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(#M40018196) ELITMUS question ELITMUS MATHS PUZZLE Keep an EYE

the interior angles of an octagon ABCDEFGH are in AP. If the largest and the second largest have an average of 153 find the average of the least two.

Asked In Elitmus pavan (12 years ago)

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Rohit  (12 years ago)

117

sum of interior angles = 4*360-360=1080 deg
if a is smallest angle and d is difference between two angles in an AP, then
average of largest and second largest angle = (a+6da+7d)/2= 153
2a+13d= 306
sum of angles = (8/2)*(2a+7d)= 1080 or 2a+7d= 270

solving, a= 114, d=6
average of least two angles= (114+120)/2= 117

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django  (12 years ago)

117...sum of interior angles=(2n-4)right angles...this is one condn.....other condn is given in question...2 eq 2 var..

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Nitin Maheshwari  (9 years ago)

117
x+(x+d)+(x+2d)+(x+3d)+(x+4d)+(x+5d)+(x+6d)+(x+7d)=1080
8x+28d=1080
2x+7d=270...................................................................................................eq1
and he largest and the second largest have an average of 153

[(x+6d)+(x+7d)]/2=153
(x+6d)+(x+7d)=306
2x+13d=306.................................eq2
solve above 2 eq
d=7
average of the least two.=117

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