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
1 (i) Volume = area of the base x height 3 1 = ah2 3
= 8.526 x 9.41 = 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 1 Now lateral surface area = perimeter of the base x 2 = 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 = 64.13 sq. cm. Example 2: A right pyramid 10m high has a square base of which the diagonal 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
a = BE = 2 BD = 2 .
= 52 Now, in the right triangle OCD, OD2 = OC2 + CD2 25 2 = 100 + = 112.50 2 = 10.6m 1 The slant surface area = perimeter of the base x 2 1 = 4a 2 = 2 10.6 = 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, a BC = = 2m 2 Angle C = 60o 348 Applied Math Mensuration of Pyramid
AB So, tan 60o = OR AB2 = AC2 – BC2 CB AB = 23m AB = 23 Now, in the right triangle OAB h tan 30o = AB 1 = x 3 h = 2m 1 Volume = area of the base x height 3 1 = x 13.86 x 2 3 = 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 To find h, we have to find and AB. n a2 180 Area of the hexagon = cot 4n 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 = 1 Therefore, volume = (54 3) x 3 3 = 162 cu. m. 3 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 10m. Find the volume and the whole surface of the pyramid. 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. 33cu.m Q2. 4330.127 cu. m Q3. 51.96 cu. m Q4. 92.376 m Q5. 61.77m Q6. 235.70 cu. m; 273.2 sq. m Q7. 30.897 sq. m Q8. 3772.296 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: 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 = (AAAA) 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 1 = sum of the perimeters of the base and top x slant height 2 (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. m. Example 6 : A square pyramid 12m high is cut 8m from the vertex to form a frustum of a pyramid with a volume of 190 cu. m. Find the side of the base of the frustum of a pyramid. Solution: Here, volume of frustum of a pyramid = 190 cu. m Height of pyramid = h = 12m Height of the frustum of a pyramid = h1 = 4m OC = 8m If ‘a’ and ‘b’ are the sides of base and top, then since the right triangles OAB and OCD’ are similar. AB OA So CD OC 1 a 12 2 1 b 8 2 a3 OR b2 2a OR b 3 Now, Volume of the frustum of a pyramid h = 1 [A A A A ] 3 1 2 1 2 352 Applied Math Mensuration of Pyramid
4 190 [a22 b ab] 3 4 4a22 2a 190 [a2 ] 3 9 3 4 19a2 190 [ ] 39 190 x 27 a2 67.5 76 a = 8.22m
Exercise 17(B)
Q.1 Find the volume and the total surface area of a frustum of a pyramid; the end being square of sides 8.6m and 4.8m respectively and the thickness of the frustum of a pyramid is 5m. Q.2 Find the lateral surface area and volume of frustum of a square pyramid. The sides of the base and top are 6m and 4m respectively and the slant height is 8m. Q.3 Find the net area of material required to make half dozen lamp shades each shaded as a hollow frustum of a square pyramid, having top and bottom sides of 10cm and 18cm respectively, and vertical height 16m. Q.4 The sides of the top and bottom ends of a frustum of a square pyramid are 6m and 15m respectively. Its height being 30m. it is capped at the top by a square pyramid 12m from the base to the apex. Find the number of cu. m in the frustum of a square pyramid and in the cape. Q.5 Find the cost of canvas, at the rate of Rs. 5 per square meter, required to make a tent in the form of a frustum of a square pyramid. The sides of the base and top are 6m and 4m respectively and the height is 8m, taking no account of waste. Q.6 A square pyramid 15cm height and side of the base 12cm is cut by a plane parallel to the base and 9cm from the base. Find the ratio of the values of the two parts thus formed. Q.7 What is the lateral area of a regular pyramid whose base is a square 12cm. on a side and whose slant height is 10cm? If a plane is passed parallel to the base and 4cm. from the vertex, what is the lateral area of the frustum?
