1.)A regular hexagon has side length $6$. If the perimeter and area of the hexagon are $p$ and $A$, respectively, what is the value of $\frac{p^4}{A^2}$?
2.)Isosceles triangle $OPQ$ has legs $OP = OQ$, base $PQ = 2$, and $\angle POQ = 45^\circ$. Find the distance from $O$ to $\overline{PQ}$.
3.)$ABCDEF$ is a regular hexagon with area $1$. The intersection of $\triangle ACE$ and $\triangle BDF$ is a smaller hexagon. What is the area of the smaller hexagon?
4.)$A,B,C,D,$ and $E$ are points on a circle of radius $2$ in counterclockwise order. We know $AB=BC=DE=2$ and $CD=EA.$ Find $[ABCDE]$.
+128475 1.)A regular hexagon has side length $6$. If the perimeter and area of the hexagon are $p$ and $A$, respectively, what is the value of $\frac{p^4}{A^2}$?
The perimeter = 6*6 = 36
The area is composed of 6 equilateral triangles and is given by :
6 * (1/2) * (6)^2 * √3/2 =
3√3 / 2 * 36 =
54√3
So
p^4 36^4 (36)(36)(36)(36) (2)(2) (36) (36) (4)(36)(36)
___ = _______ = ______________ = _____________= ________ =
A^2 54^2 * 3 (54) (54) (3) (3)(3)(3) 27
(4)(4)(36)
________ = 16 * 12 = 192
3
+128475 2.)Isosceles triangle $OPQ$ has legs $OP = OQ$, base $PQ = 2$, and $\angle POQ = 45^\circ$. Find the distance from $O$ to $\overline{PQ}$.
Bisect base PQ
Call the point of bisection M
Then angle OMQ = 90
Angle MOQ = 22.5
MQ = 1
And angle OQM = 67.5
So....by the Law of Sines
OM/sin 67.5 = 1/sin (22.5)
OM = sin 67.5 / sin 22.5 = cos 22.5 / sin22.5 = cot 22.5 = ( 1 + cos 45) / sin 45 =
1 + √2/2 1 2
______ = ___ + 1 = ___ + 1 = √2 + 1
√2/2 √2/2 √2
+128475 3.)$ABCDEF$ is a regular hexagon with area $1$. The intersection of $\triangle ACE$ and $\triangle BDF$ is a smaller hexagon. What is the area of the smaller hexagon?
For simplicity..I have constructed a hexagon with a side of 2
See...the image
A = (1, √3) and C = (1, - √3)
So...the distance from A to C = 2√3
And, by symmetry, GL is (1/3) of this = ( 2/3)√3
And this is one side of the smaller hexagon GHIJKL
So....the ratio of the area of the smaller hexagon to the larger hexagon =
(side of smaller hexagon)^2
______________________ =
(side of larger hexagon)^2
[ (2/3)√3]^2 (4/9) * 3 4 * 3 3 1
___________ = ______ = ______ = ____ = ____
2^2 4 4 * 9 9 3
So..in our case...the area of the smaller hexagon = (1/3) / 1 = 1/3 units^2
+128475 4.)$A,B,C,D,$ and $E$ are points on a circle of radius $2$ in counterclockwise order. We know $AB=BC=DE=2$ and $CD=EA.$ Find $[ABCDE]$.
Call the center of the circle , O
If AB + BC + DE = 2
Then triangles AOB, BOC and DOE must be equilateral.....and their combined area =
3 (1/2) (2)^2 * √3/2 =
3√3 units^2
Since angles AOB, BOC and DOE = 60°
And CD = EA
Then angles COD and EOA must each equal [ 360 - 3(60)] / 2 = 180/ 2 = 90°
So....the total area of triangles COD and DOE = 2 (1/2) * 2^2 = 4 units^2
So....the total area of ABCDE = 4+ 3√3 units^2
