{"id":2088,"date":"2020-09-08T09:08:47","date_gmt":"2020-09-08T03:38:47","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=2088"},"modified":"2021-07-02T15:03:44","modified_gmt":"2021-07-02T09:33:44","slug":"kseeb-solutions-class-9-maths-chapter-8-ex-8-2","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/kseeb-solutions-class-9-maths-chapter-8-ex-8-2\/","title":{"rendered":"KSEEB Solutions for Class 9 Maths Chapter 8 Heron\u2019s Formula Ex 8.2"},"content":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 8 Heron\u2019s Formula Ex 8.2<\/strong>\u00a0are part of KSEEB Solutions for Class 9 Maths<\/a>. Here we have given Karnataka Board Class 9 Maths Chapter 8 Heron\u2019s Formula Exercise 8.2.<\/p>\n

Karnataka Board Class 9 Maths Chapter 8 Heron\u2019s Formula Ex 8.2<\/h2>\n

Question 1.
\nA park, in the shape of a 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:
\nBD is joined.
\nIn Right angled ABCD, \u2220C = 90\u00b0
\n\"KSEEB
\nAs per Pythagoras Theorem,
\nBD2<\/sup> = BC2<\/sup> + CD2<\/sup>
\n= (12)2<\/sup> + (5)2<\/sup>
\n= 144 + 25
\nBD2<\/sup> = 169
\n\u2234 BD = 13m.
\n\"KSEEB
\n\"KSEEB
\nNow, in \u2206ABD,
\nLet a = 9m, b = 8m, c = 13m.
\n\"KSEEB
\n\u2234 Area of quadrilateral ABCD,
\n= Area(\u2206ABD) + Area (\u2206BCD)
\n= 35.46 + 30
\n= 65.46 sq.m.
\nArea of quadrilateral ABCD = 65.46 sq.m.<\/p>\n

Question 2.
\nFind the area of a quadrilateral ABCD which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm, and AC = 5 cm.
\n\"KSEEB
\nSolution:
\nIn quadrilateral ABCD, AC is the diagonal.
\nIn \u2206ABC, a = 5cm, b = 3 cm, c = 4 cm.
\n\u2234 Perimeter = a + b + c = 5 + 3 + 4 = 12 cm.
\n\"KSEEB
\n\"KSEEB
\n\u2234 Area(\u2206ABC) = 6 sq. cm.
\nIn \u2206ADC, a1<\/sub> = 5cm, b1<\/sub> = 5 cm, c1<\/sub> = 4cm.
\n\u2234 Perimeter = a1<\/sub> + b1<\/sub> + c1<\/sub>
\n= 5 + 5 + 4 = 14 cm.
\n\"KSEEB
\nArea of \u2206ADC = 9.16 sq.cm.
\n\u2234 Area of quadrilateral aBCD,
\n= Area(\u2206ABC) + Area(\u2206ADC)
\n= 6 + 9.16
\n= 15.16 sq. cm.<\/p>\n

Question 3.
\nRadha made a picture of an aeroplane with coloured paper as shows in Fig. Find the total area of the paper used.
\n\"KSEEB
\nSolution:
\n(I) \u2206ABC is an isosceles triangle.
\nAB = AC = 5 cm, and BC = 1 cm.
\n\u2234 a = 5 cm, b = 5 cm, c = 1 cm.
\n\u2234 Perimeter = a + b + c = 5 + 5 + 1 = 11 cm.
\n\"KSEEB
\n\u2234 Area(\u2206ABC) = 24.85 sq. cm.<\/p>\n

(II) PBCS is a rectangle.
\nl = 6.5 cm, b = 1 cm.
\nArea of rectangle PBCS = l x\u00d7 b
\n= 6.5 \u00d7 1 = 6.5 sq.cm.<\/p>\n

(III) PQRS is a trapezium PT \u22a5 QR
\nIn Right angle \u2206PTV,
\n\"KSEEB<\/p>\n

