# ABCD is a rectangle with AB = 22 units and BC = 26 units. Four congruent triangles are cut off from the corners of ABCD to get a regular octagon.What is the length of the side of the octagon?

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### 3 Answers

Since the congruent triangles are in the corners of the rectangle, then they are right angled triangles. To get a regular octagon, the sides of the two triangles which lies on a particular line, must be equal. Therefore if the sides of the triangle are a,b and c with c is the hypotenuse and b is the largest of other two, the side b lies in the BC and and the opposite line.

Then if the octagon formed is regular, the sides must be equal to each other.

`26-2b = 22-2a = c`

Therfore,

`26-2b = 22-2b`

`13-b = 11-a`

`a = b-2`

Since they form a right angle triangle,

`a^2 + b^2 = c^2`

`(b-2)^2+b^2 = (26-2b)^2`

`b^2-4b+4+b^2 = 4(13-b)^2`

`2b^2-4b+4 = 4(169-26b+b^2)`

`b^2-2b+2 = 2(169-26b+b^2)`

`b^2-2b+2 = 338 - 52b + 2b^2`

This give a quadratic equation,

`b^2-50b+336 = 0`

The answers are,

`b = 42 or b = 8`

b cannot be greater than 26, therefore b is 8.

Therefore,

The side of a octagon is `c = 26-2b`

`c = 26 -2xx8`

`c = 10`

**The answer is 10 units.**

i also got the answer through the methods explained above, but the problem is that if its a regular octagon, then each interior angle of regular octagon is 135 degrees, then the right-angle triangles cut at the four corners would be 90-45-45, then answer does not come from that approach.

Any idea, can someone elucidate?

Given ABCD is a rectangle with side AB=22 and BC=26. When Four congurent triangles are cut from the four cotners A,B,C and D, then each triangle will be a right angled triangle. Let the regular octagon obtaind after cutting the four triangles be **EFGHIJKL** with the edges E and F on the longer side BC, edges G and H on the shorter side CD, edges I and J on the longer side DA and the edges K and L on the shorter side AB. The four right right triangles on the corners be FCG, HDI, JAK and LBE. Now to find the length of the side of regular octagon EFGHIJKL let is asume it be **x**. ** i.e the side EF=FG=GH=HI=IJ=JK=KL=x , Again **the corresponding sides of the congurent triangles will also be equal. ** i.e. BE=FC=DI=AJ and the corresponding sides CG=HD=AK=BL** Since we have assume EF=x , therefore , BE=FC=DI=AK= (26-x)/2 , similarly since GH=KL=x therefore, CG=HD=AK=BL= (22-x)/2, Next We take any right angled triangle say FCG where Its three sides are Fc=(26-x)/2, CG=(22-x)/2 and the hypotenuse FG= x

Applying pythagoras theorem : h^2=p^2+b^2 we get

x^2 = ((26-x)/2)^2 + ((22-x)/2)^2

x^2 =((26)^2- 2.26.x+x^2)/4 + ((22)^2- 2.22.x+x^2)/4

4x^2= (26)^2 - 52x+ x^2 + (22)^2 - 44x+ x^2

4x^2= 676 - 96x + 484 + 2(x^2)

4x^2- 2x^2=1160 - 96x

2x^x + 96x - 1160 =0

x^2 + 48x - 580 = 0

x^2 + 58x - 10x -560 =0

x(x+58) -10(x+58)=0

(x-10)(x+58)=0

therefore x = 10 and x= -58 .

Since side cannot be negative hence we take x = 10

The side of the octagon = 10 **Answer**