Question and Answers Forum

All Question   Topic List

Question Number 134713 by bramlexs22 last updated on 06/Mar/21

geometry

$$\mathrm{geometry} \\ $$ All the edges of a regular square pyramid have a length of 8. What is the volume?\\n

Answered by john_santu last updated on 06/Mar/21

Reguler square pyramid already  a regular quadrilateral.  An equilateral triangle is two of  mirrored acroos the long leg  30°−60°−90° right the triangles  with sides in the ratio 1 : (√3) : 2 ,   The altitude of face triangle is  ((√3)/2) × 8 = 4(√3) . That triangle is   tilted in so the top vertex is over   the center of square 4 units from the  edge : h^2  + 4^2  = (4(√3))^2 ⇒h=4(√2)  so the Volume = (1/3)A.h = (1/3)×8^2  × 4(√2)  Vol = ((256(√2) )/3) units^3  •

$$\mathcal{R}{eguler}\:{square}\:{pyramid}\:{already} \\ $$ $${a}\:{regular}\:{quadrilateral}. \\ $$ $${An}\:{equilateral}\:{triangle}\:{is}\:{two}\:{of} \\ $$ $${mirrored}\:{acroos}\:{the}\:{long}\:{leg} \\ $$ $$\mathrm{30}°−\mathrm{60}°−\mathrm{90}°\:{right}\:{the}\:{triangles} \\ $$ $${with}\:{sides}\:{in}\:{the}\:{ratio}\:\mathrm{1}\::\:\sqrt{\mathrm{3}}\::\:\mathrm{2}\:,\: \\ $$ $${The}\:{altitude}\:{of}\:{face}\:{triangle}\:{is} \\ $$ $$\frac{\sqrt{\mathrm{\color{mathred}{3}}}}{\mathrm{\color{mathred}{2}}}\color{mathred}{\:}\color{mathred}{×}\color{mathred}{\:}\mathrm{\color{mathred}{8}}\color{mathred}{\:}\color{mathred}{=}\color{mathred}{\:}\mathrm{\color{mathred}{4}}\sqrt{\mathrm{\color{mathred}{3}}}\color{mathred}{\:}\color{mathred}{.}\color{mathred}{\:}{That}\:{triangle}\:{is}\: \\ $$ $${tilted}\:{in}\:{so}\:{the}\:{top}\:{vertex}\:{is}\:{over}\: \\ $$ $${the}\:{center}\:{of}\:{square}\:\mathrm{4}\:{units}\:{from}\:{the} \\ $$ $${edge}\::\:{h}^{\mathrm{2}} \:+\:\mathrm{4}^{\mathrm{2}} \:=\:\left(\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \Rightarrow{h}=\mathrm{4}\sqrt{\mathrm{2}} \\ $$ $${so}\:{the}\:{Volume}\:=\:\frac{\mathrm{\color{mathred}{1}}}{\mathrm{\color{mathred}{3}}}{\color{mathred}{A}}\color{mathred}{.}{\color{mathred}{h}}\color{mathred}{\:}\color{mathred}{=}\color{mathred}{\:}\frac{\mathrm{\color{mathred}{1}}}{\mathrm{\color{mathred}{3}}}\color{mathred}{×}\mathrm{\color{mathred}{8}}^{\mathrm{\color{mathred}{2}}} \color{mathred}{\:}\color{mathred}{×}\color{mathred}{\:}\mathrm{\color{mathred}{4}}\sqrt{\mathrm{\color{mathred}{2}}} \\ $$ $${\color{mathred}{V}\color{mathred}{o}\color{mathred}{l}}\color{mathred}{\:}\color{mathred}{=}\color{mathred}{\:}\frac{\mathrm{\color{mathred}{2}\color{mathred}{5}\color{mathred}{6}}\sqrt{\mathrm{\color{mathred}{2}}}\color{mathred}{\:}}{\mathrm{\color{mathred}{3}}}\color{mathred}{\:}{\color{mathred}{u}\color{mathred}{n}\color{mathred}{i}\color{mathred}{t}\color{mathred}{s}}^{\mathrm{\color{mathred}{3}}} \color{mathred}{\:}\color{mathblue}{\bullet}\color{mathblue}{\:} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com