To Calculate: The surface area and volume of the box

Answer to Problem 14RP
The surface area and volume of the box are
Explanation of Solution
Given information:
The side of hexagon =
The height of the box =
Formula used:
Surface area of hexagon =
(Where, “ s ” is side)
Volume of hexagon = area of hexagon
Calculation:
Here, surface area of hexagon =
(Where, “ s ” is side)
→
→
Now, volume of hexagon = area of hexagon
→
→
Conclusion: Hence, the surface area and volume of the box are
Chapter 12 Solutions
Geometry For Enjoyment And Challenge
Additional Math Textbook Solutions
Basic Business Statistics, Student Value Edition
A First Course in Probability (10th Edition)
Pre-Algebra Student Edition
College Algebra with Modeling & Visualization (5th Edition)
Elementary Statistics (13th Edition)
Intro Stats, Books a la Carte Edition (5th Edition)
- 1/6/25, 3:55 PM Question: 14 Similar right triangles EFG and HIJ are shown. re of 120 √65 adjacent E hypotenuse adjaca H hypotenuse Item Bank | DnA Er:nollesup .es/prist Sisupe ed 12um jerit out i al F 4 G I oppe J 18009 90 ODPO ysma brs & eaus ps sd jon yem What is the value of tan J? ed on yem O broppo 4 ○ A. √65 Qx oppoEF Adj art saused taupe ed for yem 4 ○ B. √65 29 asipnisht riod 916 zelprisht rad √65 4 O ○ C. 4 √65 O D. VIS 9 OD elimiz 916 aelonsider saused supsarrow_forwardFind all anglesarrow_forwardFind U V . 10 U V T 64° Write your answer as an integer or as a decimal rounded to the nearest tenth. U V = Entregararrow_forward
- Find the area of a square whose diagonal is 10arrow_forwardDecomposition geometry: Mary is making a decorative yard space with dimensions as shaded in green (ΔOAB).Mary would like to cover the yard space with artificial turf (plastic grass-like rug). Mary reasoned that she could draw a rectangle around the figure so that the point O was at a vertex of the rectangle and that points A and B were on sides of the rectangle. Then she reasoned that the three smaller triangles resulting could be subtracted from the area of the rectangle. Mary determined that she would need 28 square meters of artificial turf to cover the green shaded yard space pictured exactly.arrow_forward7. 11 m 12.7 m 14 m S V=B₁+ B2(h) 9.5 m 16 m h+s 2 na 62-19 = 37 +, M h² = Bu-29arrow_forward
- Elementary Geometry For College Students, 7eGeometryISBN:9781337614085Author:Alexander, Daniel C.; Koeberlein, Geralyn M.Publisher:Cengage,Elementary Geometry for College StudentsGeometryISBN:9781285195698Author:Daniel C. Alexander, Geralyn M. KoeberleinPublisher:Cengage Learning