(IV) Area of ADBE = Area of \u2206FCG<\/p>\n

(V)
\n\"KSEEB
\n\u2234 The total area of the paper used to make an aeroplane
\n= 2.485 + 6.5 + 1.29 + 4.5 + 4.5
\n= 19.275 sq. cm.<\/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.
\n\"KSEEB
\nSolution:
\nPerimeter of \u2206ABC = 26 + 28 + 30 = 84 cm.
\n\"KSEEB<\/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 area of grass field will each cow be getting ?
\n\"KSEEB
\nSolution:
\nABCD is a rhombus.
\nArea(\u2206ABD) = Area(\u2206BDC).
\nIn \u2206ABD, a = 30 m, b = 30 m, c = 48 m.
\n\u2234 Perimeter= a + b + c = 30 + 30 + 48 = 108 m
\n\"KSEEB
\nBut, Area of whole field = 2 \u00d7 area of \u2206ABD
\n= 2 \u00d7 432
\n= 864 sq.m.
\n\u2234 Are required for growing grass for 18 cows is 864 sq.m.
\n\u2234 Area required for growing grass for 1 cow ………. ??
\n\\(\\frac{864}{18}\\) = 48 sq.m.<\/p>\n

Question 6.
\nAn umbrella is made by stitching 10 triangular pieces of cloth of two different colours, each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?
\n\"KSEEB
\nSolution:
\nEach sides of triangular shaped cloth
\nLet a = 50 cm, b = 50 cm, c = 20 cm.
\n\u2234 Perimeter = a + b + c
\n= 50 + 50 + 20 = 120 cm.
\n\"KSEEB
\n\"KSEEB
\n\u2234 Total Area of Umbrella mad up with 10 triangular shaped cloth is
\n\"KSEEB<\/p>\n

Question 7.
\nA kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made of three different shades as shown in Fig. How much paper of each shade has been used in it ?
\n\"KSEEB
\nSolution:
\nKite is in the shape of square.
\n\"KSEEB
\n\"KSEEB
\n\u2234 Total area = I + II + III
\n= 256 + 256 + 17.92
\n= 549.92 sq.cm.<\/p>\n

Question 8.
\nA floral design of a floor is made up of 16 tiles which are triangular, the sides of the triangle being 9 cm, 28 cm and 35 cm. Find the cost of polishing the tiles at the rate of 50p per cm2<\/sup>.
\n\"KSEEB
\nSolution:
\nIn the given figure there are 16 triangles.
\nEach sides of triangle are
\na = 28 cm, b = 9 cm, c = 35 cm.
\nPerimeter of triangle = a + b + c
\n= 28 + 9 + 35 = 72 cm.
\n\"KSEEB
\n\u2234 Area of 16 triangles = 16 \u00d7 88.2 = 1411.2 sq.cm.
\nPolish expenditure for 1 sq.cm. = 50 p = Re. 0.5.
\n\u2234 Polish expenditure for 1411.2sq. cm. =?
\n= 1411.2 \u00d7 0.5
\n= Rs 705.6<\/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\"KSEEB
\nSolution:
\nABCD is a trapezium.
\nAB || DC.
\nDraw AE || BC and AF \u22a5 DC.
\n\u2234 ABCE is a parallelogram.
\nDE = DC – EC = 25 – 10
\nDE = 15 cm.
\nSides of \u2206ADE
\na = 14 cm, b = 13 cm. c = 15 cm.
\n\u2234 Perimeter= a + b + c
\n= 14 + 13 + 15 = 42 cm.
\n\u2234 s = \\(\\frac{a+b+c}{2}=\\frac{42}{2}=21 \\mathrm{cm}\\)
\n\u2234 Area of \u2206ADE, A
\n\"KSEEB<\/p>\n

(ii) Area of quadrilateral ABCE = base \u00d7 height
\n= EC \u00d7 AF
\n= 10 \u00d7 11.2
\n= 112.0
\n= 112 sq.cm.
\n\u2234 Complete area of trapezium ABCD, A
\nA = Area of \u2206ADE + Area of quadrilateral ABCE
\n= 84 + 112
\n= 196 sq.cm.<\/p>\n

We hope the KSEEB Solutions for Class 9 Maths Chapter 8 Heron\u2019s Formula Ex 8.2 help you. If you have any query regarding Karnataka Board Class 9 Maths Chapter 8 Heron\u2019s Formula Exercise 8.2, drop a comment below and we will get back to you at the earliest.<\/p>\n","protected":false},"excerpt":{"rendered":"

KSEEB Solutions for Class 9 Maths Chapter 8 Heron\u2019s Formula Ex 8.2\u00a0are part of KSEEB Solutions for Class 9 Maths. Here we have given Karnataka Board Class 9 Maths Chapter 8 Heron\u2019s Formula Exercise 8.2. Karnataka Board Class 9 Maths Chapter 8 Heron\u2019s Formula Ex 8.2 Question 1. A park, in the shape of a …<\/p>\n","protected":false},"author":6,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2088"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/6"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=2088"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/2088\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=2088"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=2088"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=2088"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}