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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

5. Solving Square Root and Other Radical Equations

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Exercise 45 Page 395

The six triangles make one hexagon.

Approximately $13.16 cm$

Practice makes perfect

Looking at the diagram, we see that the height of the triangle at the bottom is $\frac{s 3}{2},$ and one side of it is $s .$ For the hexagon to be regular, the triangles must all be equilateral and, therefore, congruent. Let's first show that the triangle at the bottom is indeed an equilateral triangle.

The measure of the interior angle of a regular hexagon must be $12 0^{\circ} .$ We see that the angles are bisected by triangles so all triangles have two $6 0^{\circ}$ angles. The Interior Angles Theorem tells us that the triangles must be equilateral then because $6 0^{\circ} + 6 0^{\circ} + 6 0^{\circ} = 18 0^{\circ} .$

Each triangle is an equilateral triangle with a side of

s .

Area

A_{T}

of an equilateral triangle with a side of

s

can be expressed as below.

A_{T} = \frac{s ^{2} 3}{4}

Since the hexagon consists of six equilateral triangles, we can find the area

A_{H}

of the hexagon by multiplying

A_{T}

by six.

A_{H} = 6 A_{T} \Leftrightarrow A_{H} = \frac{6 s ^{2} 3}{4}

Therefore, the formula for the area of the hexagon is

\frac{6 s ^{2} 3}{4} .

We know the area

A_{H}

of the hexagon, let's find the length of one side of the hexagon using the formula we found above.

A_{H} = \frac{6 s ^{2} 3}{4}

$A_{H} = 450$

450 = \frac{6 s ^{2} 3}{4}

▼

Solve for

s

$LHS \cdot 4 = RHS \cdot 4$

1800 = 6 s^{2} 3

$LHS / 63 = RHS / 63$

\frac{1 8 0 0}{6 3} = s^{2}

$\frac{a}{b} = \frac{a / 6}{b / 6}$

\frac{3 0 0}{3} = s^{2}

$\frac{a}{b} = \frac{a \cdot 3}{b \cdot 3}$

\frac{3 0 0 3}{3} = s^{2}

Calculate quotient

1003 = s^{2}

Upon reaching this state, we will need to find the square root of both sides of the equation in order to isolate

s

further.

1003 = s^{2}

$LHS = RHS$

1003 = s

▼

Solve for

s

Use a calculator

13.16074 \dots = s

Round to $2$ decimal place(s)

13.16 = s

Rearrange equation

s = 13.16

The length of each side of the hexagon is about

13.16 cm .

Subchapter links

Lesson Check p.394

Practice and Problem-Solving Exercises p.395-397

Standardized Test Prep p.397