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 + 𝑐﷮2﷯ = 5 + 5 + 1﷮2﷯ = 11﷮2﷯ cm Area of ∆ABC = ﷮𝑆(𝑆−𝑎)(𝑆−𝑏)(𝑆−𝑐)﷯ = ﷮ 11﷮2﷯ 11﷮2﷯−1﷯ 11﷮2﷯−5﷯ 11﷮2﷯−5﷯﷯ = ﷮ 11﷮2﷯ 9﷮2﷯﷯ 1﷮2﷯﷯ 1﷮2﷯﷯﷯ = ﷮11×9× 1﷮2﷯﷯﷮4﷯﷯ = 3 × 1﷮2﷯﷯﷮2﷯ ﷮11﷯ = 3﷮4﷯ ﷮11﷯ = 3﷮4﷯ (3.31) = 2.48 cm﷮2﷯ 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 cm﷮2﷯ 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 × 1﷮2﷯× 0.5 × 1 = 0.5 cm﷮2﷯ (II) is a rectangle ∴ Area (II) = Length × Breadth = 1 × 1 = 1 cm﷮2﷯ Area of III = Area of I + Area of II + Area of III = 0.5 + 1 = 1.5 cm﷮2﷯ Area of (IV) and (V) (IV) and (V) are similar right angled triangles. area of (IV) = area of (V) = 1﷮2﷯ × Base × Height = 1﷮2﷯ × 1.5 × 6 = 3 × 1.5 = 4 .5 cm﷮2﷯ 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 cm﷮2﷯ = 19.4 cm﷮𝟐﷯ (approx)