{"id":3592,"date":"2021-02-22T17:06:11","date_gmt":"2021-02-22T11:36:11","guid":{"rendered":"https:\/\/apboardsolutions.guru\/?p=3592"},"modified":"2021-09-15T11:40:42","modified_gmt":"2021-09-15T06:10:42","slug":"ap-ssc-10th-class-maths-solutions-chapter-9-ex-9-3","status":"publish","type":"post","link":"https:\/\/apboardsolutions.guru\/ap-ssc-10th-class-maths-solutions-chapter-9-ex-9-3\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 9 Tangents and Secants to a Circle Ex 9.3 Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.3<\/h2>\n

10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.3 Textbook Questions and Answers<\/h3>\n

Question 1.
\nA chord of a circle of radius 10 cm. subtends a right angle at the centre. Find the area of the corresponding: (use \u03c0 = 3.14)
\ni) Minor segment ii) Major segment
\nAnswer:
\n\"AP
\nAngle subtended by the chord = 90\u00b0 Radius of the circle = 10 cm
\nArea of the minor segment = Area of the sector POQ – Area of \u25b3POQ
\nArea of the sector = \\(\\frac{x}{360}\\) \u00d7 \u03c0r2<\/sup>
\n\\(\\frac{90}{360}\\) \u00d7 3.14 \u00d7 10 \u00d7 10 = 78.5
\nArea of the triangle = \\(\\frac{1}{2}\\) \u00d7 base \u00d7 height
\n= \\(\\frac{1}{2}\\) \u00d7 10 \u00d7 10 = 50
\n\u2234 Area of the minor segment = 78.5 – 50 = 28.5 cm2<\/sup>
\nArea of the major segment = Area of the circle – Area of the minor segment
\n= 3.14 \u00d7 10 \u00d7 10 – 28.5
\n= 314 – 28.5 cm2<\/sup>
\n= 285.5 cm2<\/sup><\/p>\n

\"AP<\/p>\n

Question 2.
\nA chord of a circle of radius 12 cm. subtends an angle of 120\u00b0 at the centre. Find the area of the corresponding minor segment of the circle.
\n(use \u03c0 = 3.14 and \u221a3 = 1.732)
\nAnswer:
\nRadius of the circle r = 12 cm.
\nArea of the sector = \\(\\frac{x}{360}\\) \u00d7 \u03c0r2<\/sup>
\nHere, x = 120\u00b0
\n\"AP
\n\\(\\frac{120}{360}\\) \u00d7 3.14 \u00d7 12 \u00d7 12 = 150.72
\nDrop a perpendicular from \u2018O\u2019 to the chord PQ.
\n\u25b3OPM = \u25b3OQM [\u2235 OP = OQ \u2220P = \u2220Q; angles opp. to equal sides OP & OQ; \u2220OMP = \u2220OMQ by A.A.S]
\n\u2234 \u25b3OPQ = \u25b3OPM + \u25b3OQM = 2 . \u25b3OPM
\nArea of \u25b3OPM = \\(\\frac{1}{2}\\) \u00d7 PM \u00d7 OM
\n\"AP
\n= 18 \u00d7 1.732 = 31.176 cm
\n\u2234 \u25b3OPQ = 2 \u00d7 31.176 = 62.352 cm2<\/sup>
\n\u2234 Area of the minor segment
\n\"AP = (Area of the sector) – (Area of the \u25b3OPQ)
\n= 150.72 – 62.352 = 88.368 cm2<\/sup><\/p>\n

Question 3.
\nA car has two wipers which do not overlap. Each wiper has a blade of length 25 cm. sweeping through an angle of 115\u00b0. Find the total area cleaned at each sweep of the blades. (use \u03c0 = \\(\\frac{22}{7}\\))
\nAnswer:
\nAngle made by the each blade = 115\u00b0
\nTotal area swept by two blades
\n= Area of the sector with radius 25 cm and angle 115\u00b0+ 115\u00b0 = 230\u00b0
\n= Area of the sector = \\(\\frac{x}{360}\\) \u00d7 \u03c0r2<\/sup>
\n= \\(\\frac{230}{360}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 25 \u00d7 25
\n= 1254.96
\n\u2243\u00a0 1255 cm2<\/sup><\/p>\n

