{"id":44190,"date":"2022-05-29T02:30:40","date_gmt":"2022-05-28T21:00:40","guid":{"rendered":"https:\/\/cbselibrary.com\/?p=44190"},"modified":"2023-11-09T16:08:54","modified_gmt":"2023-11-09T10:38:54","slug":"ml-aggarwal-class-8-solutions-for-icse-maths-chapter-15-check-your-progress","status":"publish","type":"post","link":"https:\/\/cbselibrary.com\/ml-aggarwal-class-8-solutions-for-icse-maths-chapter-15-check-your-progress\/","title":{"rendered":"ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 15 Circle Check Your Progress"},"content":{"rendered":"
Question 1.
\nDraw a circle of radius 2\u00b77 cm. Draw a chord PQ of length 4 cm of this circle. Shade the major segment of this circle.
\nSolution:
\n(i) Draw a circle of radius = 2\u00b77 cm.
\n(ii) Take a point P anywhere on the circle.
\n(iii) With P as centre and 4 cm as radius,
\ndraw an arc which cuts the circle at Q.
\n
\n(iv) Join PQ which is the required chord.
\n(v) Shade the major segment.<\/p>\n
Question 2.
\nDraw a circle of radius 3\u00b72 cm and in it make a sector of angle.
\n(i) 30\u00b0
\n(ii) 135\u00b0
\n(iii) \\(2 \\frac{2}{3}\\) right angles
\nDraw separate diagrams and shade the sectors.
\nSolution:
\n(i) 30\u00b0
\nSteps :
\n(a) Draw a circle with centre C and radius CB = 3.2 cm
\n(b) From C, make an angle of 30\u00b0.
\n(c) Shade the region enclosed in ABC.
\n
\n(ii) 135\u00b0
\nSteps :
\n(a) Draw a circle with centre C and radius CB = 3.2 cm.
\n(b) From C, make an angle of 135\u00b0.
\n(c) Shade the region enclosed in ACB.
\n
\n(iii) \\(2 \\frac{2}{3}\\) right angles
\nSteps :
\n(a) Draw a circle with centre C and radius 3.2 cm.
\n(b) From C, make an \u2220, \\(2 \\frac{2}{3}\\) of right angle
\n= \\(2 \\frac{2}{3}\\) \u00d7 90\u00b0
\n= \\(\\frac{8}{3}\\) \u00d7 90\u00b0 = 240\u00b0
\n(c) Shade the region enclosed in ACB.
\n
\n= \\(\\frac{8}{3}\\) \u00d7 90\u00b0 = 240\u00b0<\/p>\n
Question 3.
\nDraw a line segment PQ = 6\u00b74 cm. Construct a circle on PQ as diameter. Take any point R on this circle and measure \u2220PRQ.
\nSolution:
\n(i) Draw a line segment PQ = 6\u00b74 cm.
\n(ii) Draw \u22a5 bisector of PQ.
\n(iii) With O as centre and OP or OQ as radius draw a circle
\nwhich passes through P as well as through Q.
\n
\n(iv) Take point R on the circle.
\n(v) Join PR and QR.
\n(vi) Measure \u2220PRQ, we get \u2220PRQ = 90\u00b0.<\/p>\n
Question 4. Question 5. (ii) Similarly \u2220AQB = 90\u00b0 (Angle in a semicircle) ML Aggarwal Class 8 Solutions for ICSE Maths Chapter 15 Circle Check Your Progress Question 1. Draw a circle of radius 2\u00b77 cm. Draw a chord PQ of length 4 cm of this circle. Shade the major segment of this circle. Solution: (i) Draw a circle of radius = 2\u00b77 cm. (ii) Take a point … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[3034],"tags":[],"yoast_head":"\n
\nIn the adjoining figure, the tangent to a circle of radius 6 cm from an external point P is of length 8 cm. Calculate the distance of the point P from the nearest point of the circumference.
\n
\nSolution:
\nC is the centre of the circle
\nPT is the tangent to the circle from P.
\nCT is the radius
\n\u2234 CT \u22a5 PT
\nCT = CR = 6 cm, PT = 8 cm
\nNow in right \u2206CPT (By Pythagoras Theorem)
\nCP2<\/sup> = PT2<\/sup> + CT2<\/sup> = (8)2<\/sup> + (6)2<\/sup>
\n= 64 + 36 = 100 = (10)2<\/sup>
\n\u2234 CP = 10 cm
\nNow PR = CP – CR = 10 – 6 = 4 cm<\/p>\n
\nIn the given figure, O is the centre of the circle. If \u2220ABP = 35\u00b0 and \u2220BAQ = 65\u00b0, find
\n(i) \u2220PAB
\n(ii) \u2220QBA
\n
\nSolution:
\nIn the figure,
\nAB is the diameter of the circle with centre O
\n\u2220ABP = 35\u00b0 and \u2220BAQ = 65\u00b0
\n(i) \u2220APB = 90\u00b0 (Angle in a semicircle)
\nIn \u2206APB, By \u2220sum property of \u2206
\n\u2220PAB + \u2220P + \u2220ABP = 180\u00b0
\n\u2220PAB + 90\u00b0 + \u2220ABP = 180\u00b0
\n\u2234 \u2220PAB + \u2220ABP = 90\u00b0
\n\u21d2 \u2220PAB + 35\u00b0 = 90\u00b0 \u21d2 \u2220PAB = 90\u00b0 – 35\u00b0 \u2220PAB = 55\u00b0<\/p>\n
\nIn \u2206AOB, By angle sum property of \u2206
\n\u2220BAQ + \u2220Q + \u2220QBA = 180\u00b0
\n\u2220BAQ + \u2220QBA + 90\u00b0 = 180\u00b0
\n\u2234 \u2220BAQ + \u2220QBA = 90\u00b0
\n\u21d2 65\u00b0 + \u2220QBA = 90\u00b0
\n\u21d2 \u2220QBA = 90\u00b0- 65\u00b0 = 25\u00b0
\nHence \u2220PAB = 55\u00b0 and \u2220QBA = 25\u00b0<\/p>\nML Aggarwal Class 8 Solutions for ICSE Maths<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"