Areas of Parallelograms and Triangles LESSON 10-1 Additional Examples

Find the area of the parallelogram.

You are given two sides with lengths 12 m and 10.5 m and an altitude that measures 8 m to the side that measures 12 m. Choose the side with a corresponding height to use as a base.

A

=

b h A

= 12 ( 8 ) Area of a parallelogram Substitute 12 for

b

and 8 for

h

.

A

= 96 Simplify.

The area of the parallelogram is 96 m 2 .

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Areas of Parallelograms and Triangles LESSON 10-1 Additional Examples

A parallelogram has 9-in. and 18-in. sides. The height corresponding to the 9-in. base is 15 in. Find the height corresponding to the 18-in. base.

Find the area of the parallelogram using the 9-in. base and its corresponding 15-in. height.

A A A

=

b h

= 9 ( 15 ) = 135 Area of a parallelogram Substitute 9 for

b

and 15 for

h

.

Simplify.

The area of the parallelogram is 135 in.

2

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Areas of Parallelograms and Triangles LESSON 10-1 Additional Examples (continued)

Use the area 135 in.

2 to find the height to the 18-in. base.

A

=

b h

135 = 18

h

135 18 =

h

7.5 =

h

Area of a parallelogram Substitute 135 for

A

Simplify.

and 18 for

b

.

Divide each side by 18.

The height corresponding to the 18-in. base is 7.5 in.

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Areas of Parallelograms and Triangles LESSON 10-1 Additional Examples

Find the area of

XYZ

.

1

A

=

b h

2

A

1 = ( 30 )( 13 ) 2

A

= 195 Area of a triangle

XYZ

has area 195 cm 2 .

Substitute 30 for Simplify.

b

and 13 for

h

.

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Areas of Parallelograms and Triangles LESSON 10-1 Additional Examples

The front of a garage is a square 15 ft on each side with a triangular roof above the square. The height of the triangular roof is 10.6 ft. To the nearest hundred, how much force is exerted by an 80 mi/h wind blowing directly against the front of the garage? Use the formula

F

= 0.004

Av

2 .

Draw the front of the garage, and then use the area formulas for rectangles and triangles to find the area of the front of the garage.

Area of the square:

bh

= 15 2 Area of the triangular roof: 1 2 = 79.5

ft 2 =

bh

225 2 ft 2 = (15)(10.6) The total area of the front of the garage is 225 + 79.5

= 304.5 ft 2 .

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Areas of Parallelograms and Triangles LESSON 10-1 Additional Examples (continued)

Find the force of the wind against the front of the garage.

F

= 0.004

A v

2

F

= 0.004( 304.5

)( 80 ) 2 Use the formula for force.

Substitute 304.5 for

A

and 80 for

v

.

A A

= 7795.2

7800 Simplify.

Round to the nearest hundred.

An 80 mi/h wind exerts a force of about 7800 lb against the front of the garage.

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