99 Chapter 13 13.1 Circle A circle is a plane figure bunded by ne line which is called the circumference, and is such that all straight lines drawn frm a certain pint within the figure t the circumference are equal t anther. The pint C is called the center f the circle. Radius A radius f a circle is a straight line drawn frm the center t the circumference. Diameter A diameter f a circle is a straight line drawn thrugh the center and terminated bth ways by the circumference. In figure AC is a radius, and AB a diameter. Circumference f a circle = πd r π r Chrd A chrd f a circle is a straight line jining tw pints n the circumference. ED is chrd. 13. Area f the Circle: If r is the radius f the circle, d is the diameter f the circle. π Then Area f circle = πr r d 4 A Hence r = π 13.3 Cncentric Circles: Cncentric circles are such as have the same center.
300 Area f the Annulus (Ring) Area between tw cncentric circles is knwn as annulus, fr example, area f a washer, the area f crss-sectin f a cncrete pipe. Area f the annulus = Area f uter circle area f inner circle = = π 4 D d 4 π π (D d ) 4 Where D is the diameter f uter circle and d is the diameter f inner circle. If R and r dente the radius f the uter and inner circles respectively. Then,Area f ring = πr πr = π(r r ) square units Example 1: Find the radius and the perimeter f a circle the area f which is 9.319 sq. cm. Area f circle = 9.319 sq. cm. Radius r =? Area f circle = πr 9.319 = (3.14)r r = 9.319 3.14 r =.96 r = 1.7 cm Perimeter f the circle = πr = (3.14)(1.7) = 10.80cm. Example : A path 14 m wide surrunds a circular lawn whse diameter is 10 m. find the area f the path. Diameter f inner circle = 10m Radius f inner circle = r = 60m Radius f uter circle = R = 60 + 14 = 74m Area f path = ( R r )
301 = 7 (74-60 ) = 7 (1876) = 5896 sq. m Example 3: A hllw shaft with 5 m internal diameter is t have the same crss-sectinal area as the slid shaft f 11m diameter. Find the external diameter f the hllw shaft. Let D = diameter f slid shaft = 11m π 11 Area f the slid shaft = (11) π 4 4 Let, d = Internal diameter f hllw shaft, d = 5m Let, D = External diameter f hllw shaft =? Area f annulus = π (D d ) 4 π π = [(D (5) ] (D 5) 4 4 But, Area f annulus = Area f slid shaft π 11 (d 5) π 4 4 D 5 = 11 D = 146 D = 1.1m which is external diameter 13.4 Sectr f a Circle: A sectr f a circle is a figure bunded by tw radii and the arc intercepted between them. The angle cntained by the tw radii is the angle f the sectr. In figure <PCQ is called the angle f sectr PCQ 1. Area f sectr When angle is given in degree. Area f circle fr angle 1 = πr 360 Hence, Area f sectr fr angle N = πr 360 x N Length f arc = l = πr 360. N. If angle <POQ is given in radian say θ radian.
30 Area f circle fr angle π rad = πr πr 1 Area f circle fr angle 1 rad = r π Hence, 1 Area f sectr fr angle θ rad = r θ --------(1) 3. Area f sectr when arc and the radius f the circle r are given. 1 Since, Area f sectr = r θ 1 = r (because θ = /r ) r A = 1 rl Example 4: Find the area f the sectr f the circle whse radius is 4cm and length f the arc is 9cm. Let, AOB be the sectr f the circle in which OA = OB = r = 4cm AB = l = 9cm Area f the sectr = 1 lr = 1 x 9 x 4 = 18 sq. cm Example 5: Find the area f the sectr f the circle when the radius f the circle is 15cm and the angle at the center is 60. Since, r = 15cm, and angle θ = 60 πr Area f the sectr = x 60 360 1 = (15) x 117.8 sq. cm 7 6 Example 6:
303 Find the expense f paving a circular curt 60cm in diameter at Rs. 3.37 per square cm. If a space is left in the center fr a funtain in the shape f a hexagnal each side f which is ne cm. π Area f the circle = d, but d = 60cm 4 Area = 3.14 x 60 x 60 4 = 88.5714 sq. cm na 180 Area f the hexagn = Ct 4 n But n = 4, a = 1cm 6 x (1) 180 Area f the hexagn = Ct 4 6 = 3 Ct 60 Area =.5980 sq. cm Area f the plt which is t be paved: = Area f the circle area f the hexagn = 88.5714.5981 = 86 sq. cm Expense = 86 x 3.37 = 953.5 rupees 13.5 Area f Segment: A segment is a prtin f circle which is cut ff by a straight line nt passing thrugh the centre. The straight line AB is called the chrd f the circle. The segment smaller than a semicircle is called a minr segment and a segment greater than a semi-circle is called a majr segment. Area f segment = Area f sectr AOB area f AOB = 1 1 r θ r Sinθ