\"AP<\/p>\n

Question 4.
\nFind the area of the shaded region in figure, where ABCD is a square of side 10 cm. and semicircles are drawn with each side of the square as diameter (use \u03c0 = 3.14).
\n\"AP
\nAnswer:
\nLet us mark the four unshaded regions as I, II, III and IV.
\n\"AP
\nArea of I + Area of II
\n= Area of ABCD – Areas of two semicircles with radius 5 cm
\n= 10 \u00d7 10 – 2 \u00d7 \\(\\frac{1}{2}\\) \u00d7 \u03c0 \u00d7 52<\/sup>
\n= 100 – 3.14 \u00d7 25
\n= 100 – 78.5 = 21.5 cm2<\/sup>
\nSimilarly, Area of II + Area of IV = 21.5 cm2<\/sup>
\nSo, area of the shaded region = Area of ABCD – Area of unshaded region
\n= 100 – 2 \u00d7 21.5 = 100 – 43 = 57 cm2<\/sup><\/p>\n

Question 5.
\nFind the area of the shaded region in figure, if ABCD is a square of side 7 cm. and APD and BPC are semicircles. (use \u03c0 = \\(\\frac{22}{7}\\))
\n\"AP
\nAnswer:
\nGiven,
\nABCD is a square of side 7 cm.
\nArea of the shaded region = Area of ABCD – Area of two semicircles with radius \\(\\frac{7}{2}\\) = 3.5 cm
\nAPD and BPC are semicircles.
\n= 7 \u00d7 7 – 2 \u00d7 \\(\\frac{1}{2}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 3.5 \u00d7 3.5
\n= 49 – 38.5
\n= 10.5 cm2<\/sup>
\n\u2234 Area of shaded region = 10.5 cm<\/p>\n

\"AP<\/p>\n

Question 6.
\nIn figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm., find the area of the shaded region, (use \u03c0 = \\(\\frac{22}{7}\\)).
\nAnswer:
\nGiven, OACB is a quadrant of a Circle.
\nRadius = 3.5 cm; OD = 2 cm.
\nArea of the shaded region = Area of the sector – Area of \u25b3BOD
\n\"AP
\n= 9.625 – 3.5 = 6.125 cm2<\/sup>
\n\u2234 Area of shaded region = 6.125 cm2<\/sup>.<\/p>\n

Question 7.
\nAB and CD are respectively arcs of two concentric circles of radii 21 cm. and 7 cm. with centre O (See figure). If \u2220AOB = 30\u00b0, find the area of the shaded region. (use \u03c0 = \\(\\frac{22}{7}\\)).
\n\"AP
\nAnswer:
\nGiven, AB and CD are the arcs of two concentric circles.
\nRadii of circles = 21 cm and 7 cm and \u2220AOB = 30\u00b0
\nWe know that,
\nArea of the sector = \\(\\frac{x}{360}\\) \u00d7 \u03c0r2<\/sup>
\nArea of the shaded region = Area of the OAB – Area of OCD
\n\"AP
\n\u2234 Area of shaded region = 102.66 cm2<\/sup><\/p>\n

\"AP<\/p>\n

Question 8.
\nCalculate the area of the designed region in figure, common between the two quadrants of the circles of radius 10 cm each, {use \u03c0 = 3.14)
\nAnswer:
\nMark two points P, Q on the either arcs.
\nLet BD be a diagonal of ABCD
\nNow the area of the segment
\n\"AP
\n= 28.5 + 28.5 = 57 cm2<\/sup><\/p>\n

Side of the square = 10 cm
\nArea of the square = side \u00d7 side
\n= 10 \u00d7 10 = 100 cm2<\/sup>
\nArea of two sectors with centres A and C and radius 10 cm.
\n= 2 \u00d7 \\(\\frac{\\pi r^{2}}{360}\\) \u00d7 x = 2 \u00d7 \\(\\frac{x}{360}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 10 \u00d7 10
\n= \\(\\frac{1100}{7}\\)
\n= 157.14 cm2<\/sup>
\n\u2234 Designed area is common to both the sectors,
\n\u2234 Area of design = Area of both sectors – Area of square
\n= 157 – 100 = 57 cm2<\/sup>
\n(or)
\n\\(\\frac{1100}{7}\\) – 100 = \\(\\frac{1100-700}{7}\\)
\n= \\(\\frac{400}{7}\\)
\n= 57.1 cm2<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 9 Tangents and Secants to a Circle Ex 9.3 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 9th Lesson Tangents and Secants to a Circle Exercise 9.3 10th Class Maths 9th Lesson Tangents and Secants to a Circle Ex 9.3 …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/3592"}],"collection":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/comments?post=3592"}],"version-history":[{"count":0,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/posts\/3592\/revisions"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/media?parent=3592"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/categories?post=3592"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.guru\/wp-json\/wp\/v2\/tags?post=3592"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}