Exercise 53 Page 628

The area of a triangle is half of the product of its base and its height. The given triangle is regular and for any regular polygon it is true that all sides are congruent and all angles are congruent.

We have been given that the triangle's height is 10 meters. Next we want to find m∠ A. We will use the Interior Angles Theorem and that we know that all angles are congruent. By the definition of congruence, the angles have the same measure.

Let's consider the right triangle with the acute angle A.

By using the sine ratio we can calculate the length of the hypotenuse of this triangle. Let's do it!

sin(θ)=Opp./Hyp.

SubstituteValues

Substitute values

sin(60^(∘))=10/b

MultEqn

LHS * b=RHS* b

sin(60^(∘))(b)=10

DivEqn

.LHS /sin(60^(∘)).=.RHS /sin(60^(∘)).

b=10/sin(60 ^(∘))

We can write the length of b in radical form by using the properties of special right triangles. b=10/sin(60 ^(∘)) ⇕ b=10/sqrt(3)2=20/sqrt(3) Let's now use that the sides of the triangle are congruent. By the definition of congruence they have the same length.

The triangle's base is 20sqrt(3) meters long. We now know both the base and the height of the triangle. We can substitute these two values in the formula for the area of a triangle and simplify.

A=1/2bh

SubstituteII

b= 20/sqrt(3), h= 10

A=1/2( 20/sqrt(3) )(10)

▼

Evaluate right-hand side

CommutativePropMult

Commutative Property of Multiplication

A=1/2(10)(20/sqrt(3) )

Multiply

Multiply

A=(5)(20/sqrt(3) )

MoveLeftFacToNum

a*b/c= a* b/c

A=(5)20/sqrt(3)

Multiply

Multiply

A=100/sqrt(3)

The area of the triangle is 100sqrt(3)m^2.