Exercise 24 Page 626

The area of a rhombus is half the product of its diagonals. Recall that the diagonals of a rhombus are perpendicular and that they bisect each other. Therefore, we know that the horizontal diagonal's right half also is 8 inches.

Let's consider the right triangle in the upper right-hand side of the rhombus.

We can see that the hypotenuse is c=10 and that one of the legs is b=8. We will substitute these values into the Pythagorean Theorem to obtain the value of the missing leg a.

a^2+b^2=c^2

SubstituteII

b= 8, c= 10

a^2+ 8^2= 10^2

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Solve for a

CalcPow

Calculate power

a^2+64=100

SubEqn

LHS-64=RHS-64

a^2=36

SqrtEqn

sqrt(LHS)=sqrt(RHS)

a=6

We found that a=6. Note that when solving the equation we kept the principal root, because a represents the length of a side and therefore is a positive number. Again we will use that the diagonals bisect each other.

By using the Segment Addition Postulate we can find the lengths of the diagonals. d_1: 8+8= 16in. d_2: 6+6= 12in. We now know that the lengths of the diagonals are 16 meters and 12 inches. Finally, to find the area of the rhombus we will substitute these values in the formula for the area of a rhombus.

A=1/2d_1d_2

SubstituteII

d_1= 16, d_2= 12

A=1/2( 16)( 12)

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Evaluate right-hand side

Multiply

Multiply

A=1/2(192)

MoveRightFacToNumOne

1/b* a = a/b

A=192/2

CalcQuot

Calculate quotient

A=96

We found that the area of the rhombus is 96in^2.