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“Origami” is the art of paper folding, which is often associated with Japanese culture. Gurmeet is trying to learn origami using proper cutting and folding technique. A square base is sometimes referred to as a “preliminary” base or preliminary fold.Here is a 20 cm × 20 cm square. Gurmeet wants to first cut the squares of integral length from the corners and by folding the flaps along the sides.(a) How many squares of dimensions 2cm × 2cm can Gurmeet make? i.5 ii.7 iii.10 iv.20 (b) What is the max radius of hemisphere which can be cut out from the square ? i.5cm ii.10cm iii.12cm iv.7cm (c) What is the area of remaining left out paper if cone of radius 1 cm and height 10 cm is cut out from the square? i.365.303 cm2 ii.371.583 cm2 iii.300 cm2iv.360 cm2 (d) How many folds are possible if final dimension of fold has to be 4cm × 4cm? i.25 ii.20 iii.15 iv.30 (e) How many triangles can be cut out, each of area 80 cm2? i.6 ii.3 iii.8 iv.5 |
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Answer» Dimension of square (given paper) is 20cm×20cm. Therefore, Area of given paper is 400 cm2 (a) Option : (iii) Gurmeet want to make squares of 2cm × 2cm from the given paper (square of 20 cm). Therefore, The area of required square is 4cm2 Hence, Total number of required squares = \(\frac{area\, given\,square}{ area\,of\,required\,squre}\) = \(\frac{400}{40}\) = 10. Hence, 10 squares of dimensions 2cm × 2cm can Gurmeet make. Hence, Option (iii) is correct. (b) Option : (ii) The maximum diameter of hemisphere which can be cut out from the square = Side length of square = 20 cm. Hence, Maximum radius of hemisphere = \(\frac{20}{2}\) = 10cm. Hence, Option (ii) is correct. (c) Option : (ii) Radius of cone is r = 1 cm And height of the cone is h = 10cm. The slant height of the cone is l = \(\sqrt{r^2+h^2}\) = \(\sqrt{1^2+10^2}\) = \(\sqrt{101}\) cm. Total surface area of the cone = πr(r+l) = 3.14 ×1(1+√101) = 3.14 × (1+10.05) = 3.14(11.05) = 34.697 cm2. Area of base of the cone = πr2 = 3.14 × 12 = 3.14 cm2. Remaining area = Area of square –Total surface area of cone +2×Area of base of the cone = 400 – 34.697 + 2×3.14 cm2 = 365.303 + 6.28 = 371.583 cm2. Hence, Option (ii) is correct. (d) Option : (i) Area of square of side length 4 cm = 42 = 16 cm2 Hence, Total square of 4 cm = \(\frac{Area\,of\,square\,of side\,length\,20\,cm}{Area\,of\,square\,of\,side\,length\,4\,cm}\) = \(\frac{400}{16}\) = 25. Hence, Option (i) is correct. (e) Option : (iv) Given, Area of triangle = 80 cm2. Total number of such triangle = \(\frac{Area\,of\,given\,paper}{ Area\,of\,one\,triangle}\) = \(\frac{400}{80}\) = 5. Hence, 5 triangles can be cut out such that each triangle is having area 80 cm2. Hence, Option (iv) is correct. |
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