# Ex 12.2, 3 - Chapter 12 Class 9 Herons Formula

Last updated at Sept. 24, 2018 by Teachoo

Last updated at Sept. 24, 2018 by Teachoo

Transcript

Ex 12.2, 3 Radha made a picture of an aeroplane with coloured paper as shown in Fig. Find the total area of the paper used. Area of I Area of ∆ABC area of triangle = 𝑆(𝑆−𝑎)(𝑆−𝑏)(𝑆−𝑐) S → semi − perimeter a, b, c → sides of triangle Here a = 1 cm, b = 5 cm, c = 5 cm S = 𝑎 + 𝑏 + 𝑐2 = 5 + 5 + 12 = 112 cm Area of ∆ABC = 𝑆(𝑆−𝑎)(𝑆−𝑏)(𝑆−𝑐) = 112 112−1 112−5 112−5 = 112 92 12 12 = 11×9× 124 = 3 × 122 11 = 34 11 = 34 (3.31) = 2.48 cm2 area of II (BCQP) It is a triangle with length = 6.5 cm Breadth = 1 cm Area of rectangle = Length × Breadth = 6.5 × 1 = 6.5 cm2 Area of III (PQRS) Draw PM ⊥ SR and QN ⊥ SR Hence we have the following Area of III = Area of I + Area of II + Area of III I and III are right angled triangles. area (I) +area (II) = 2 area (I) = 2 × 12× 0.5 × 1 = 0.5 cm2 (II) is a rectangle ∴ Area (II) = Length × Breadth = 1 × 1 = 1 cm2 Area of III = Area of I + Area of II + Area of III = 0.5 + 1 = 1.5 cm2 Area of (IV) and (V) (IV) and (V) are similar right angled triangles. area of (IV) = area of (V) = 12 × Base × Height = 12 × 1.5 × 6 = 3 × 1.5 = 4 .5 cm2 Thus, Area = Area (I) + Area (II) + Area (III) + Area (IV) + Area (V) = 2.48 + 6.5 + 1.5 + 4.5 + 4.5 = 2.48 + 8.0 + 9.0 = 17.0 + 2.48 = 19.48 cm2 = 19.4 cm𝟐 (approx)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.