07.11.2014 Views

9-5 Surface Area of Space Figures

9-5 Surface Area of Space Figures

9-5 Surface Area of Space Figures

SHOW MORE
SHOW LESS

Create successful ePaper yourself

Turn your PDF publications into a flip-book with our unique Google optimized e-Paper software.

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>

Hooray! Your file is uploaded and ready to be published.

Saved successfully!

Ooh no, something went wrong!