Answer to Question #92947 in Geometry for MECHELLE

Question

Two regular quadrilateral vinyl tiles each of 1ft side overlap each other such that overlapping region is a regular octagon. What is the area of the overlapping region?

Answer 1

Two regular quadrilaterals (squares) ABCD and EFGH each of 1 ft side: AB=BC=CD=DA=EF=FG=GH=HE=1 ft, angle A=B=C=D=E=F=G=H=90^o. Squares ABCD and EFGH overlap each other such that overlapping region is a regular octagon IJKLMNPR (see Pic.).

The angle $\angle$ AKL is the external angle for the angle $\angle$ LKJ. Angle $\angle$ ALK is the external angle for the angle $\angle$ KLM. External angle + angle = 180 °. Since all angles (in particular, the angle $\angle$ LKJ = $\angle$ KLM=$\angle$ LMN=135 ° ) in the regular octagon are equal, then the external angles $\angle$ AKL=$\angle$ ALK =$\angle$ FLM= $\angle$FML= 45 ° are equal.

In the triangle △ ALK, two angles are equal, and the third angle is right angle. Triangle △ ALK is a isosceles right triangle.  In the triangle △ FLM, two angles are equal, and the third angle is right angle. Triangle △ FLM is a isosceles right triangle. All sides (in particular, KL=LM) in the regular octagon are equal, Triangles $\triangle$ ALK and $\triangle$ FLM are congruent by ASA theorem, hence AL=FL, AK=FM. The resulting triangles △ ALK, LFM, MBN, NGP, PCR, RHI, IDJ, JEK are isosceles right triangles: 

sides KA=AL=LF=FM=MB=BN=NG=GP=PC=CR=RH=HI=ID=DJ=JE=EK=leg.

The formula for finding the area of a regular octagon:

area of a regular octagon = 8 * base * height / 2.

OS is a height of the regular octagon. The height OS is half the side of the square ABCD:

OS=AB/2=1/2=0.5 ft.

NP is the base of a regular octagon. 

From the isosceles right triangle NGP, we express the base NP using the side NG (leg), applying the Pythagorean theorem:

NP is hypotenuse of triangle NGP;

leg²+leg²=hypotenuse²

2*leg²=hypotenuse²

$leg=\frac{1}{\sqrt{2}} *hypotenuse$

Express the side of square BC using the base of octagon NP and the sides of the triangles BN, PC (BN = PC = NG=leg):

BN+NP+PC=1 ft

leg+hypotenuse+leg=1

2*leg+hypotenuse=1.

Substitute the resulting expression obtained for leg into the previous equation:

$2*\frac{1}{\sqrt{2}} *hypotenuse+hypotenuse=1$

$hypotenuse*({\sqrt{2}}+1)=1$

$hypotenuse=\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}-1}{(\sqrt{2}-1)*(\sqrt{2}+1)}=$

$=\frac{\sqrt{2}-1}{2-1}=\sqrt{2}-1$=0.414 ft is the base of a regular octagon.

The formula for finding the area of a regular octagon:

Area of a regular octagon = 8 * base * height / 2 = 8 * 0.414 * 0.5/2 =0.828 ft².

Answer: the area of the overlapping region is 0.828 ft².