a Let's start with calculating the area of the base of the larger pan. Then, we will find the scale factor of the hexagonal bases and use it to determine the side length of the smaller pan.

Area of the Larger Pan

It is given that the base of the larger pan is a regular hexagon with a side length of

4.5

inches.

To find its area, let's consider the diagram below.

As we can see, the hexagon is inscribed in a circle, so $\overline{O K}$ and $\overline{O M}$ are the radii of the circle. They have the same length, which allows us to conclude that $△ K O M$ is an isosceles triangle. Hence, by the Isosceles Triangle Theorem, angles $∠ O K M$ and $∠ O M K$ are congruent.

Now we will calculate the measure of the central angle

∠ K O M .

Let's recall that the measure of each central angle of a regular

n -

gon is

\frac{3 6 0}{n} .

Substituting

n

with

6,

we can calculate

m ∠ K O M .

m ∠ K O M = \frac{3 6 0}{6} = 6 0^{\circ}

Next, we can use the fact that the sum of the angle measures in any triangle is

18 0^{\circ}

to determine the measures of

∠ O K M

and

∠ O M K .

m ∠ O K M + m ∠ O M K + m ∠ K O M = 18 0^{\circ} ⇓ m ∠ O K M + m ∠ O M K + 6 0^{\circ} = 18 0^{\circ}

Since

∠ O K M

and

∠ O M K

are congruent their measures are the same, so we can substitute

m ∠ O K M

with

m ∠ O M K .

m ∠ O M K + m ∠ O M K + 6 0^{\circ} = 18 0^{\circ}

Let's solve this equation!

m ∠ O M K + m ∠ O M K + 6 0^{\circ} = 18 0^{\circ}

▼

Solve for

m ∠ O M K

AddTerms

Add terms

2 m ∠ O M K + 6 0^{\circ} = 18 0^{\circ}

SubEqn

$LHS - 6 0^{\circ} = RHS - 6 0^{\circ}$

2 m ∠ O M K = 12 0^{\circ}

DivEqn

$LHS / 2 = RHS / 2$

m ∠ O M K = 6 0^{\circ}

Angle

∠ O K M

also measures

6 0^{\circ} .

The measure of each angle in

△ O K M

6 0^{\circ},

which allows us to conclude that the triangle is equilateral. Hence, we can calculate its area using the following formula.

A = \frac{s ^{2} 3}{4}

Here

s

represents the side length of the triangle, which in our case is

4.5

inches. Let's substitute

s

with

4.5

and solve the equation for

A .

A = \frac{s ^{2} 3}{4}

▼

Substitute

4.5

for

s

and evaluate

Substitute

$s = 4.5$

A = \frac{4 . 5 ^{2} 3}{4}

CalcPow

Calculate power

A = \frac{2 0 . 2 5 3}{4}

UseCalc

Use a calculator

A = 8.768507 \dots

RoundDec

Round to $2$ decimal place(s)

A \approx 8.77

We got that the area of

△ O K M

is approximately

8.8

square inches.

From the diagram we can see that the hexagon consists of six identical triangles, each with an area of

8.76

square inches. Therefore, by multiplying this value by

6

we can find the area of the whole hexagon.

A_{hexagon} = 6 \cdot 8.77 = 52.62 \approx 52.6 in^{2}

The Side Length of the Smaller Pan

We are told that the area of the base of the smaller pan is $41.6$ square inches. To find its side length we will use Theorem $11 - 1,$ which states the following.

Theorem $11$ - $1$
If two polygons are similar, then their areas are proportional to the square of the scale factor between them.

By the theorem, the ratio of the area of the small and large pan is equal to the square of the scale factor

k

between them.

\frac{A _{small}}{A _{large}} = k^{2}

Let's substitute

A_{small}

with

41.6

and

A_{large}

with

52.6

and find the value of

k .

\frac{A _{small}}{A _{large}} = k^{2}

SubstituteII

$A_{small} = 41.6$ , $A_{large} = 52.6$

\frac{4 1 . 6}{5 2 . 6} = k^{2}

▼

Solve for

k

RearrangeEqn

Rearrange equation

k^{2} = \frac{4 1 . 6}{5 2 . 6}

CalcQuot

Calculate quotient

k^{2} = 0.790874 \dots

SqrtEqn

$LHS = RHS$

k = 0.889311 \dots

RoundDec

Round to $2$ decimal place(s)

k = 0.88

Let's recall that a scale factor is equal to the ratio of the corresponding sides of the polygons.

\frac{s _{small}}{s _{large}} = k

We are given that the side length of the larger pan is

4.5

inches. Let's substitute this value for

s_{large}

and

k

with

0.88

and calculate the side length of the smaller pan.

\frac{s _{small}}{s _{large}} = k

SubstituteII

$s_{large} = 4.5$ , $k = 0.88$

\frac{s _{small}}{4 . 5} = 0.88

MultEqn

$LHS \cdot 4.5 = RHS \cdot 4.5$

s_{small} = 3.96

RoundInt

Round to nearest integer

s_{small} \approx 4

We conclude that the side length of the smaller pan is approximately

4

inches.

Exercise

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