Exercise 51 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. Let's use this in our triangle.

Now we know that the triangle's base is 10 cm. Next we want to find m∠ C. We will use the Interior Angles Theorem and the fact that we know that all angles are congruent.

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

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

sin(θ)=Opp./Hyp.

SubstituteValues

Substitute values

sin(60^(∘))=h/10

MultEqn

LHS * 10=RHS* 10

sin(60^(∘))(10)=h

RearrangeEqn

Rearrange equation

h=sin(60^(∘))(10)

We can write the length of h in radical form by using the properties of special right triangles. h=sin(60^(∘))(10) ⇕ h=sqrt(3)/2(10)= 5sqrt(3) 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= 10, h= 5sqrt(3)

A=1/2(10)( 5sqrt(3))

▼

Evaluate right-hand side

MoveRightFacToNumOne

1/b* a = a/b

A=(10)(5sqrt(3))/2

SimpQuotProd

Simplify quotient and product

A=25sqrt(3)

The area of the triangle is 25 sqrt(3)cm^2.