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Chapter 17 Mensuration of Pyramid 344 Applied Math Mensuration of Pyramid Chapter 17 Mensuration of Pyramid 17.1 Pyramid: A pyramid is a solid whose base is a plane polygon and sides are triangles that meet in a common vertex. The triangular sides are called lateral faces. The common vertex is also called Apex. A pyramid is named according to the shape of its base. If the base is a triangle, square, hexagon etc. the pyramid is called as a triangular pyramid, a square pyramid, a hexagonal pyramid etc. respectively. Altitude (or height): The altitude of a pyramid is the perpendicular distance from the vertex to the base. Axis: The axis of a pyramid is the distance from the vertex to the centre of the base. 17.2 Right or Regular Pyramid: A pyramid whose base is a regular polygon and congruent isosceles triangles as lateral faces. In a regular pyramid the axis is perpendicular to the base. Thus in a regular pyramid th axis and the altitude are identical. Slant Height: The slant height of a regular pyramid is the length of the median through the apex of any lateral face. In the Fig 17.1 OG is the slant height. It is denoted by . Fig. 17.1 Fig. 17.1 Lateral edge: It is the common side where the two, faces meet. In the OA is the lateral edge. 17.3 The surface area and Volume of a Regular Pyramid: If a is the side of the polygon base, h is the height and is the slant height of a regular pyramid, then (i) Lateral surface area = Sum of the triangular sides forming the pyramid, which are all equal in areas = n (area of one triangle of base a and slant height ) 345 Applied Math Mensuration of Pyramid 1 = n a 2 1 = (n a) x 2 1 = Perimeter of the base x slant height 2 (ii) Total surface area = Lateral surface area + area of the base (iii) Volume: The volume of a pyramid may be easily derived from the volume of a cube. By joining the centre O of the cube with all vertices, six equal pyramids are formed. The base of each pyramid is one of the faces of the cube. Hence the volume of each pyramid is one-sixth of the volume of the cube. The height of each pyramid is a h = 2 1 Volume of each pyramid = the volume of the cube 6 1 = a3 6 1a = a.2 32 1 = ah2 3 1 Volume of the pyramid = area of the base x height 3 Example 1: Find the volume, the lateral surface area and the total surface area of the square pyramid of perpendicular height 9.41cm and the length of the side of base 2.92cm. Solution: Here h = 9.41cm, a = 2.92cm 346 Applied Math Mensuration of Pyramid (i) Volume = area of the base x height = = 1 8.526 x 9.41 2 = 26.74 cu. cm (ii) For the lateral surface area we first calculate the slant height . In the right triangle OAB, OA = a h = 9.41 cm, AB = = 1.45 cm 2 By Pythagorean theorem, 2h 2 AB 2 = 85.55 + 2.13 2 = 90.68 = 9.52 Now lateral surface area = perimeter of the base x = 4a x = 2 x 2.92 x 9.52 = 55.60 sq. cm. (iii) Total surface area= Lateral surface area + Area of the base = 55.60 + (2.92)2 = 55.60 + 8.53 1 2 = 64.13 sq. cm.ah Example 2: 3 A right pyramid 10m high1 has a square base of which the diagonal3 is 10m. Find its slant surface. Solution: Here h = 10m AB = d = 10m d BC = = 5m 2 In the right triangle BCD, 347 Applied Math Mensuration of Pyramid CD = BD So BC2 = CD2 + BD2 25 = 2 CD2 5 Or CD = m 2 Side of the base 1 2 a = BE = 2 BD = 2 . = 52 Now, in the right triangle OCD, OD2 = OC2 + CD2 25 = 100 + = 112.50 2 = 10.6m The slant surface area = perimeter of the base x = 2 4a a = 2 10.6 2 = 150 sq. m. Example 3: The base of a right pyramid is a regular hexagon of side 4m and its slant surfaces are inclined to the horizontal at an angle of 30o. Find the volume. Solution: Here, a = 4m θ = B = 30o n a2 180 Area of the base = cot 4n 6 x 42 = cot 60o 4 1 = 24 13.86 sq. m 3 In the right triangle ABC, BC = = 2m Angle C = 60o 348 Applied Math Mensuration of Pyramid