Exercise 12 Page 642

We are asked to find the area of the given figure. Since it has one pair of parallel sides, we know it is a trapezoid.

Recall that the area of a trapezoid is one half the product of the height and the sum of the bases. Notice we are given the bases lengths but we are missing the height.

Height

We can see that the height divides the trapezoid into two figures, a rectangle and a right triangle. Recall that opposite sides in a rectangle are the same length. We can use that fact to find the length of one of the legs in the triangle.

We found that one of the triangle's legs is 3 ft. We also know that its hypotenuse is 5ft. Keep in mind that the other missing leg is the trapezoid's height.

We can find the length of the other leg by substituting a= 3 ft, b= h, and c= 5 ft into the Pythagorean Theorem. Let's do it!

a^2+b^2=c^2

SubstituteValues

Substitute values

3^2+ h^2= 5^2

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Solve for h

CalcPow

Calculate power

9+h^2=25

SubEqn

LHS-9=RHS-9

h^2=16

SqrtEqn

sqrt(LHS)=sqrt(RHS)

h=sqrt(16)

CalcRoot

Calculate root

h=4

Recall that h represents a side of a triangle and the height of the trapezoid. Therefore, we only kept the principal root when solving the equation because h must be positive. The height of the trapezoid is 4 ft.

Area

From the diagram, we can see that the base lengths are 5 ft and 8 ft. We also found that the height is 4 ft. Having the height and the lengths of the bases, we can substitute them into the formula for the area of a trapezoid.

A=1/2h(b_1+b_2)

SubstituteValues

Substitute values

A=1/2( 4)( 5+ 8)

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Evaluate right-hand side

AddTerms

Add terms

A=1/2(4)(13)

Multiply

Multiply

A=1/2(52)

MoveRightFacToNumOne

1/b* a = a/b

A=52/2

CalcQuot

Calculate quotient

A=26

The area of the trapezoid is 26ft^2.