Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 111284 by Aina Samuel Temidayo last updated on 03/Sep/20

Find the maximum area of a triangle whose vertices lie on a regular hexagon of unit area.

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{area}\:\mathrm{of}\:\mathrm{a}\:\mathrm{triangle} \\ $$$$\mathrm{whose}\:\mathrm{vertices}\:\mathrm{lie}\:\mathrm{on}\:\mathrm{a}\:\mathrm{regular} \\ $$$$\mathrm{hexagon}\:\mathrm{of}\:\mathrm{unit}\:\mathrm{area}. \\ $$

Commented by mr W last updated on 03/Sep/20

Commented by mr W last updated on 03/Sep/20

max. triangle =((hexagon)/2)=(1/2)

$${max}.\:{triangle}\:=\frac{{hexagon}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

Just explain please.

$$\mathrm{Just}\:\mathrm{explain}\:\mathrm{please}. \\ $$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

What about a well explanatory solution? Thanks.

$$\mathrm{What}\:\mathrm{about}\:\mathrm{a}\:\mathrm{well}\:\mathrm{explanatory} \\ $$$$\mathrm{solution}?\:\mathrm{Thanks}. \\ $$

Commented by Her_Majesty last updated on 03/Sep/20

what is it you do not understand???

$${what}\:{is}\:{it}\:{you}\:{do}\:{not}\:{understand}??? \\ $$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

Isn′t there a proof? How do we know definitely that the area of that triangle is half of that of the hexagon?

$$\mathrm{Isn}'\mathrm{t}\:\mathrm{there}\:\mathrm{a}\:\mathrm{proof}?\:\mathrm{How}\:\mathrm{do}\:\mathrm{we}\:\mathrm{know} \\ $$$$\mathrm{definitely}\:\mathrm{that}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{that} \\ $$$$\mathrm{triangle}\:\mathrm{is}\:\mathrm{half}\:\mathrm{of}\:\mathrm{that}\:\mathrm{of}\:\mathrm{the}\:\mathrm{hexagon}? \\ $$

Commented by mr W last updated on 03/Sep/20

i just don′t want to waste my time for obvious things.

$${i}\:{just}\:{don}'{t}\:{want}\:{to}\:{waste}\:{my}\:{time}\:{for} \\ $$$${obvious}\:{things}. \\ $$

Commented by Her_Majesty last updated on 03/Sep/20

it′s plain to see that the red lines cut the small triangles in two, isn′t it?

$${it}'{s}\:{plain}\:{to}\:{see}\:{that}\:{the}\:{red}\:{lines}\:{cut}\:{the} \\ $$$${small}\:{triangles}\:{in}\:{two},\:{isn}'{t}\:{it}? \\ $$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

Not really. We can′t just assume that.

$$\mathrm{Not}\:\mathrm{really}.\:\mathrm{We}\:\mathrm{can}'\mathrm{t}\:\mathrm{just}\:\mathrm{assume}\:\mathrm{that}. \\ $$

Commented by Aina Samuel Temidayo last updated on 03/Sep/20

How can it be proven and solved mathematically?

$$\mathrm{How}\:\mathrm{can}\:\mathrm{it}\:\mathrm{be}\:\mathrm{proven}\:\mathrm{and}\:\mathrm{solved} \\ $$$$\mathrm{mathematically}? \\ $$

Commented by Her_Majesty last updated on 03/Sep/20

two equilateral squares form a rhombus the diagonals each bisect the rhombus the red line is the longer diagonal we don′t need more

$${two}\:{equilateral}\:{squares}\:{form}\:{a}\:{rhombus} \\ $$$${the}\:{diagonals}\:{each}\:{bisect}\:{the}\:{rhombus} \\ $$$${the}\:{red}\:{line}\:{is}\:{the}\:{longer}\:{diagonal} \\ $$$${we}\:{don}'{t}\:{need}\:{more} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com