Exercise 35 Page 627

The area of a kite is half the product of its diagonals. Recall that a kite is a quadrilateral with exactly two pairs of consecutive congruent sides and that the diagonals of a kite are perpendicular.

Let's consider the right triangle with an acute angle of 45^(∘).

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(45^(∘))=a/9sqrt(2)

▼

Solve for a

MultEqn

LHS * 9sqrt(2)=RHS* 9sqrt(2)

sin(45^(∘))* (9sqrt(2))=a

UseCalc

Use a calculator

9=a

RearrangeEqn

Rearrange equation

a=9

We will also calculate the length of the adjacent side. For this we will make use of the cosine ratio.

cos(θ)=Adj./Hyp.

SubstituteValues

Substitute values

cos(45^(∘))=b/9sqrt(2)

▼

Solve for b

MultEqn

LHS * 9sqrt(2)=RHS* 9sqrt(2)

cos(45^(∘))* (9sqrt(2))=b

UseCalc

Use a calculator

9=b

RearrangeEqn

Rearrange equation

b=9

For kites it is true that one of the diagonals bisects the other. Therefore, we know that the two halves of the horizontal diagonal must be congruent. By the definition of congruence they have the same length. Let's use this in our kite.

By using the Segment Addition Postulate we can now find the lengths of the diagonals. d_1: 9+9= 18m d_2: 9+6=15m Using the diagonal lengths, let's find the area of the kite.

A=1/2d_1 d_2

SubstituteII

d_1= 18, d_2= 15

A=1/2( 18)(15)

▼

Evaluate right-hand side

MoveRightFacToNumOne

1/b* a = a/b

A=(18)(15)/2

CalcQuotProd

Calculate quotient and product

A=135

The area of the kite is 135m^2. Therefore, the correct area is C.