Chapter 5.5, Problem 39E

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Chapter
Section

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

# For equilateral ∆ R S T , R T = 6 . Squares and triangles on the sides form hexagon A B C D E F . Find the perimeter of A B C D E F .

To determine

To find:

The perimeter of the hexagon ABCDEF formed by squares and triangles on sides with an equilateral triangle RST, RT=6.

Explanation

Definition:

The perimeter of any figure is given by the sum of the total length of its boundary

Calculation:

Given,

RST is an equilateral triangle with RT=6

STDE, TRBC and RSFR are squares

RAB, TCD and SEF are triangles.

Since, RST is an equilateral triangle all the three sides of the triangle are congruent.

RS=ST=RT=6 units.

As such each of the squares STDE, TRBC and RSFR are formed with one of ifs sides in common with the equilateral triangle each side of all the three squares measures 6 units.

Further, two of the adjacent sides of the triangle âˆ†RAB, âˆ†TCD and âˆ†SEF are formed by the sides of the square which in turn makes the triangle as an isosceles triangle with equal sides measure 6 units.

The interior angles formed at the vertex R, S and T of the triangle âˆ†RAB, âˆ†TCD and âˆ†SEF is give by

=180Â°- the interior angle of the equilateral âˆ†

=180Â°-60Â°

=120Â°

Now, consider one of the isosceles âˆ†RAB and draw RLÂ¯âŠ¥ABÂ¯, which bisects AB and also the angle âˆ R

Thus, the obtained two right triangle is of the type 30Â°-60Â°-90Â° type with

AR=RB=6= Hypotenuse (Opposite to 90Â°)

AL=BL= Longer leg (Opposite to 60Â°)

RL= shorter leg (Opposite to 30Â°)

By the application of 30-60-90 theorem

30Â°-60Â°-90Â° theorem

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