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Ex 12.2, 7 - A kite in the shape of a square with a diagonal - Finding area of quadrilateral

Ex 12.2, 7 - Chapter 12 Class 9 Herons Formula - Part 2
Ex 12.2, 7 - Chapter 12 Class 9 Herons Formula - Part 3 Ex 12.2, 7 - Chapter 12 Class 9 Herons Formula - Part 4 Ex 12.2, 7 - Chapter 12 Class 9 Herons Formula - Part 5

 


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Ex 12.2, 7 A kite in the shape of a square with a diagonal 32 cm and an isosceles triangles of base 8 cm and sides 6 cm each is to be made of three different shades as shown in the given figure. How much paper of each shade has been used in it? Given ABCD is a square AB = BC = CD = DA Hence, AB = DB = CD = AD = a Now, there are 3 triangles, Δ ABD, Δ BCD, Δ DCE In Δ ABD AD = a AB = a BD = 32cm Also, ∠ BAD = 90° Hence, Δ BAD is right triangle Using Pythagoras Theorem (Hypotenuse)2 = (Height)2 + (Base)2 BD2 = AD2 + AB2 (32)2 = a2 + a2 (32)2 = 2a2 a2 = 〖32〗^2/2 a = √(〖32〗^2/2) Hence, Side of square = a = √(〖32〗^2/2) Area of square ABCD = Side × Side = a2 = (√(〖32〗^2/2))^2 = 〖32〗^2/2 = 1024/2 = 512 cm2 Area of Δ ABD = Area of Δ BCD ∴ Area of Δ ABD = Area of Δ BCD = 1/2 × Area of square ABCD Area of shade I = Area of shade II = 1/2 × 512 Area of shade I = Area of shade II = 256 cm2 Area Δ CEF Area of triangle = √(s(s−a)(s−b)(s −c)) Here, s is the semi-perimeter, and a, b, c are the sides of the triangle Let a = 6 cm, b = 8cm, c = 6 cm s = (𝑎 + 𝑏 + 𝑐)/2 Area of ∆CEF = √(10(10 −6)(10 −6)(10 −8)) = √(10×4×4×2) cm2 = √((5×2)×(4×4)×2) cm2 = √((2×2)×(4×4)×5) cm2 = √((22)×(42)×5) cm2 = √42 ×√22×√5 cm2 = (4×2×√5 " " ) cm2 = 8 √5 cm2 = (8 × 2.24) cm2 = 17.92 cm2 Thus, Area Δ CEF = 17.92 cm2 So, Area of shade III = 17.92 cm2

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.