{"id":25971,"date":"2023-03-24T15:51:53","date_gmt":"2023-03-24T10:21:53","guid":{"rendered":"https:\/\/ncertsolutions.guru\/?p=25971"},"modified":"2023-03-24T17:38:16","modified_gmt":"2023-03-24T12:08:16","slug":"ncert-solutions-for-class-9-maths-chapter-12-ex-12-2","status":"publish","type":"post","link":"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-12-ex-12-2\/","title":{"rendered":"NCERT Solutions for Class 9 Maths Chapter 12 Heron\u2019s Formula Ex 12.2"},"content":{"rendered":"

These NCERT Solutions for Class 9 Maths<\/a> Chapter 12 Heron\u2019s Formula Ex 12.2 Questions and Answers are prepared by our highly skilled subject experts.<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 12 Heron\u2019s Formula Exercise 12.2<\/h2>\n

Question 1.
\nA park in the shape of quadrilateral ABCD, has \u2220C = 90\u00b0, AB = 9 m, BC = 12 m, CD = 5 m and AD = 8 m. How much area does it occupy?
\nSolution:
\nIn ABCD,
\nCD = 5 cm, BC = 12 and \u2220C = 90\u00b0
\n\"NCERT
\n\u2234 By pythagoras theorem we know that
\nh2<\/sup> = P2<\/sup> + b2<\/sup>
\n\u21d2 BD2<\/sup> = CD2<\/sup> + BC2<\/sup>
\n\u21d2 BD2<\/sup> = 52<\/sup> + 122<\/sup>
\n\u21d2 BD2<\/sup>\u00a0= 169
\n\u21d2 BD = \u221a169 = 13 cm.
\nArea of triangle BCD = \\(\\frac {1}{2}\\) \u00d7 12 \u00d7 5 = 30 cm.
\nIn \u2206ABD,
\nAD = 8 m, AB = 9 m, and BD = 13 cm
\n\u2234 s = \\(\\frac{8+9+13}{2}\\) = 15
\n\u2234 By Heron’s formula we know that
\nArea of triangle
\n\"NCERT
\n= 35.5 m2<\/sup> (approx)
\nNow,
\nArea perpendicular ABCD = Ar \u0394BCD + Ar \u0394ABD
\n= 30 + 35.5
\n= 65.5 m2<\/sup> (approx)
\nTherefore, quadrilateral ABCD occupy 65.5 m2<\/sup> area.<\/p>\n

\"NCERT<\/p>\n

Question 2.
\nFind the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm.
\nSolution:
\nIn \u0394ABC,
\nAB = 3 cm, BC = 4 cm and AC = 5 cm
\n\u2234 s = \\(\\frac{3+4+5}{2}\\) = 6
\n\u2234 By Heron’s formula we know that
\nArea of \u0394ABC
\n\"NCERT
\nAgain in \u0394ACD,
\nAC = 5 cm, AD = 5 cm, and CD = 4 cm
\n\u2234 s = \\(\\frac{5+5+4}{2}\\) = 7
\n\u2234 By Heron’s formula we know that,
\n\"NCERT
\n= 9.2 cm2<\/sup> (approx)
\n\u2234 Area of quadrilateral ABCD = Ar \u0394ABC + Ar \u0394ACD
\n= 6 + 9.2
\n= 15.2 cm2<\/sup> (approx)<\/p>\n

\"NCERT<\/p>\n

Question 3.
\nRadha made a picture of an aeroplane with coloured paper .is shown in Fig. 12.15. Find the total area of the paper used.
\n\"NCERT
\nSolution:
\nIn part I, sides of triangle are 5 cm, 5 cm and 1 cm
\n\u2234 s = \\(\\frac{5+5+1}{2}\\) = 5.5
\n\u2234 By Heron’s formula
\n\"NCERT
\nThe area of the IInd part which is in the form of the rectangle is
\n6.5 \u00d7 1 = 6.5 cm2<\/sup> (Area of rectangle = l \u00d7 b)
\nArea of IIIrd part = 3 \u00d7 \\(\\frac{\\sqrt{3}}{4}\\) cm = 1.3 cm2<\/sup> (approx)
\nArea of IVth part, which is in the form of triangle is = \\(\\frac {1}{2}\\) \u00d7 15 \u00d7 6 = 4.5 cm2<\/sup>
\nArea of Vth part, which is also in Hie form of triangle is = \\(\\frac {1}{2}\\) \u00d7 15 \u00d7 6 = 4.5 cm2<\/sup>
\nTherefore, total area of the paper used = 2.5 + 6.5 +1.3 + 4.5 + 4.5 = 19.3 cm2<\/sup> (approx)<\/p>\n

Question 4.
\nA triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm, and 30 cm, and the parallelogram stands on the base 28 cm, find the height of the parallelogram.
\nSolution:
\nSides of triangle are 26 cm, 28 cm and 30 cm.
\n\u2234 s = \\(\\frac{26+28+30}{2}\\) = 42
\n\u2234 By Heron’s formula we know that
\n\"NCERT
\nAccording to question,
\nArea of parallelogram = Area of triangle
\nBase \u00d7 altitude = 336
\n28 \u00d7 altitude = 336
\n\u2234 Altitude = \\(\\frac{336}{28}\\) = 12
\n\u2234 Height of the parallelogram stands on the base 28 cm is 12 cm.<\/p>\n

