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Geometry 16 Online
OpenStudy (anonymous):

Is there a ploygon which has only 2 types of interior angles 120 & 60 degree? Is so how many sides such a polygon has ?

ganeshie8 (ganeshie8):

120*2 + 60*2 = 240 + 120 = 360 so a quadrilateral is possible ? there can be many other polygons, just setup equations and solve them

OpenStudy (anonymous):

But y to multiply 2 ?

ganeshie8 (ganeshie8):

120*2 + 60*2 this expression tells, i hav picked a polygon wid two 120 angles, and two 60 angles

ganeshie8 (ganeshie8):

|dw:1383575072479:dw|

ganeshie8 (ganeshie8):

^^ above quadrilateral, i see oly two type of angles : 60 and 120

OpenStudy (anonymous):

thank u

ganeshie8 (ganeshie8):

np :)

OpenStudy (anonymous):

120*2 + 60*2 = 240 + 120 = 360 so a quadrilateral is possible Can we solve only this much ?

ganeshie8 (ganeshie8):

question is asking us, if a polygon is possible. we showed that a quadrilateral is possible. so we're done. ofcourse we can pursue analyzing wat all kinds of polygons are possible...

OpenStudy (anonymous):

Is so how many sides such a polygon has ?

OpenStudy (anonymous):

Sorry that was little mistaken this is the actual question Is there a ploygon which has only 2 types of interior angles 120 & 60 degree? Is so how many sides such a polygon has ? Please solve it completely sir .I am completely confused ?

ganeshie8 (ganeshie8):

here is the equation to solve :- 60x + 120y = 180(n-2) you need to solve x, y, and n in positive integers n = x+y

ganeshie8 (ganeshie8):

60x + 120y = 180(n-2) x+ 2y = 3(n-2) x+2y = 3(x+y-2) simplify

OpenStudy (anonymous):

Hint :-If n is the no of sides & if k angles are of 60 degree,then n-k angles are of 120 degree and (n-2)180=60k+120(n-k)

ganeshie8 (ganeshie8):

did u get my previous reply ?

OpenStudy (anonymous):

Really no

ganeshie8 (ganeshie8):

60x + 120y = 180(n-2) x+ 2y = 3(n-2) x+2y = 3(x+y-2) x + 2y = 3x + 3y - 6 2x + y = 6

ganeshie8 (ganeshie8):

so, the possible positive values for x, y are :- x = 0, y = 6 (hexagon) x = 1, y = 4 (pentagon) x = 2, y = 2 (quadrilateral) x = 3, y = 0 (triangle)

OpenStudy (anonymous):

answer is 4 or 5. But teach me that how to solve that ?

ganeshie8 (ganeshie8):

answer is 4 or 5, if we want both types of angles to exist in our polygon - as can be seen in above table.

ganeshie8 (ganeshie8):

go thru the above posts, let it sink a bit.... and then come back if u still get stuck :)

OpenStudy (anonymous):

I am really happy because i really got ot by u.. THANK U

ganeshie8 (ganeshie8):

Sure ?

OpenStudy (anonymous):

how n=x+y

ganeshie8 (ganeshie8):

good question :) msg me when you're online

OpenStudy (anonymous):

ok

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