304 Fr majr segment, psitive sign is taken and fr minr segment, negative sign is used. Area f the segment in terms f Height and Length f the Chrd f the Segment: If h is the maximum height and c is the length f the chrd f the segment, then area f the segment is given by. h Area = (3h +4c ). (1) 6c Length f Chrd and Maximum height f arc: Let ACD is an arc f a circle with center O and radius r and ADB is the chrd f length c and CD is maximum height h f the segment. (a) If h and r are given then c can be calculated. In right OAD, by Pythagras therem, (OA) = (AD) + (OD) (AD) = (OA) (OD) (AD) = r (r h) = AD = C = 1 (ADB) (b) hr h r r rh h hr h because AD = C = hr h. (i) If r and c are given, then h can be calculated. Frm (i) Squaring (i) bth sides C = 4(hr h ) c = 8hr 4h Or 4h 8hr + c = 0 Which is the quadratic equatin in h a = 4, b = 8r, c = c h = h = b ± b 4ac 8r ± 64r 16c a 8 c 8r ± 8 r 4 c r ± r 8
305 +ve sign is used fr majr segment and ve sign t be taken fr minr segment. Example 7: Find the area f a segment the chrd f which 8cm with a height f cm. Since, h = cm, c = chrd f segment = 8cm Area f segment = h (3h 4c ) 6c = (3() 4(8) ) 11.16 sq. cm 6(8) Example 8: The span f a circular arch f 90 is 10cm. Find the area f the segment. Slutin Let O be the centre and r be the radius f the circle. Span AB = 10cm. In right angle AOB, (OA) + (OB) = (AB), r + r = (10) r = 14400 r = 700 (i) r = 84.85 cm Nw area f AOB = 1 (OB)(OA) = 1 (r) (r) = 1 r Area f AOB = 1 (700) by (i) = 3600 sq. cm. πr Area f sectr = x N 360 = 3.14 x 700 x 90 5656 sq. cm 360 Area f segment = Area f sectr Area f AOB = 5656 3600 = 056 sq. cm Example 9: The chrd f an arc is 5cm and the diameter f the circle is 7cm. Find the height f the arc. Here c = 5cm d = 7cm r = 3.5cm
306 height f arc = h = 3.5 ± (3.5) r ± r 5 c h = 3.5 ± 1.5 6.5 h = 3.5 ±.45 h = 3.5 ±.45, h = 3.5.45 h = 5.59 cm, h = 105cm Example 10: Find the chrd f arc whse height is 4 cm, in a circle f radius 15 cm. Here, h = 4cm r = 15cm Chrd f arc = C = hr h (4)(15) (4) C = 70 576 144 (1) C = 4cm 13.6 Ellipse: An ellipse is defined as the lcus f a pint which mves such that the sum f its distance frm tw fixed pints remains cnstant. The fixed pints are the fci f the ellipse. i.e. PF + PF = cnstant In figure, F and F are the tw fci and D is the centre f the ellipse. AA is the majr axis and BB is the minr axis, OA and OB are the semiaxes. Als a is the length f majr axis and b is the length f minr axis. Area f an ellipse Area f an ellipse = π ab Where a = semi-majr axis b = semi-minr axis and perimeter f an ellipse = π (a + b)
307 Example 11: It is desired t lay ut a plt in the frm f an ellipse. The area is 3100 sq. cm. The axes are in the rati 3:. Find the length f the fence required fr this plt. Given area f plt in the frm f ellipse = 3100 sq. cm. Since axes are in the rati 3: a = 3 b Area f plt = π ab = 3100 = 4.71b b = 3100 4.71 = 4904.46 b = 70.03cm 3 x b x b 7 a = 3 (70.30) = 105cm Perimeter f plt = π (a + b) = (70 + 105) = 550 cm 7 Example 1: An elliptical pipe has a majr axis f 16cm and minr axis f 10cm. Find the diameter f a circular pipe that has the same area f crss-sectin. Majr axis a = 8cm Minr axis = b = 10cm b = 5cm A = π ab = a = 16cm A = (3.14)(8)(5) = 15.64 sq. cm. Area f circle = π r D Area f circle = π 4 By given cnditin Area f circle = Area f ellipse D π = 15.64 4 but D = r
D = 15.64 x 4 = 159.97 3.14 D = 1.64 cm 308 Exercise 13 Q.1 The area f a semi-circle is 130 sq. cm. Find its ttal perimeter. Hint P = r + r. Q. A rad 10m wide is t be made arund a circular plt f 75m diameter. Find the cst f the grund needed fr the rad at Rs 4.00 per square meter. Q.3 The areas f tw cncentric circles are 1386 sq. cm and 1886.5 sq. cm respectively. Find the width f the ring. Q.4 The area f a circle is 154 sq. cm. Find the length f the side f the inscribed squares. Q.5 A circular arc has a base f 4cm and maximum height 1.6cm. Find radius, length f arc and area f segment. Q.6 The height f an arc is 7cm and its chrd 4cm. Find the diameter f the circle. Q.7 In a circle f diameter 5cm, the chrd f an arc in 10cm, find its height. Q.8 The radius f a circle is 33.5cm. Find the area f a sectr enclsed by tw radii and an arc 133.74 cm in length. Q.9 The inner diameter f a circular building is 54m and the base f the wall ccupies a space f 35 sq. m. Find the thickness f the wall. Q.10 The axis f an ellipse are 40 cm and 60 cm. Find its perimeter and area. Q.11 The sides f the triangle are 8, 1 and 5 m. find the radius f the circle whse area is equal t the area f triangle. Q.1 The area f a sectr is 76 sq.cm and angle f the sectr is 70 0. Find radius f the circle. Answers 13 Q1. 46.8cm Q. 10684 rupees Q3. 3. sq. cm Q4. 9.89 cm Q5. 4.771 sq. cm;.05cm;.77cm Q6. 70cm Q7. 1.04 cm Q8. 40.14 sq.cm Q9. m Q10. 157.1cm; 1885.0sq.cm Q11. 4.99m Q1. 19.77cm.