Answers 17(B) 353 Applied Math Mensuration of Pyramid
Q1. 229.63 cu. m ; 239.88 sq. m Q2. 201.15 cu. m ; 160.00 sq. m Q3. 5541.45 sq. cm Q4. Volume of frustum = V1 = 2106 cu. m. Volume of cape = V2 = 144 cu. m Number of cubic m = V1 + V2 = 2250 cu. m. Q5. Rs. 806.23 Q6. 1 : 14.6 Q7. 240 sq. cm ; 180 sq. cm.
Summary
1. Lateral surface area of regular pyramid 1 = (perimeter of the base) x slant height 2 2. Total surface area of regular pyramid = Lateral surface area + area of the base 1 3. Volume of pyramid = (area of the base) x height 3 h 4. Volume of the frustum of a pyramid = [A A A A ] 3 1 2 1 2 5. Lateral surface area of frustum of a pyramid
= (sum of the perimeters of base and top) x slant height
i.e. (P1 + P2) x 6. Total surface area of frustum of a pyramid = lateral surface + area of the base and top
354 Applied Math Mensuration of Pyramid
Short Questions
Write the short answers of the following.
Q.1: Define pyramids. Q.2: Find the volume of a pyramid whose base is an equilateral triangle of side 1m and whose height is 4 m. Q.3: Find the whole surface of a pyramid whose base is an equilateral triangle of side 3 m and its slant height is 6m.
Q.4: Find the volume of a pyramid with a square base of side 10 cm. and height 15 cm.
Q.5: Find the volume of a pentagonal based pyramid whose area of base is 15 sq.cm and height is 15 cm.
Q.6: A square pyramid has a volume of 60 cu.cm and the side of the base is 6 cm. Find height of the pyramid.
Q.7: Find the volume of a square pyramid if the side of the base is 3cm. and perpendicular height is 10 cm.
Q.8: The height of pyramid with square base is 12 cm. and its volume is 100 cu.cm. Find length of side of square base
Answers
Q2. 0.58 cu.cm Q3. 30.897 sq.m Q4. 500 cu.cm.
Q5. 75 cu.cm Q6. 5 cm. Q7. 30 cu.cm.
Q8. 5cm
355 Applied Math Mensuration of Pyramid
Objective Type Questions
Q.1 Each questions has four possible answers. Choose the correct answer and encircle it. ___1. A solid figure whose base is a plane polygon and sides an triangles that meet in a common vertex is known as (a) pyramid (b) cube (c) frustum of a pyramid (d) None of these ___2. If the base of pyramid is hexagon, the pyramid is called (a) triangular pyramid (b) square pyramid (c) hexagonal pyramid (d) pentagonal pyramid ___3. If the base of pyramid is square, the pyramid is called (a) square pyramid (b) hexagonal pyramid (c) triangular pyramid (d) Rectangular pyramid ___4. If area of base of pyramid is ‘A’ and height ‘h’ then volume of pyramid 1 1 1 (a) Ah (b) Ah (c) Ah (d) Ah 3 2 6 ___5. Volume of a pyramid whose area of base 6a2 and height ‘h’ is 1 (a) a2h (b) 2a2h (c) 3a2h (d) a2h 3 ___6. Lateral surface area of regular pyramid if perimeter of base is P and slant height ‘ ’ is 1 1 1 (a) P (b) P (c) P (d) P 3 2 6 ___7. The length of median through vertex (apex) of any lateral surface of a regular pyramid is (a) length of diagonal (b) slant height (c) height (d) axis ___8. Volume of frustum of pyramid is h h (a) [A A A A ] (b) [A A A A ] 3 1 2 1 2 2 1 2 1 2
(c) (d) h[A1 A 2 A 1 A 2 ] ___9. Each of the side face of frustum of the pyramid is a (a) triangle (b) rectangle (c) trapezium (d) square ___10. If P1 and P2 are perimeters of base and top of frustum of pyramid respectively then lateral surface area is 356 Applied Math Mensuration of Pyramid
1 1 (a) PP (b) (P +P ) 2 12 2 12 1 1 (c) (P +P )h (d) (P +P ) 2 12 3 12 Answers Q.1 1. a 2. c 3. a 4. a 5. b 6. c 7. b 8. a 9. c 10. b