\"NCERT<\/p>\n

Question 5.
\nA rhombus-shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much rea of grass field will each cow be getting.
\n\"NCERT
\nSolution:
\nIn \u2206ABD
\nAB = 30 m, AD = 30 m and BD = 48 m.
\n\u2234 s = \\(\\frac{30+30+48}{2}\\) = 54
\n\u2234 By Heron’s formula we know that
\n\"NCERT
\n\u2234 Area of rhombus ABCD = 2 \u00d7 432 = 864 m2<\/sup>
\nTherefore, area of grass getting by each cow = \\(\\frac{864}{18}\\) = 48 m2<\/sup><\/p>\n

Question 6.
\nAn umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see Fig. 12.16), each piece measuring 20 cm, 50 cm, and 50 cm. How much cloth of each colour is required for the umbrella?
\n\"NCERT
\nSolution:
\nFor one triangular piece,
\n\u2234 s = \\(\\frac{20+50+50}{2}\\) = 60
\n\u2234 By Heron’s formula
\nArea of one triangular piece
\n\"NCERT
\nTherefore,
\nArea of five triangular piece = 5 \u00d7 200\u221a6 = 1000\u221a6 cm2<\/sup>
\nHence, clothes of each colour are required for the umbrella is 1000\u221a6 cm2<\/sup>.<\/p>\n

\"NCERT<\/p>\n

Question 7.
\nA kite in the shape of a square with a diagonal 32 cm and art isosceles triangle of base S cm and sides 6 cm each is lobe made of three diffident shades as shown in Fig. 12.17. How much paper of each shade has been used in it?
\n\"NCERT
\nSolution:
\nWe have given that, the shape of a kite is square and length of its diagonal is 32 cm.
\nArea of 1 triangular region
\n= \\(\\frac {1}{2}\\) \u00d7 base \u00d7 height
\n= \\(\\frac {1}{2}\\) \u00d7 32 \u00d7 16
\n= 256 cm2<\/sup>
\n(\u2235 We know that diagonals of square bisects each other at 90\u00b0)
\nNow, Area of IInd triangular region
\n= \\(\\frac {1}{2}\\) \u00d7 base \u00d7 height
\n= \\(\\frac {1}{2}\\) \u00d7 32 \u00d7 16
\n= 256 cm2<\/sup>
\nAgain, we have given that IIIrd part of the given kite is in the shape of an isosceles triangle whose equal sides are 6 cm and the base is 8 cm.
\ns = \\(\\frac{a+b+c}{2}\\)
\n= \\(\\frac{6+6+8}{2}\\)
\n= 10
\n\u2234 By Heron’s formula,
\nArea of IIIrd triangular region
\n\"NCERT<\/p>\n

\"NCERT<\/p>\n

Question 8.
\nA floral design on a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm, and 35 cm (see Fig. 12.18). Find the cost of polishing the tiles at the rate of 50 p per cm2<\/sup>.
\n\"NCERT
\nSolution:
\nThe sides of one triangular tile are 9 cm, 28 cm, and 35 cm
\ns = \\(\\frac{9+28+35}{2}\\) = 36 cm
\nBy Heron’s formula, we know that
\nArea of one triangular tile
\n\"NCERT
\nTherefore, area of 16 triangular tiles = 88.18 \u00d7 16 = 1410.88 cm2<\/sup>
\nHence, the cost of polishing at the rate of 50 p per cm2<\/sup> is 1410.88 \u00d7 0.50 = Rs. 705.44 (approx)<\/p>\n

\"NCERT<\/p>\n

Question 9.
\nA field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field.
\n\"NCERT
\nSolution:
\nDraw a line through A and parallel to BC cut CD at E. Again draw AM \u22a5 CD.
\nNow, sides of \u2206ADE are 13 m, 15 m and 14 m
\ns = \\(\\frac{13+14+15}{2}=\\frac{42}{2}\\) = 21 m
\nBy Heron’s formula, we know that
\n\"NCERT
\nBut, Area of \u2206ADE = \\(\\frac {1}{2}\\) \u00d7 DE \u00d7 AM
\nor, 84= \\(\\frac {1}{2}\\) \u00d7 15 \u00d7 AM
\nor, AM = \\(\\frac{84 \\times 2}{15}\\) = 11.2 m
\nWe know that
\nArea of trapezium
\n= \\(\\frac {1}{2}\\) \u00d7 (25 + 10) \u00d7 11.2
\n= \\(\\frac {1}{2}\\) \u00d7 35 \u00d7 11.2
\n= 196 m2<\/sup><\/p>\n","protected":false},"excerpt":{"rendered":"

These NCERT Solutions for Class 9 Maths Chapter 12 Heron\u2019s Formula Ex 12.2 Questions and Answers are prepared by our highly skilled subject experts. NCERT Solutions for Class 9 Maths Chapter 12 Heron\u2019s Formula Exercise 12.2 Question 1. A park in the shape of quadrilateral ABCD, has \u2220C = 90\u00b0, AB = 9 m, BC …<\/p>\n

NCERT Solutions for Class 9 Maths Chapter 12 Heron\u2019s Formula Ex 12.2<\/span> Read More »<\/a><\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"spay_email":""},"categories":[6],"tags":[],"yoast_head":"\nNCERT Solutions for Class 9 Maths Chapter 12 Heron\u2019s Formula Ex 12.2 - NCERT Solutions<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/ncertsolutions.guru\/ncert-solutions-for-class-9-maths-chapter-12-ex-12-2\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"NCERT Solutions for Class 9 Maths Chapter 12 Heron\u2019s Formula Ex 12.2 - NCERT Solutions\" \/>\n<meta property=\"og:description\" content=\"These NCERT Solutions for Class 9 Maths Chapter 12 Heron\u2019s Formula Ex 12.2 Questions and Answers are prepared by our highly skilled subject experts. 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NCERT Solutions for Class 9 Maths Chapter 12 Heron\u2019s Formula Exercise 12.2 Question 1. 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