309 Summary 1. Area f circle A = πr. Perimeter r circumferences f circle = πr 3. Area f Annulus (ring) A = π(r r ) Where R = radius f uter circle, r = radius f inner circle 4. (a) Area f the sectr if angle is given in degree πr Area f the sectr = x N 360 πr Length f the arc = x N 360 (b) If the angle in radian, say θ radians, then 1 Area f sectr = r θ (c) If l is the length f an arc and r, radius f the circle, then area f sectr. A = 1 lr 5. Area f segment = Area f sectr AOB 1 r sinθ +ve sign is taken fr majr axis -ve sign is taken fr majr axis 6. Area f the segment in terms f Height and Length f the chrd f the segment h Area = (3h +4c ) 6c Where h = maximum height, c = length f the chrd.
310 Q.1: Q.: Q.3: Define a circle. Define diameter f a circle Define chrd f a circle. Shrt Questins Q.4: What is the area and circumference f circle. Q.5: Q.6: Q.7: Find the radius f a circle the area f which is 9.319 sq. cm. What are cncentric circle. Define area f the Annulus (Ring). Q.8: A path 14cmwide, surrunds a circular lawn whse diameter is 360 cm. Find the area f the path. Q.9: Define a sectr f the circle. Q.10: Write the area f the sectr. Q.11: The minute hand f a clck is 1 cm lng. Find the area which is described n the clck face between 6 A.M.t6.0A.M. Q.1: Define a segment. Q.13: Write the frmula f Area f the minr segment and majr segment when angle and radius r are given. Q.14: Write the area f the segment in terms f Height and length f the chrd f the segment. Q.15: Find the area f a segment the chrd f which 8 cm with a height f cm. Q.16: The area f a semi-circle is 130 sq. cm. Find its ttal perimeter. A r r B Answers Q5. r= 1.7 cm. Q8. 16456 sq.cm Q11. 150.7 sq.cm. Q13. Area f minr segment = 1 r ( sin ) Area f a majr segment = 1 r ( + sin ) Q15. 11.16 sq. cm. Q16. 46.7 cm.
311 Objective Type Questins Q.1 Each questins has fur pssible answers. Chse the crrect answer and encircle it. 1. Area f a circle whse radius is a cm is: (a) πr (b) πa (c) πa (d) π a. Circumference f a circle whse radius is 1 cm is equal t πr r (a) π (b) πr (c) (d) 3. Area f a sectr f 60 in a circle f radius 6 cm is: (a) 6π (b) 6π (c) 36π (d) 3π 4. The space enclsed between tw cncentric circles is called (a) cne (b) ellipse (c) annulus (d) nne f these 5. Arc f circle with diameter d is: π (a) r π (b) d π (c) d (d) Nne f these 4 6. If R and r dente the radii f the uter and inner circles, then Area f annulus (ring) is: π (a) π(r r ) (b) (R r ) (c) π(r + r ) (d)(r r ) 7. If a and b are the majr and minr axis f the ellipse, then area f ellipse is: π (a) ab (b) πab (c) ab (d) π ab 8. If a and b are the majr and minr axis f the ellipse, the circumference f the ellipse is: (a) π(a + b) (b) π(a b) (c) π(a + b) (d) π (a b) 9. If the area f circle is, then radius r is: (a) 4 (b) (c) 16 (d) 8 10. If x and y the majr and minr axis f the ellipse then area f ellipse is: πxy πxy (a) (b) (c) πxy (d) π xy 4
31 11. The circumference f a pulley is 440 cm, its diameter is: 7 (a) 0cm (b) 40cm (c) 80cm (d) 10cm Answers Q.1 (1) b () a (3) a (4) c (5) c (6) a (7) b (8) a (9) a (10) b (11) a