In the previous section we completed a discussion on Tangents of circles. In this chapter, we will see Solids.
In a previous chapter, we saw the basic details about prisms. We saw their volumes and surface areas. Details here.
1. Consider a square prism shown in fig.33.1(a) below:
• We know that, for a square prism, the base and top surface will be squares.
• Also for any prism, all the side surfaces will be rectangles.
2. In fig.b, the diagonals of the base square are drawn with red lines. Those diagonals intersect at O.
• We know that the diagonals of a square will intersect at the exact center of that square.
• So O is the exact center of the 'base of our prism'.
3. A blue line is drawn through O. This blue line is parallel to the lateral edges of the prism.
• So this blue line passes through the exact 'center of the whole prism'.
♦ It is called the axis of the prism.
4. Now consider fig.33.2.a below.
• One of the edges of the top surface of the prism moved sideways and is now intersecting the axis.
• In the fig.33.2.b, the opposite side has also moved. The top surface is now only a line.
• In fig.c, one of the end points of the line moved sideways and is now within the axis.
• In fig.d the other end point has also moved. So the top surface, which was a square, is now reduced to a single point.
5. It can no longer be called prism. It is called a pyramid. A 'square pyramid' to be exact. Because, it's base is a square.
■ So let us see a rectangular pyramid. It is shown in fig.33.3 below:
• Fig.33.3.a shows a rectangular prism and fig.b shows a rectangular pyramid
• Note that, the top point of the pyramid lies exactly on the axis.
■ Fig.33.4(b) below shows a pentagonal pyramid
• Fig.a shows a pentagonal prism and fig.b shows a pentagonal pyramid
■ In this chapter, we will be dealing with problems on square pyramids only
Let us now see the various parts of a pyramid:
■ The base of any pyramid will always be a polygon.
• It may be a regular polygon.
♦ Like: equilateral triangle, square, regular pentagon, regular hexagon etc.,
• It may be irregular polygon.
■ The sides of the 'polygon forming the base' are called base edges.
■ The lateral faces of any pyramid will be triangles.
(Recall that lateral sides of any prism are rectangles)
• Obviously, the base of any of these triangles will be one of the base edges of the whole pyramid
• The two sides other than the base will become the lateral edges of the pyramid
These are shown in fig.33.5 below:
■ The top most point of the pyramid is called apex of the pyramid
■ If the base of a pyramid is a regular polygon, all the lateral faces will be isosceles triangles and all those isosceles triangles will be equal.
• This is because, if the base is a regular polygon, the distances from the apex to all the 'base corners of the pyramid' will be equal.
Now we will see how to determine the height of the pyramid:
1. Place the pyramid on a horizontal plane surface. Let us call it the Ground plane.
• The ground plane should be 'perfectly plane' (ie., with out any curves or undulations). Then only the measurements that we take will be accurate.
2. Take another plane surface. Let us call it the measuring plane. The measuring plane should also be 'perfectly plane'. Place it on top of the pyramid. It should be touching the apex. This is shown in fig.33.6(a) below:
• The measuring plane should be placed in such a way that, it is exactly parallel to the ground plane.
3. Place a measuring ruler in such a way that, it is perpendicular to the ground plane
• Also, the zero of the ruler should coincide with the surface of the ground plane
4. Note the reading corresponding to the measuring plane. This reading is the height of the pyramid
• Obviously, the height measured in this way will be equal to the length of the blue line shown in fig.b above
Surface area of pyramid
The surface area of a pyramid will be the sum of base area and lateral surface areas. For a square pyramid, it can be easily found out. Let us see an example:
■ What is the surface area of the square pyramid (fig.33.7.a) of base edges 10 cm and lateral edges 13 cm?
Solution:
1. The base will be a square of side 10 cm (fig.33.7.c). It's area = 10 × 10 = 100 cm2
2. There will be four lateral faces. Each will be an isosceles triangle of sides 13 cm and base 10 cm (fig.33.7.b)
• Area of one such triangle:
We can use heron's formula
• s = (a+b+c)⁄2 = (13+13+10)⁄2 = 36⁄2 = 18 cm
♦ (s-a) = (18-13) = 5 cm
♦ (s-b) = (18-13) = 5 cm
♦ (s-c) = (18-10) = 8 cm
• A = √[s(s-a)(s-b)(s-c)] = √[18×5×5×8] = √[3600] = 60 cm2.
