Area Overview
Define:
Area:
Kite:
Parallelogram:
Rectangle:
Rhombus:
Square:
Trapezoid:
Postulates/Theorems:
Every closed region has an area.
If 2 closed figures are congruent, then their areas are equal.
If 2 closed regions intersect only along a common boundary, then the area of their union is equal to the sum of their individual areas. (This means break it up into pieces, find area of each piece and then add areas together.)
Formulas:
Figure Formula Notes
Circle A = πr2 r is the radius
Kite
A = 1
2d d1 2 d’s are the lengths of the diagonals of the kite Parallelogram A = bh h is the perpendicular height to the base Rectangle A = lw or A = bh Length x width or base x height
Rhombus
A = 1
2d d1 2 or A = bh
Square A = s2
Trapezoid
A = 1
2h b( 1b2) Or
A = Mh
h is the altitude and the b’s are the bases.
M is the median of the Trapezoid Triangle
A = 1
2bh h is the altitude to the base Regular Polygon
A = 1
2Pa P is the perimeter of polygon & a is the apothem Sector of Circle
A = N r 360
2 N is the degree of the arc
Ratios of Areas
Theorems
If two figures are similar, then the ratio of their areas will equal the square of the ratio of the corresponding sides.2
2 1 2
1
s s A A
A median of a triangle divides the triangle into two triangles with equal areas.Examples
1. If ∆ABC ~ ∆DEF, find the ratio of the areas of the two triangles.
4 9 8
12 2
2
1
A A
2. If the ratio of the areas of two similar parallelograms is 49:121, find the ratio of their bases.
11 7 121
49
3. AM is the median of ∆ABC. Find the ratio of AABM :AACM 1: 1
Practice
4. Given the similar pentagons shown, find the ratio of their areas.
5. The ratio of corresponding medians of two similar triangles is 5:2. Find the area of the smaller triangle if the larger triangle has an area of 250.
6. Two similar ∆s have areas of 16 and 36. The length of a side of the smaller ∆ is 10ft. find the length
Coordinate Geometry: Areas
Shoelace Method:
Select a vertex to start with. List all coordinates starting and ending with the initial vertex. Be sure to go either clockwise or counterclockwise.
Multiply diagonally in both directions (ie x-coordinate of 1st x y-coordinate of 2nd, etc)
Add each column
A = ½ (sum of column 1 – sum of column 2)
Note: be sure to take the absolute value.
Examples/Practice:
1. Find the area of the following polygon
2. Find the area of a polygon with the following vertices: {(2, 1), (4, 5), (7, 8)}
3. Find the area of the quadrilateral with the following vertices: A(-4,-3), B(2,-3), C(4,-6) and D(-2,-6).
4. Find the area of the quadrilateral with the following vertices: A(0,0), B(4,0), C(5,5) and D(1,5).
Area of Rectangle
Area of Rectangle: bh or lw *
Note: h is the altitude to the baseExamples: Area of Rectangle
1. Given that the area of a rectangle is 20cm2 and the altitude is 5cm, find the base.
2. A rectangle is 4m longer than it is wide. If the area = 252m2, find the length.
Area of Parallelogram
Area of Parallelogram: bh *
Note: h is the altitude to the base (it is not the slanted line)Examples: Area of Parallelogram
3. Find the area of parallelogram whose sides are 14cm and 6cm and whose acute angle = 60º.
Dealing with 30-60-90 ∆, therefore h = ( )6
2 3= 3√3 Area of Parallelogram = bh = 14(3√3) = 42√3 cm2
4. Find the area of parallelogram whose sides are 15cm and 8cm and whose acute angle = 45º.
Dealing with 45-45-90 ∆, therefore h = 8
2 2= 4√2 Area of Parallelogram = bh = 15(4√2) = 60√2 cm2
Practice: Area of Parallelogram
Find the area of the following parallelograms.
5. Sides are 12cm and 18cm and acute angle = 60º.
6. Sides are 8cm and 10cm and acute angle = 45º.
Area of Irregular Shapes
When asked to find the area of irregular shapes, there are two approaches to take
Break it down into smaller shapes and find area of each. Then add all small areas together
Find the area of the whole and subtract the area that is missing.
Find the area of the following
1. 2.
3. 4.
5.
6.
Area of Triangle
Area of Triangle:
A = ½ (bh) *Note: h is altitude to base
Area of equilateral triangle s2 3 4
Note: s is the side of the triangle
Hero’s Formula: s(sa)(sb)(sc) Note: s = 2
) (abc
Examples: Area of Triangle
1. Find the area of each ∆.
a. b. c.
