{"id":3432,"date":"2023-02-07T02:21:27","date_gmt":"2023-02-06T20:51:27","guid":{"rendered":"https:\/\/apboardsolutions.com\/?p=3432"},"modified":"2023-03-29T15:45:26","modified_gmt":"2023-03-29T10:15:26","slug":"ap-ssc-10th-class-maths-solutions-chapter-10-ex-10-1","status":"publish","type":"post","link":"https:\/\/apboardsolutions.com\/ap-ssc-10th-class-maths-solutions-chapter-10-ex-10-1\/","title":{"rendered":"AP SSC 10th Class Maths Solutions Chapter 10 Mensuration Ex 10.1"},"content":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions<\/a> Chapter 10 Mensuration Ex 10.1 Textbook Questions and Answers.<\/p>\n

AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.1<\/h2>\n

10th Class Maths 10th Lesson Mensuration Ex 10.1 Textbook Questions and Answers<\/h3>\n

Question 1.
\nA joker\u2019s cap is in the form of right circular cone whose base radius is 7 cm and height is 24 cm. Find the area of the sheet required to make 10 such caps.
\nAnswer:
\n\"AP
\nRadius of the cap (r) = 7 cm
\nHeight of the cap (h) = 24 cm
\nSlant height of the cap (l) = \\(\\sqrt{r^{2}+h^{2}}\\)
\n= \\(\\sqrt{7^{2}+24^{2}}\\)
\n= \\(\\sqrt{49+576}\\)
\n= \u221a625
\n= 25
\n\u2234 l = 25 cm.
\nLateral surface area of the cap = Cone = \u03c0rl
\nL.S.A. = \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 25 = 550 cm2<\/sup>.
\n\u2234 Area of the sheet required for 10 caps = 10 x 550 = 5500 cm2<\/sup>.<\/p>\n

\"AP<\/p>\n

Question 2.
\nA sports company was ordered to prepare 100 paper cylinders without caps for shuttle cocks. The required dimensions of the cylinder are 35 cm length \/ height and its radius is 7 cm. Find the required area of thin paper sheet needed to make 100 cylinders.
\nAnswer:
\n\"AP
\nRadius of the cylinder, r = 7 cm
\nHeight of the cylinder, h = 35 cm
\nT.S.A. of the cylinder with lids at both ends = 2\u03c0r(r+h)
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 (7 + 35)
\n= 2 \u00d7 \\(\\frac{22}{7}\\) \u00d7 7 \u00d7 42 = 1848 cm2<\/sup>.
\nArea of thin paper required for 100 cylinders = 100 \u00d7 1848
\n= 184800 cm2<\/sup>
\n= \\(\\frac{184800}{100 \\times 100}\\) m2<\/sup>
\n= 18.48 m2<\/sup>.<\/p>\n

Question 3.
\nFind the volume of right circular cone with radius 6 cm. and height 7 cm.
\nAnswer:
\n\"AP
\nBase radius of the cone (r) = 6 cm.
\nHeight of the cone (h) = 7 cm
\nVolume of the cone = \\(\\frac{1}{3}\\)\u03c0r2<\/sup>h
\n= \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 6 \u00d7 6 \u00d7 7
\n= 264 c.c. (Cubic centimeters)
\n\u2234 Volume of the right circular cone = 264 c.c.<\/p>\n

\"AP<\/p>\n

Question 4.
\nThe lateral surface area of a cylinder is equal to the curved surface area of a cone. If their base be the same, find the ratio of the height of the cylinder to slant height of the cone.
\nAnswer:
\nBase of cylinder and cone be the same.
\n\"AP
\nCSA \/ LSA of cylinder = 2\u03c0rh
\nCSA of cone = \u03c0rl
\nThe lateral surface area of a cylinder is equal to the curved surface area of cone.
\n\u2234 2\u03c0rh = \u03c0rl
\n\u21d2 \\(\\frac{h}{l}=\\frac{\\pi r}{2 \\pi r}\\)
\n\u21d2 \\(\\frac{h}{l}\\) = \\(\\frac{1}{2}\\)
\n\u2234 h : l = 1 : 2<\/p>\n

Question 5.
\nA self help group wants to manufacture joker’s caps (conical caps) of 3 cm radius and 4 cm height. If the available colour paper sheet is 1000 cm2<\/sup>, then how many caps can be manufactured from that paper sheet?
\nAnswer:
\n\"AP
\nRadius of the cap (conical cap) (r) = 3 cm
\nHeight of the cap (h) = 4 cm
\nSlant height l = \\(\\sqrt{r^{2}+h^{2}}\\)
\n(by Pythagoras theorem)
\n= \\(\\sqrt{3^{2}+4^{2}}\\)
\n= \\(\\sqrt{9+16}\\)
\n= \u221a25
\n= 5 cm
\nC.S.A. of the cap = \u03c0rl
\n= \\(\\frac{22}{7}\\) \u00d7 3 \u00d7 5
\n\u2243 47.14 cm2<\/sup>
\nNumber of caps that can be made out of 1000 cm2<\/sup> = \\(\\frac{1000}{47.14}\\) \u2243 21.27
\n\u2234 Number of caps = 21.<\/p>\n