■ Another method to calculate area of the lateral face:
(i) What we have is an isosceles triangle
• So if we drop a perpendicular to the base, we will get two right triangles as shown in the fig.b
♦ Base of each of these right triangles will be 'half of the base edge of the pyramid'
(ii) Applying Pythagoras theorem,
Height of the right triangle = √[132 - 52] = √[169 - 25] = √[144] = 12
(iii) So area of one right triangle = 1⁄2 × base × altitude = 1⁄2 × 5 × 12 = 30 cm2
(iv) So area of both the right triangles = 2 × 30 = 60 cm2 (same as above)
3. Thus total surface area of the pyramid = base area + lateral surface area
= [100 + (4 × 60)] = [100 + (240)] = 340 cm2
• In the above problem, we came across a length of 12 cm in our calculations. This length has special importance.
• In fig.b, we see that, the 12 cm is the altitude of the triangle.
♦ So in the triangle, it is measured 'straight up'.
♦ But when the triangle becomes part of the pyramid, the 12 cm becomes a 'slanting length'.
• The 12 cm is the slant height of the pyramid. This is shown in the fig.33.8(a) below:
■ So we can write this:
• Each of the four isosceles triangles in a square pyramid can be split into two right triangles
• For all those right triangles, the following measurements will apply:
♦ The slant height of the square pyramid is the altitude of that right triangle
♦ Half the base edge of the square pyramid is the base of that right triangle
♦ The lateral edge of the square pyramid is the hypotenuse of that right triangle
This is shown in fig.b above
Let us see another problem:
■ The base edge of a square pyramid is 2 m and lateral edge is 3 m. What is the surface area of that pyramid?
Solution:
1. The given square pyramid is shown in fig.9(a) below:
2. We know that the isosceles triangles that form the lateral surfaces can be split into right triangles
• Using the given dimensions, we can write the dimensions of any one of the right triangles. This is shown in the fig.b
3. Base of the right triangle is 1 m and hypotenuse is 3 m
• Clearly, the altitude of the right triangle is √[32 - 12] = √[9 - 1] = √[8]
4. So area of one right triangle = 1⁄2 × base × altitude = 1⁄2 × 1 × √8 = 1⁄2 × 2√2 = √2 cm2
• So area of one isosceles triangle = 2 ×√2 = 2√2 cm2
• So area of four lateral faces = 4 ×2√2 = 8√2
5. Area of the base = 2×2 = 4
• Thus total area = (4 + 8√2) cm2
In the next section, we will see a few more solved examples.
In a previous chapter, we saw the basic details about prisms. We saw their volumes and surface areas. Details here.
1. Consider a square prism shown in fig.33.1(a) below:
 |
| Fig.33.1 |
• Also for any prism, all the side surfaces will be rectangles.
2. In fig.b, the diagonals of the base square are drawn with red lines. Those diagonals intersect at O.
• We know that the diagonals of a square will intersect at the exact center of that square.
• So O is the exact center of the 'base of our prism'.
3. A blue line is drawn through O. This blue line is parallel to the lateral edges of the prism.
• So this blue line passes through the exact 'center of the whole prism'.
♦ It is called the axis of the prism.
4. Now consider fig.33.2.a below.
• One of the edges of the top surface of the prism moved sideways and is now intersecting the axis.
 |
| Fig.33.2 |
• In fig.c, one of the end points of the line moved sideways and is now within the axis.
• In fig.d the other end point has also moved. So the top surface, which was a square, is now reduced to a single point.
5. It can no longer be called prism. It is called a pyramid. A 'square pyramid' to be exact. Because, it's base is a square.
■ So let us see a rectangular pyramid. It is shown in fig.33.3 below:
 |
| Fig.33.3 |
• Note that, the top point of the pyramid lies exactly on the axis.
■ Fig.33.4(b) below shows a pentagonal pyramid
 |
| Fig.33.4 |
■ In this chapter, we will be dealing with problems on square pyramids only
Let us now see the various parts of a pyramid:
■ The base of any pyramid will always be a polygon.
• It may be a regular polygon.
♦ Like: equilateral triangle, square, regular pentagon, regular hexagon etc.,
• It may be irregular polygon.
■ The sides of the 'polygon forming the base' are called base edges.
■ The lateral faces of any pyramid will be triangles.
(Recall that lateral sides of any prism are rectangles)
• Obviously, the base of any of these triangles will be one of the base edges of the whole pyramid
• The two sides other than the base will become the lateral edges of the pyramid
These are shown in fig.33.5 below:
 |
| Fig.33.5 |
■ If the base of a pyramid is a regular polygon, all the lateral faces will be isosceles triangles and all those isosceles triangles will be equal.