2. Find the base of a ∆ with altitude 15 and area 60u2.
3. Find the area of a triangle with sides 3, 6 and 7.
Practice: Area of Triangle
4. Find the area of each triangle.
a. b. c. Sides = 8cm
5. Area of a triangle is 72in2. If the height is 8in, find the length of the base.
6. Right ∆ has perimeter of 36m, hypotenuse of 15m and a leg of 9m. Find the area of the ∆.
Area of Trapezoid
Area of Trapezoid: ½ h(b
1+ b
2) or Mh
Note: h is altitude to the base Median of Trapezoid (M) = ½(b1 + b2) (average of 2 bases)
Examples: Area of Trapezoid
9. Given a trapezoid with height 7, lower base 18 and upper base 12. Find the area of the trapezoid.
10. The height of a trapezoid is 12. The bases are 6 and 14.
a. find median b. find area
Practice: Area of Trapezoid
11. Find shorter base of a trapezoid if the trapezoid’s area is 52u2, altitude is 8 and its longer base is 10.
12. The bases of a trapezoid are 8 and 22 and the trapezoid’s area is 135u2. Find the height.
Area of Rhombus
Area of Rhombus:
A = 12d d1 2 or A = bh
Examples: Area of Rhombus
13. Given diagonals of 50 and 30, what is the area of the rhombus?
14. Find the area of a rhombus whose perimeter is 20 and whose longer diagonal is 8.
Practice: Area of Rhombus
15. Find the area of a rhombus with diagonals of 9 and 14.
16. A rhombus has a perimeter of 80cm and the length of 1 of the diagonals is 24m. Find the area of the rhombus.
Area of Circles, Sectors, Segments
Area of Circle: πr
2Area of Sector: A =
N r 360 2
Finding area of segment
1. Draw radii to endpoints of segment 2. Find area of sector you’ve created 3. Find area of triangle portion of sector
4. Subtract results from 2 & 3 ( Area of sector – Area of triangle)
Examples: Area of Circles, Sectors & Segments
1. Find area of circle whose diameter = 10. 2. Find circumference of circle whose area is 49π sq units.
3. Find area of sector with radius of 12 and 45º arc.
4. Find area of sector with radius of 2 and 90º arc.
5. The measure of arc of segment is 90º. The radius of circle is 10. Find the area of the segment.
Area of segment = area of sector AOB – area of ∆AOB Area of sector =
Area of triangle =
Area of segment
=
6. Find the area of the segment.
Area of sector = Area of triangle = Area of segment
=
Area of Regular Polygons
Define:
Apothem:
Radius of polygon:
Observations about apothems
1. All apothems of regular polygon are congruent 2. Only regular polygons have apothems
3. An apothem is radius of circle inscribed in polygon 4. An apothem is perpendicular bisector of side
5. Radius of regular polygon is radius of circle circumscribed about polygon 6. Radius of regular polygon bisects angle of polygon
Formulas
Area of equilateral triangle: S2 3
4 where s is length of side of triangle
Area of regular polygon = 1
2Pawhere a is apothem and P is perimeter
Examples: Area of Regular Polygons
1. A regular polygon has perimeter of 40 and apothem of 5. Find the area.
A = 1
2 Pa = 1
2(40 5)( )= 100u2
2. An equilateral triangle has side 10cm long. Find triangle’s area.
A = 10
4 3
2
= 25√3 cm2
3. Find area of regular hexagon with sides 18 units long.
AF = 18 AP = 9
∆OAP is 30-60-90 triangle Apothem = 9√3 (60 side) perimeter = 6(18) = 108 Area = ½ (108)(9√3) = 486√3
Practice: Area of Regular Polygons
4. Find the area of a regular hexagon with perimeter 36.
5. Find the area of a regular pentagon with an apothem 6.2m long and a side 9m long.
Area of Inscribed/Circumscribed Polygons
When finding the area of a shaded region, do the following:
1. Find the area of the circle 2. Find the area of the polygon 3. Subtract the two areas
Examples
1. Find the area of the shaded region.
Since radius = 4, Area of circle = 42π= 16π Each side of square = 4(2) = 8
Area of square = 8 * 8 = 64
Area of shaded region = 64 - 16π u2
2. Find the area of the shaded region.
Since it is an equilateral triangle, Area of Triangle = 4
3 102
=
25√3The apothem of the triangle is the radius of the circle.
600 side = 10/2 = 5. Need the 300 side. Apothem = 3
3 5
Area of circle =
3 25 3
3
5 2
Area of shaded region = 25√3- 3 25
u2
3. Find the area of the shaded region.
Area of square = 10 * 10 = 100
The hypotenuse of the square is the diameter of the circle. Since dealing with square, dealing with 45-45-90 triangle.
Diameter = hypotenuse = 10√2 Radius = 5√2
Area of circle = (5√2)2π = 50π
Area of shaded region = 50π
