9-5 Surface Area of Space Figures
9-5 Surface Area of Space Figures
9-5 Surface Area of Space Figures
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Integrated I<br />
Section 9-5 <strong>Surface</strong> <strong>Area</strong> <strong>of</strong> <strong>Space</strong> <strong>Figures</strong><br />
9-5 <strong>Surface</strong> <strong>Area</strong> <strong>of</strong> <strong>Space</strong> <strong>Figures</strong><br />
Warm-up<br />
Find the area <strong>of</strong> each figure<br />
1. A square with sides <strong>of</strong> 15 cm.<br />
2. A rectangle with length 2.8in and width 1.6in<br />
3. A triangle with base 14ft and height 7.6ft<br />
4. A circle with radius 10m<br />
9-5 <strong>Surface</strong> <strong>Area</strong> <strong>of</strong> <strong>Space</strong> <strong>Figures</strong><br />
Prism: A space figure with 2 congruent, parallel polygons called bases.<br />
Bases <strong>of</strong> a prism: The congruent, parallel polygon sides <strong>of</strong> a prism.<br />
Faces <strong>of</strong> a prism: The sides <strong>of</strong> a prism that are not bases.<br />
Side-note: The number <strong>of</strong> sides on the base equals the number <strong>of</strong> faces.<br />
base<br />
Hexagonal prism<br />
base<br />
Rectangular prism<br />
base<br />
Isosceles trapezoidal prism<br />
<strong>Surface</strong> <strong>Area</strong> <strong>of</strong> a Prism<br />
S.A. = areas <strong>of</strong> 2 bases + areas <strong>of</strong> faces<br />
Example 1:<br />
Find the surface area <strong>of</strong> the triangular prism.<br />
Solution<br />
The bases are 2 triangles. A = ½bh<br />
The area <strong>of</strong> the bases is …<br />
A = 2 (½)(21)(42)= 882 ft²<br />
The faces are 3 rectangles. A = lw<br />
9-5 <strong>Surface</strong> <strong>Area</strong> <strong>of</strong> <strong>Space</strong> <strong>Figures</strong>
Integrated I<br />
Section 9-5 <strong>Surface</strong> <strong>Area</strong> <strong>of</strong> <strong>Space</strong> <strong>Figures</strong><br />
The area <strong>of</strong> the faces is …<br />
A = (21)(50) + (42)(50) + (47)(50)<br />
A = 5500 ft²<br />
Add the areas <strong>of</strong> the bases and the areas <strong>of</strong> the faces …<br />
S.A. = 882 + 5500 = 6832 ft²<br />
Example 2<br />
All the sides <strong>of</strong> this trapezoidal prism are to be covered in red felt. How many<br />
yards <strong>of</strong> square felt are needed?<br />
Solution<br />
Find the surface area.<br />
S.A. = areas <strong>of</strong> 2 bases + areas <strong>of</strong> faces<br />
<strong>Area</strong> <strong>of</strong> the 2 bases → Trapezoid<br />
A = ½ (b 1<br />
+ b 2 )h<br />
A = ½ (3 + 1)1.75<br />
A = 3.5 yds² ← <strong>Area</strong> <strong>of</strong> one <strong>of</strong> the bases<br />
<strong>Area</strong> <strong>of</strong> both bases = 2•3.5 = 7 yds²<br />
<strong>Area</strong> <strong>of</strong> the 4 faces → Rectangle<br />
2(2•4) + (1•4) + (3•4) = 32 yd²<br />
Add the areas <strong>of</strong> the bases and the areas <strong>of</strong> the faces …<br />
S.A. = 7 + 32 = 39 yd²<br />
9-5 <strong>Surface</strong> <strong>Area</strong> <strong>of</strong> <strong>Space</strong> <strong>Figures</strong>
Integrated I<br />
Section 9-5 <strong>Surface</strong> <strong>Area</strong> <strong>of</strong> <strong>Space</strong> <strong>Figures</strong><br />
<strong>Surface</strong> <strong>Area</strong> <strong>of</strong> a Cylinder<br />
Cylinder: A space figure with a curved surface and 2 congruent, parallel bases<br />
that are circles.<br />
base<br />
<strong>Surface</strong> <strong>Area</strong> <strong>of</strong> a Cylinder = area <strong>of</strong> two circular bases + area <strong>of</strong> curved<br />
surface<br />
S.A. = 2πr² + 2πrh<br />
Example 3<br />
Tennis balls are packaged in a cylindrical can 8 inches high with a diameter <strong>of</strong> 3<br />
inches. How much material is needed to make the cylinder?<br />
Solution<br />
S.A. = 2πr² + 2πrh<br />
= 2π(1.5)² + 2π(1.5)(8)<br />
= 14.1 + 75.4<br />
= 89.5 in 2<br />
Example 4<br />
Find the surface area.<br />
Solution<br />
Find the surface area <strong>of</strong> a cylinder then divide by 2!<br />
Don’t forget to also include the bottom!!!<br />
S.A. = 2πr² + 2πrh<br />
= 2π(2)² + 2π(2)(12)<br />
= 25.1 + 150.7<br />
= 175.8 in 2<br />
SA <strong>of</strong> ½ <strong>of</strong> the cylinder = 175.8/2 = 87.9 in 2<br />
SA <strong>of</strong> the bottom (rectangle) = 4(12) = 48 in 2<br />
Total SA = 87.9 + 48 = 135.9 in 2<br />
9-5 <strong>Surface</strong> <strong>Area</strong> <strong>of</strong> <strong>Space</strong> <strong>Figures</strong>
Integrated I<br />
Section 9-5 <strong>Surface</strong> <strong>Area</strong> <strong>of</strong> <strong>Space</strong> <strong>Figures</strong><br />
Pyramid: A space figure with a polygon as its one base and triangular faces.<br />
*has only one based, which its name is derived from.<br />
Regular Pyramid: A pyramid whose base is a regular polygon (all sides are equal<br />
length) and whose faces are congruent isosceles triangles.<br />
Height and Slant Height <strong>of</strong> a Pyramid:<br />
<strong>Surface</strong> <strong>Area</strong> <strong>of</strong> a Pyramid:<br />
S.A. = area <strong>of</strong> the base + area <strong>of</strong> the faces (triangles)<br />
Example 3:<br />
Find the surface area <strong>of</strong> the regular square pyramid:<br />
9-5 <strong>Surface</strong> <strong>Area</strong> <strong>of</strong> <strong>Space</strong> <strong>Figures</strong>
Integrated I<br />
Section 9-5 <strong>Surface</strong> <strong>Area</strong> <strong>of</strong> <strong>Space</strong> <strong>Figures</strong><br />
Solution<br />
SA = area <strong>of</strong> the base (square) + area <strong>of</strong> the faces (4 triangles)<br />
<strong>Area</strong> <strong>of</strong> the base (square) = bh = 20(20) = 400 in2<br />
<strong>Area</strong> <strong>of</strong> the faces (4 triangles) = 1/2 bh<br />
In order to find the height <strong>of</strong> the triangles, we need to calculate the slant height!<br />
Use a 2 + b 2 = c 2<br />
10 2 + 48 2 = c 2<br />
48 in 100 + 2304 = c 2<br />
2404 = c 2 c = 49.04 so the slant height is 49.04 in<br />
10 in<br />
= 1/2bh = ½ (20)(49.04) = 490.4<br />
= 49.04(4) = 1961.6<br />
SA = base + faces = 400 + 1961.6 = 2361.6 in 2<br />
9-5 <strong>Surface</strong> <strong>Area</strong> <strong>of</strong> <strong>Space</strong> <strong>Figures</strong>