 4 x 77/2 = 196 - 154 = 42 cm²

**Question-8 :-** In Fig. depicts a racing track whose left and right ends are semicircular. (Take π = 22/7).
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.

Distance between the two inner parallel Line segments = 60 m So, radius is = 60/2 = 30 m length of segment lines = 106 m Distance around the track along its inner edge = AB + arc BEC + CD + arc DFA = 106 + 1/2 x 2πr + 106 + 1/2 x 2πr = 212 + 2πr = 212 + 2 x 22/7 x 30 = 212 + 1320/7 = (1484 + 1320)/7 = 2804/7 m Area of the track = (Area of GHIJ − Area of ABCD) + (Area of semi-circle HKI − Area of semi-circle BEC) + (Area of semi-circle GLJ − Area of semi-circle AFD) = (106 x 80 - 106 x 60) + (1/2 x 22/7 x 40 x 40 - 1/2 x 22/7 x 30 x 30) + (1/2 x 22/7 x 40 x 40 - 1/2 x 22/7 x 30 x 30) = 106(80 - 60) + 22/7 x 40 x 40 - 22/7 x 30 x 30 = 106 x 20 + 22/7(1600 - 900) = 2120 + 22/7 x 700 = 2120 + 2200 = 4320 m²

**Question-9 :-** In Fig., 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. (Take π = 22/7).

Radius (r₁) of larger circle = 7 cm Radius (r₂) of smaller circle = 7/2 cm Area of smaller circle = πr₁² = 22/7 x 7/2 x 7/2 = 77/2 cm² Area of semi-circle AECFB of larger circle = 1/2 x πr₂² = 1/2 x 22/7 x 7 x 7 = 77 cm² Ara of ΔABC = 1/2 x AB x OC = 1/2 x 14 x 7 = 49 cm² Area of the shaded region = Area of smaller circle + Area of semi-circle AECFB − Area of ΔABC = 77/2 + 77 - 49 = 28 + 77/2 = 28 + 38.5 = 66.5 cm²

**Question-10 :-** The area of an equilateral triangle ABC is 17320.5 cm². 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.). Find the area of the shaded region. (Use π = 3.14 and 3 = 1.73205).

Let the side of the equilateral triangle be a. Area of equilateral triangle = 17320.5 cm² √3/4 x a² = 17320.5 1.73205/4 x a² = 17320.5 a² = (4 x 17320.5)/1.73205 a² = 4 x 10000 a = √40000 a = 200 cm Each sector is of measure 60°. Area of sector ADEF = θ/360 x πr² = 60/360 x 3.14 x 100 x 100 = 1/6 x 3.14 x 10000 = 15700/6 cm² Area of shaded region = Area of equilateral triangle − 3 × Area of each sector = 17320.5 - 3 x 15700/3 = 17320.5 - 15700 = 1620.5 cm²

**Question-11 :-** On a square handkerchief, nine circular designs each of radius 7 cm are made (see Fig.). Find the area of the remaining portion of the handkerchief. (Take π = 22/7).

Here, side of the square is = 42 cm Radius of each circular design = 7 cm Area of square = (Side)² = (42)² = 1764 cm² Area of each circle = πr² = 22/7 x 7 x 7 = 154 cm² Area of 9 circles = 9 × 154 = 1386 cm² Area of the remaining portion of the handkerchief = 1764 − 1386 = 378 cm²

**Question-12 :-** In Fig., 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 (Take π = 22/7).

(i) Since OACB is a quadrant, it will subtend 90° angle at O. Radius of circle = 3.5 cm Area of quadrant OACB = θ/360 x πr² = 90/360 x 22/7 x 3.5 x 3.5 = 1/4 x 22 x 0.5 x 3.5 = 1/2 x 11 x 5/10 x 35/10 = 1/2 x 11 x 1/2 x 7/2 = 77/8 cm² (ii) Area of ΔOBD = 1/2 x OB x OD = 1/2 x 3.5 x 2 = 3.5 = 35/10 = 7/2 cm² Area of the shaded region = Area of quadrant OACB − Area of ΔOBD = 77/8 - 7/2 = (77 - 28)/2 = 49/2 cm²

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

A square OABC is inscribed in a quadrant OPBQ. OA = 20 cm In ΔOAB, OB² = OA² + AB² = (20)² + (20)² OB² = 2 x (20)² OB = √2 x 20 OB = 20√2 cm So, Radius (r) of circle = 20√2 cm Area of quadrant OPBQ = θ/360 x πr² = 90/360 x 3.14 x 20√2 x 20√2 = 1/4 x 3.14 x 400 x 2 = 3.14 x 100 x 2 = 3.14 x 200 = 628 cm² Area of OABC = (Side)² = (20)² = 400 cm² Area of shaded region = Area of quadrant OPBQ − Area of OABC = (628 − 400) cm² = 228 cm²

**Question-14 :-** AB and CD are respectively arcs of two concentric circles of radii 21 cm and 7 cm and centre O (see Fig.). If ∠ AOB = 30°, find the area of the shaded region. (Take π = 22/7).

Two concentric circles of radii AB and CD are 21 cm and 7 cm respectively. ∠ AOB = 30° Area of the shaded region = Area of sector OAEB − Area of sector OCFD = 30/360 x 22/7 x 21 x 21 - 30/360 x 22/7 x 7 x 7 = 1/12 x 22/7(441 - 49) = 1/6 x 11/7 x 392 = 1/6 x 11 x 56 = 308/3 cm²

**Question-15 :-** In Fig., 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. (Take π = 22/7).

As ABC is a quadrant of the circle, ∠BAC will be of measure 90º. In ΔABC, BC² = AC² + AB² = (14)² + (14)² BC² = 2 x (14)² BC = √2 x 14 BC = 14√2 cm So, Radius (r1) of semi-circle drawn on BC = 14√2/2 = 7√2 cm Area of ΔABC = 1/2 x AB x AC = 1/2 x 14 x 14 = 98 cm² Area of sector ABDC = θ/360 x πr² = 90/360 x 22/7 x 14 x 14 = 1/4 x 22 x 2 x 14 = 11 x 14 = 154 cm² Area of semi circle drawn on BC = 1/2 x πr² = 1/2 x 22/7 x 7√2 x 7√2 = 1/2 x 22/7 x 98 = 11 x 14 = 154 cm² Area of shaded region = Area of semi-circle - (Area of sector ABDC - Area of ΔABC) = 154 - (154 - 98) = 154 - 56 = 98 cm²

**Question-16 :-** Calculate the area of the designed region in Fig. common between the two quadrants of circles of radius 8 cm each. (Take π = 22/7)

The designed area is the common region between two sectors BAEC and DAFC. Radius of circle = 8 cm Area of sector BAEC = θ/360 x πr² = 90/360 x 22/7 x 8 x 8 = 1/4 x 22/7 x 64 = 22/7 x 16 = 352/7 cm² Area of ΔBAC = 1/2 x BA x BC = 1/2 x 8 x 8 = 32 cm² Area of the designed portion = 2 × (Area of segment AEC) = 2 × (Area of sector BAEC − Area of ΔBAC) = 2 x (352/7 - 32) = (2 x 128)/7 = 256/7 cm²

CLASSES