\"AP<\/p>\n

Question 6.
\nA cylinder and cone have bases of equal radii and are of equal heights. Show that their volumes are in the ratio of 3 : 1.
\nAnswer:
\nGiven dimensions are:
\nCone:
\nRadius = r
\nHeight = h
\nVolume (V) = \\(\\frac{1}{3}\\)\u03c0r2<\/sup>h<\/p>\n

Cylinder:
\nRadius = r
\nHeight = h
\nVolume (V) = \u03c0r2<\/sup>h<\/p>\n

Ratio of volumes of cylinder and cone = \u03c0r2<\/sup>h : \\(\\frac{1}{3}\\)\u03c0r2<\/sup>h
\n= 1 : \\(\\frac{1}{3}\\)
\n= 3 : 1
\nHence, their volumes are in the ratio = 3 : 1.<\/p>\n

Question 7.
\nA solid iron rod has cylindrical shape. Its height is 11 cm. and base diameter is 7 cm. Then find the total volume of 50 rods?
\nAnswer:
\n\"AP
\nDiameter of the cylinder (d) = 7 cm
\nRadius of the base (r) = \\(\\frac{7}{2}\\) = 3.5 cm
\nHeight of the cylinder (h) = 11 cm
\nVolume of the cylinder V = \u03c0r2<\/sup>h
\n= \\(\\frac{22}{7}\\) \u00d7 3.5 \u00d7 3.5 \u00d7 11 = 423.5 cm3<\/sup>
\n\u2234 Total volume of 50 rods = 50 \u00d7 423.5 cm3<\/sup> = 21175 cm3<\/sup>.<\/p>\n

\"AP<\/p>\n

Question 8.
\nA heap of rice is in the form of a cone of diameter 12 m. and height 8 m. Find its volume? How much canvas cloth is required to cover the heap? (Use \u03c0 = 3.14)
\nAnswer:
\n\"AP
\nDiameter of the heap (conical) (d) = 12 cm
\n\u2234 Radius = \\(\\frac{d}{2}\\) = \\(\\frac{12}{2}\\) = 6 cm
\nHeight of the cone (h) = 8 m
\nVolume of the cone, V = \\(\\frac{1}{3}\\)\u03c0r2<\/sup>h
\n= \\(\\frac{1}{3}\\) \u00d7 \\(\\frac{22}{7}\\) \u00d7 6 \u00d7 6 \u00d7 8
\n= 301.71 m3<\/sup>.<\/p>\n

Question 9.
\nThe curved surface area of a cone is 4070 cm2<\/sup> and its diameter is 70 cm. What is its slant height?
\nAnswer:
\nC.S.A. of a cone = \u03c0rl = 4070 cm2<\/sup>
\nDiameter of the cone (d) = 70 cm
\nRadius of the cone = r = \\(\\frac{d}{2}\\) = \\(\\frac{70}{2}\\) = 35 cm
\nLet its slant height be ‘l’.
\nBy problem,
\n\u03c0rl = 4070 cm2<\/sup>
\n\\(\\frac{22}{7}\\) \u00d7 35 \u00d7 l = 4070
\n110 l = 4070
\nl = \\(\\frac{4070}{110}\\) = 37 cm
\n\u2234 Its slant height = 37 cm.<\/p>\n","protected":false},"excerpt":{"rendered":"

AP State Board Syllabus\u00a0AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Ex 10.1 Textbook Questions and Answers. AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.1 10th Class Maths 10th Lesson Mensuration Ex 10.1 Textbook Questions and Answers Question 1. A joker\u2019s cap is in the form of right … Read more<\/a><\/p>\n","protected":false},"author":5,"featured_media":35214,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[2],"tags":[],"jetpack_featured_media_url":"https:\/\/apboardsolutions.com\/wp-content\/uploads\/2023\/02\/Chapter-10-Mensuration-Ex-10.1.png","_links":{"self":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/3432"}],"collection":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/comments?post=3432"}],"version-history":[{"count":1,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/3432\/revisions"}],"predecessor-version":[{"id":34128,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/posts\/3432\/revisions\/34128"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media\/35214"}],"wp:attachment":[{"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/media?parent=3432"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/categories?post=3432"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/apboardsolutions.com\/wp-json\/wp\/v2\/tags?post=3432"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}