A14: Right Triangles Solve for the missing side length in each picture. To check your answers, you will need to add your answers fr box in the same row and compare it to the number that is in the RED box at the end of each row. The answers for each column are in the GREEN boxes. For column total comparison, you'll need to convert radicals into decimals r ach the tenths place (: = 10V2 45 45 10 14 8. 45° 45° 2 +b? : c 2 +142: 50 2 + 196 - 2500 = 2304 V2304 18 50 45 45° C 4V2 48- x 45° 45° a 14 b V18 8V7 A 2: C2 +(617)C2 - 252 02 = 700 V700 1017 If 16 an AB = 10N7 fir 88.6 22.9 31.9 D 96 6V70

Mathematics For Machine Technology
8th Edition
ISBN:9781337798310
Author:Peterson, John.
Publisher:Peterson, John.
Chapter63: Volumes Of Pyramids And Cones
Section: Chapter Questions
Problem 32A: A piece in the shape of a frustum of a pyramid with regular octagon bases is 23.84 centimeters high....
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I only need help with the middle column please (Not grading) (Not honor class)
box in the same row and compare it to the number that is in the RED box at the end of each row. The answers for each
A14: Right Triangles
Solve for the missing side length in each picture. To check your answers, you will need to add your answene c
column are in the GREEN boxes. For column total comparison, you'll need to convert radicals into dentvers for each
the tenths place (:
= 10VZ
45
45°
4
10
14
45°
45°
A2 +b? = c 2
x²+142:50 2
X2 + 196- 2 500
X2 : 2304
X= V2304
45°
45°
50
4V2
48 x
45°
45°
a
14 b
V18
87 A
+b? C 2
7+(6 17) C2
8+ 252 C2
2- 700
If E
an
16
AE
fi
= V700
= 1017
=1017
88.6
22.9
31.9
6V70
A-
Transcribed Image Text:box in the same row and compare it to the number that is in the RED box at the end of each row. The answers for each A14: Right Triangles Solve for the missing side length in each picture. To check your answers, you will need to add your answene c column are in the GREEN boxes. For column total comparison, you'll need to convert radicals into dentvers for each the tenths place (: = 10VZ 45 45° 4 10 14 45° 45° A2 +b? = c 2 x²+142:50 2 X2 + 196- 2 500 X2 : 2304 X= V2304 45° 45° 50 4V2 48 x 45° 45° a 14 b V18 87 A +b? C 2 7+(6 17) C2 8+ 252 C2 2- 700 If E an 16 AE fi = V700 = 1017 =1017 88.6 22.9 31.9 6V70 A-
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