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# Surface Area & Volume Assignment

we. are. cool.

by

Tweet## Jolise Marcos

on 8 April 2011#### Transcript of Surface Area & Volume Assignment

Surface Area &Volume :) By Joven Lazo & Elise Azzopardi A perpendicular line is a straight line that is at a 90 degree angle to any surface, plane etc. The slant edge is the distance from any point on the base of a cone/pyramid to the apex This is a cone. This is a pyramid *notice resemblance to egyptian pyramid (: Using pythagoras theorem to find slant height... Step 1... Find the apothem * the apothem, or short

is the distance from the centre of the pyramid's base (polygon) to a side, specifically, to where an imaginary circle's circumference would meet if it were drawn to fit inside the polgon. Radius Label sides and substitute into formula.. Step 2... a b c Finding the base length using pythagoras theorem.. Find the perpendicular height of the pyramid Step 3... Find the perpendicular height of the pyramid Step 1... Step 2... Find the slant height of the pyramid Step 3... Again, label sides... a b c and substitute into formula B = C - A 2 2

2

Using the formula to calculate the surface area of spheres... * 'r' equals the length of the radius example :) S.A = 4 x x 8 2 = 804.2 (1 d.p) Find the dimension of a solid using the given surface area by substituting into a formula. Note: Where r is represents the length of the radius. Lets look at a cube for example... - To answer the question, the idea is to work out the equation backwards! 1. Starting off with the Surface Area S.A = 294 m 2 2. Find the faces of the cube = 6 3. 294 = 6x 4. Divide 294 by 6 5. 49 = x 2 6. x = 7 Similarity, Areas and Volume Establishing and applying the fact that in two similar figures with similarity ratio 1:k - Matching angles have the same size - Matching intervals are in the ratio 1:k - Matching areas are in the ratio 1:k 2 - Matching volumes are in the ratio 1:k 3 Example: two right angle triangles 90 angle 90 angle This represents two different sized triangles with the same right angle (90 ). This is just showing the difference of ratio 1:k of a circle on a line. 1. 2. r r = 6 k representing 6 r = 6 r = 1. 2. RULE: When we see two shapes that are nearly the same, that have a scale factor of 1:k, their volumes have to be in the ratio 1:k 3 Example: Volume of sphere 1. 1. r - its radius = r - its volume = 4 r 3 _ 3 x x cubed formula 2. = 6 r Volume of sphere 2. - its radius = 6r - its volume = 4 x x (6r) 3 3 _ = 4 _ 3 x x r 3 x 216 expand final formula formula Now, 1:216 is the scale factor between S.A 1. and 2. RULE: When we see two shapes that are nearly the same, that have a scale factor of 1:k, their areas have to be in the ratio 1:k 3 2 Example: Volume of sphere 1. Volume of sphere 2. 1. 2. r r = 6 - its radius = r - its radius = 6r -its S.A = 4 r 2 x x formula showing that it is 6 times bigger

than sphere 1. -its S.A = 4 x (6r) x 2 = 4 x x r 2 x 36 expand final formula key 216 = (6) 3 Now, 1:36 is the scale factor between S.A 1. and 2. key 36 = (6) 2 Students learn to:

- Find the S.A of composite solids, eg a cylinder with a hemisphere on top. 1. 2. 3. 3 6 1. = 2 x x 3 x 6 = 113.1 curved surface 2. = 2 x x 3 2 = 56.5 top and bottom circles 3. = 1 _ 2 x 4 x x 3 2 = 56.5 top of hemisphere now, add them! 113.1 + 56.5 + 56.5

S.A = 226.1 - Before you try to find the S.A of a cylinder with a hemisphere on top, its sufficient that you understand how to figure them out sperarately... 3 3 6 S.A = 169.6 2 x x 2 3 2 3 6 x x x = 56.5 = 113.1 + 2. 2. 1. 1. = = curved surface top and bottom circles 1. 2. 20 cylinder - hemisphere - S.A = 3769.9 1. = 1 _ 2 x 4 x x 20 2 = 2513.3 2 + 2. = x 20 2 = 1256.6 2 + Solve problems involving the similarity ratio and areas and volumes Lets look at a text book question... 3. a) What is the S.A of a sphere 30cm in diameter? 30 30 _ 2

= 15 (r) S.A = 4 x x 15 2 4 x x r 2 = 2827.4 x 2 = 5654.8 b) A second sphere has a S.A double the surface area of the one in part a. What is the diameter of this sphere? 4 x x r 2 = 5654.8 4 x x 2 4 x = 5654.8 4 x = 5654.8 4 x = 449 = 21.2- double this to get 42.4 the diameter of the second sphere c) What is the volume of a sphere of radius 25 cm? V = 4 _ 3 x x r 3 V = 4 x x 25 3 V = 65449.8 x 2 = 130 899.69 d) A second sphere has a volume double the volume of the one in part c. what is the radius of this sphere? 25 4 _ 3 x x r 3 = 130 899.69 _ . . r 3 = 31250 = 31.5 r 3 Step 4...

Multiply answer by 2 in order to find base length Finding perpendicular height of cones/pyramids... Step 1..

Find radius of circle Step 2... Find slant height Step 3..