AB So, tan 60o = OR AB2 = AC2 – BC2 CB AB = 23m AB = Now, in the right triangle OAB h tan 30o = AB 1 = x 3 h = 2m Volume = area of the base x height = x 13.86 x 2 = 9.25 cu. m Example 4: The area of the base of a hexagonal pyramid is 54 3 sq. m. and the area of one of its face is 96sq. m. Find the volume of the pyramid. Solution: Here, area of the base = sq. m Area of one side face = sq. m Volume = area of the base x h = ( ) x h n a2 180 cot To find h, we have to4n find and AB. 1 Area of the hexagon = 3 6 x a2 54 3 cot 30o 4 6 54 3 x 3 a2 4 a2 = 36 a = 6m 349 Applied Math Mensuration of Pyramid 1 Area of one triangle, say, OCD = a 2 1 9 6 x 6 x 2 = 3 6 m In the right triangle ABC, AC = 6m, BC = 3m So AB2 = AC2 – BC2 = 36 – 9 = 27 AB = 33m Now, in the right triangle OAB OA2 = OB2 – AB2 h2 = 2 – AB2 = 54 – 27 = 27 h = Therefore, volume = (54 3) x 3 3 = 162 cu. m. Exercise 17(A) Q.1: Find the volume of a pyramid whose base is an equilateral triangle of side 3m and height 4m. Q.2: Find the volume of a right pyramid whose base is a regular hexagon each side of which is 10m and height 50m. Q.3: A regular hexagonal pyramid has the perimeter of its base 12cm and its altitude is 15cm. Find its volume. Q.4; A pyramid with a base which is an equilateral triangle each side of which is 3m and has a volume of 120 cu. m find its height. Q.5: A pyramid on a square base has every edge 100m long. Find the edge of a cube of equal volume. Q.6: The faces of a pyramid on a square base are equilateral triangles. If each side of the base is 110m. Find the volume and the whole surface of the pyramid. 3 Q.7: Find the whole surface of a pyramid whose base is an equilateral triangle of side 3m and its slant height is 6m. Q.8: The slant edge of a right regular hexagonal pyramid is 65 cm and the height is 56cm. Find the area of the base. Q.9: Find the slant surface of a right pyramid whose height is 65m and whose base is a regular hexagon of side 48 3 m. 350 Applied Math Mensuration of Pyramid Q.10: The sides of the base of a square pyramid are each 12.5cm and height of the pyramid is 8.5cm. Find its volume and lateral surface. Q.11: Find volume of a square pyramid whose every edge is 100cm long. Answers 17(A) Q1. cu.m Q2. 4330.127 cu. m Q3. 51.96 cu. m Q4. 92.376 m Q5. 61.77m1 Q6. 235.70 cu. m; 273.2 sq. m Q7. 30.897 sq. m Q8. 3772.2962 sq. cm Q9. 13968 3 sq. m Q10. 1328.125 cu. cm, 263.76 sq. cm Q11. 235702.26 cu.cm 17.4 Frustum of a Pyramid: 33 When a pyramid is cut through by a plane parallel to its base, the portion of the pyramid between the cutting plane and the base is called a frustum of the pyramid. Each of the side face of the frustum of the pyramid is a trapezium. Slant height: The distance between the mid points of the sides of base and top. It is denoted by . 17.5 Volume and surface area of Frustum of a Regular Pyramid: Let, A1 by the area of the base, and A2 be the area of the top, a is the side of the base and b is the side of the top, is the slant height and h is the height of the frustum of a pyramid, then (i) Volume of the frustum of a pyramid h = (A A A A ) 3 1 2 1 2 (ii) Lateral surface area = Sum of the areas of all the trapezium faces, which are equals = n (area of one trapezium, say, ABA’B’) a + b 1 = nx = (na + nb) x 2 2 = sum of the perimeters of the base and top x slant height (iii) Total surface area= Lateral surface area + area of the base and the top Example 5 : 351 Applied Math Mensuration of Pyramid A frustum of a pyramid has rectangular ends, the sides of the base being 20m and 32m. If the area of the top face is 700 sq. m. and the height of the frustum is 50m; find its volume. Solution: Here A1 = 20 x 32 = 640 sq. m , A2 = 700 sq. m h Volume = [A A A A ] 3 1 2 1 2 50 = [640 700 640 x 700] 3 50 50 = [1340 + 669.33] = (2009.33) 3 3 = 33488.80 cu.
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