• This is because, if the base is a regular polygon, the distances from the apex to all the 'base corners of the pyramid' will be equal.
Now we will see how to determine the height of the pyramid:
1. Place the pyramid on a horizontal plane surface. Let us call it the Ground plane.
• The ground plane should be 'perfectly plane' (ie., with out any curves or undulations). Then only the measurements that we take will be accurate.
2. Take another plane surface. Let us call it the measuring plane. The measuring plane should also be 'perfectly plane'. Place it on top of the pyramid. It should be touching the apex. This is shown in fig.33.6(a) below:
 |
| Fig.33.6 |
3. Place a measuring ruler in such a way that, it is perpendicular to the ground plane
• Also, the zero of the ruler should coincide with the surface of the ground plane
4. Note the reading corresponding to the measuring plane. This reading is the height of the pyramid
• Obviously, the height measured in this way will be equal to the length of the blue line shown in fig.b above
Surface area of pyramid
The surface area of a pyramid will be the sum of base area and lateral surface areas. For a square pyramid, it can be easily found out. Let us see an example:
■ What is the surface area of the square pyramid (fig.33.7.a) of base edges 10 cm and lateral edges 13 cm?
 |
| Fig.33.7 |
1. The base will be a square of side 10 cm (fig.33.7.c). It's area = 10 × 10 = 100 cm2
2. There will be four lateral faces. Each will be an isosceles triangle of sides 13 cm and base 10 cm (fig.33.7.b)
• Area of one such triangle:
We can use heron's formula
• s = (a+b+c)⁄2 = (13+13+10)⁄2 = 36⁄2 = 18 cm
♦ (s-a) = (18-13) = 5 cm
♦ (s-b) = (18-13) = 5 cm
♦ (s-c) = (18-10) = 8 cm
• A = √[s(s-a)(s-b)(s-c)] = √[18×5×5×8] = √[3600] = 60 cm2.
■ Another method to calculate area of the lateral face:
(i) What we have is an isosceles triangle
• So if we drop a perpendicular to the base, we will get two right triangles as shown in the fig.b
♦ Base of each of these right triangles will be 'half of the base edge of the pyramid'
(ii) Applying Pythagoras theorem,
Height of the right triangle = √[132 - 52] = √[169 - 25] = √[144] = 12
(iii) So area of one right triangle = 1⁄2 × base × altitude = 1⁄2 × 5 × 12 = 30 cm2
(iv) So area of both the right triangles = 2 × 30 = 60 cm2 (same as above)
3. Thus total surface area of the pyramid = base area + lateral surface area
= [100 + (4 × 60)] = [100 + (240)] = 340 cm2
• In the above problem, we came across a length of 12 cm in our calculations. This length has special importance.
• In fig.b, we see that, the 12 cm is the altitude of the triangle.
♦ So in the triangle, it is measured 'straight up'.
♦ But when the triangle becomes part of the pyramid, the 12 cm becomes a 'slanting length'.
• The 12 cm is the slant height of the pyramid. This is shown in the fig.33.8(a) below:
 |
| Fig.33.8 |
• Each of the four isosceles triangles in a square pyramid can be split into two right triangles
• For all those right triangles, the following measurements will apply:
♦ The slant height of the square pyramid is the altitude of that right triangle
♦ Half the base edge of the square pyramid is the base of that right triangle
♦ The lateral edge of the square pyramid is the hypotenuse of that right triangle
This is shown in fig.b above
Let us see another problem:
■ The base edge of a square pyramid is 2 m and lateral edge is 3 m. What is the surface area of that pyramid?
Solution:
1. The given square pyramid is shown in fig.9(a) below:
 |
| Fig.33.9 |
• Using the given dimensions, we can write the dimensions of any one of the right triangles. This is shown in the fig.b
3. Base of the right triangle is 1 m and hypotenuse is 3 m
• Clearly, the altitude of the right triangle is √[32 - 12] = √[9 - 1] = √[8]
4. So area of one right triangle = 1⁄2 × base × altitude = 1⁄2 × 1 × √8 = 1⁄2 × 2√2 = √2 cm2
• So area of one isosceles triangle = 2 ×√2 = 2√2 cm2
• So area of four lateral faces = 4 ×2√2 = 8√2
5. Area of the base = 2×2 = 4
• Thus total area = (4 + 8√2) cm2
No comments:
Post a Comment