We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website.
Please click on "Accept cookies" if you agree to the setting of cookies. Cookies that do not require consent remain unaffected by this, see
cookie policy and privacy policy.
DECLINE COOKIES

1: How many sides would there be in a convex polygon if the sum of all but one of its interior angles is 1070 degrees?

2: Given that BDEF is a square and AB = BC = 1, find the number of square units in the area of the regular octagon.

Thanks!

AnonymousConfusedGuy Apr 4, 2018

#1**+1 **

2)

In triangle ABC, Hypotenuse =1^2 + 1^2 =sqrt(2)

Area of Octagon=2[1 + sqrt(2)]*S^2, where S =One side =Sqrt(2)

A =2*2[1 + sqrt(2)]

**A =4[1 + sqrt(2)] units^2 in octagon.**

Guest Apr 4, 2018

#2**+1 **

1) I will take a crack at this one!!

If we try using 1070 degrees as the sum of the interior angles of a convex polygon, we get:

1070 =[n - 2] x 180, where n = number of sides

1070 =180n - 360

180n =1070+360

180n =1430

n = 1430 / 180

n = 7.94 number of sides. Since this is not a whole number, will simply round it up to** 8.**

So, the convex polygon is an **octagon** with sum of its interior angles:

[8 - 2] x 180 =**1080 degrees.**

Guest Apr 4, 2018