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 />
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>
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