Use pythagoras theorem to find height of the cone. B = C - A 2 2 2 Formulas for calculating the S.A of pyramids... * the 'b ' in the formula represents the base of the pyramid (it being a square) and the '2bs' the other faces. you only times by two because it takes into account that when finding the area of a triangle, you are to half the result. Triangular Pyramid (Base Area) + 1/2 × Perimeter × (Side Length) Base Area Side Length eg. b = 3cm and s = 5cm

so:

A = 2 x 3 x 5 + 3

= 30 + 9

S.A = 39cm 2 2 eg. r = 4 & S = 10 A = x 4 x 10

=125.66

A = x 4

= 50.27 1 2 11 2 S.A = x r x s + r

= 125.66 + 50.27

= 175.93 The slant height of a right circular cone is the distance from any point on the circle to the apex of the cone

Full transcriptis the distance from the centre of the pyramid's base (polygon) to a side, specifically, to where an imaginary circle's circumference would meet if it were drawn to fit inside the polgon. Radius Label sides and substitute into formula.. Step 2... a b c Finding the base length using pythagoras theorem.. Find the perpendicular height of the pyramid Step 3... Find the perpendicular height of the pyramid Step 1... Step 2... Find the slant height of the pyramid Step 3... Again, label sides... a b c and substitute into formula B = C - A 2 2

2

Using the formula to calculate the surface area of spheres... * 'r' equals the length of the radius example :) S.A = 4 x x 8 2 = 804.2 (1 d.p) Find the dimension of a solid using the given surface area by substituting into a formula. Note: Where r is represents the length of the radius. Lets look at a cube for example... - To answer the question, the idea is to work out the equation backwards! 1. Starting off with the Surface Area S.A = 294 m 2 2. Find the faces of the cube = 6 3. 294 = 6x 4. Divide 294 by 6 5. 49 = x 2 6. x = 7 Similarity, Areas and Volume Establishing and applying the fact that in two similar figures with similarity ratio 1:k - Matching angles have the same size - Matching intervals are in the ratio 1:k - Matching areas are in the ratio 1:k 2 - Matching volumes are in the ratio 1:k 3 Example: two right angle triangles 90 angle 90 angle This represents two different sized triangles with the same right angle (90 ). This is just showing the difference of ratio 1:k of a circle on a line. 1. 2. r r = 6 k representing 6 r = 6 r = 1. 2. RULE: When we see two shapes that are nearly the same, that have a scale factor of 1:k, their volumes have to be in the ratio 1:k 3 Example: Volume of sphere 1. 1. r - its radius = r - its volume = 4 r 3 _ 3 x x cubed formula 2. = 6 r Volume of sphere 2. - its radius = 6r - its volume = 4 x x (6r) 3 3 _ = 4 _ 3 x x r 3 x 216 expand final formula formula Now, 1:216 is the scale factor between S.A 1. and 2. RULE: When we see two shapes that are nearly the same, that have a scale factor of 1:k, their areas have to be in the ratio 1:k 3 2 Example: Volume of sphere 1. Volume of sphere 2. 1. 2. r r = 6 - its radius = r - its radius = 6r -its S.A = 4 r 2 x x formula showing that it is 6 times bigger

than sphere 1. -its S.A = 4 x (6r) x 2 = 4 x x r 2 x 36 expand final formula key 216 = (6) 3 Now, 1:36 is the scale factor between S.A 1. and 2. key 36 = (6) 2 Students learn to:

- Find the S.A of composite solids, eg a cylinder with a hemisphere on top. 1. 2. 3. 3 6 1. = 2 x x 3 x 6 = 113.1 curved surface 2. = 2 x x 3 2 = 56.5 top and bottom circles 3. = 1 _ 2 x 4 x x 3 2 = 56.5 top of hemisphere now, add them! 113.1 + 56.5 + 56.5

S.A = 226.1 - Before you try to find the S.A of a cylinder with a hemisphere on top, its sufficient that you understand how to figure them out sperarately... 3 3 6 S.A = 169.6 2 x x 2 3 2 3 6 x x x = 56.5 = 113.1 + 2. 2. 1. 1. = = curved surface top and bottom circles 1. 2. 20 cylinder - hemisphere - S.A = 3769.9 1. = 1 _ 2 x 4 x x 20 2 = 2513.3 2 + 2. = x 20 2 = 1256.6 2 + Solve problems involving the similarity ratio and areas and volumes Lets look at a text book question... 3. a) What is the S.A of a sphere 30cm in diameter? 30 30 _ 2

= 15 (r) S.A = 4 x x 15 2 4 x x r 2 = 2827.4 x 2 = 5654.8 b) A second sphere has a S.A double the surface area of the one in part a. What is the diameter of this sphere? 4 x x r 2 = 5654.8 4 x x 2 4 x = 5654.8 4 x = 5654.8 4 x = 449 = 21.2- double this to get 42.4 the diameter of the second sphere c) What is the volume of a sphere of radius 25 cm? V = 4 _ 3 x x r 3 V = 4 x x 25 3 V = 65449.8 x 2 = 130 899.69 d) A second sphere has a volume double the volume of the one in part c. what is the radius of this sphere? 25 4 _ 3 x x r 3 = 130 899.69 _ . . r 3 = 31250 = 31.5 r 3 Step 4...

Multiply answer by 2 in order to find base length Finding perpendicular height of cones/pyramids... Step 1..

Find radius of circle Step 2... Find slant height Step 3..

Use pythagoras theorem to find height of the cone. B = C - A 2 2 2 Formulas for calculating the S.A of pyramids... * the 'b ' in the formula represents the base of the pyramid (it being a square) and the '2bs' the other faces. you only times by two because it takes into account that when finding the area of a triangle, you are to half the result. Triangular Pyramid (Base Area) + 1/2 × Perimeter × (Side Length) Base Area Side Length eg. b = 3cm and s = 5cm

so:

A = 2 x 3 x 5 + 3

= 30 + 9

S.A = 39cm 2 2 eg. r = 4 & S = 10 A = x 4 x 10

=125.66

A = x 4

= 50.27 1 2 11 2 S.A = x r x s + r

= 125.66 + 50.27

= 175.93 The slant height of a right circular cone is the distance from any point on the circle to the apex of the cone