Chapter 11

11.1 Understanding Area

11.2 Areas of Parallelograms and Triangles

11.3 Area of a Trapezoid

11.4 Areas of kites and related figures

11.5 Areas of Regular Polygons

11.6 Areas of Circles, Sectors and Arc Segments.

Review - 11.1-6 Test - 2002

Extra Review per Fritz request

11.7 Ratios of Areas

11.8 Hero's and Brahmagupta's Formulas

Review- 11.1-4 Quiz

Review Chapter 11.5-8 Test

11. 1 Understanding Area

Distance or length is measured in linear units - centimeters, inches etc.

Area -is measured in square units to find the surface region bounded by segments or curves (square centimeters, square inches etc.)

We will be using a lot of formulas in this chapter to figure out the number of square units that can be found in the various polygonal regions that we have studied so far.

Our first formula is for the area of a Rectangle- Postulate

The area of a rectangle is equal to the product of its base and height.

Area of a rectangle = bh

Since a square is a rectangle and both the base and height of a square are the same measure, we can add a formula based on the rectangle

The area of a square = s²

Two other postulates are important for understanding area.

If two closed figures are congruent, then their areas are equal.

This does not reverse. If the areas of two figures are equal it does not lead to the conclusion that the figures are congruent. For example a rectangle that is 8 x 2 has an area of 16 and a square 4 x 4 has an area of 16, but the figures are not congruent as they do not have sides of equal measures.

Area addition - To find the area of a closed region that is irregular in shape, divide the area into non-overlapping regions and add the individual areas.

11.2 Areas of Parallelograms and Triangles

If a height is dropped from a vertex of a parallelogram a triangle is formed. If the area of the triangle is shifted to the other end of the parallelogram, a rectangle is formed. Notice that the height of the parallelogram is the same as the height of the rectangle and the length of the base of the parallelogram is the same as the length of the base of the rectangle. So the area of a parallelogram's formula is the same as that of a rectangle with the only difference between them is that the height of the parallelogram is not a side of the parallelogram/

Area = 20 x 12 = 240

The area of a parallelogram = base x height. (make sure that you use the perpendicular and not the side of the parallelogram.

Every triangle can be made into a parallelogram by constructing a rotated congruent triangle on a side.
Since the area of the parallelogram is twice that of the triangle the formula for a triangle is 1/2 of that of a parallelogram

Area of a triangle = 1/2 base x height

Since a right triangles legs form a base and height of the triangle, the formula for finding the area of a right triangle is

Area of a right triangle = 1/2 leg x leg

The application of 30-60-90 and 45-45-90 triangles are used when finding the areas of parallelograms and triangles.

Example: Find the area of the parallelogram shown with a 60 degree angle

The height of the parallelogram would be and the area of the parallelogram would be

Sometimes you are asked to find a second height of either a triangle or parallelogram given two sides and one height. Remember the area of the figure doesn't change so the base times the height in one direction must equal the base times the height in the other direction.

Example:

11.3 The area of a Trapezoid.

Since a trapezoid can be divided into two non congruent triangles with the same heights, the area formula is a combined form of the areas of the two triangles.

Example of finding area of trapezoid

A median of a trapezoid is the segment that joins the midpoints of the two non parallel sides (legs) of the trapezoid. The median of a trapezoid has two properties.

1. It is parallel to the bases

2. Is is half the sum of the bases.

Since the formula for finding the area of a trapezoid can be rearranged to

11.4 Areas of Kites and related figures.

The areas of quadrilaterals with perpendicular diagonals can be found using the following formula.

Ex. If ABCD is a rhombus find its area.

Since the diagonals of a rhombus are perpendicular and bisect each other, use the pythag. theorem on the triangle and find the 3rd side. AX = 3 AB = 5 then BX = 4. The diagonals are then 6 and 8 and the area of the rhombus is 1/2(6)(8)= 24

REVIEW

1. A rectangle's length is 3 cm longer than twice its width. The area of the rectangle is

119 cm². Find the length and width of the rectangle.

ans. Let x = the width and 2x+3 represent the length. The product of the length and width gives the area of the rectangle. (x) (2x+3) = 119. 2x²+3x=119 Set your quadratic equal to zero and factor. You are looking for factors of 119 which are 7 and 17. 2x²+3x-119= 0 (2x+17)(x - 7) = 0 x= -17/2 or x =7 Since a negative number won't do for a length our solution is the width is 7 cm and 2(7)+3 = 17 cm which is our length.

2. One side of a triangle is 12 inches long and its corresponding altitude is 9 inches. Find the corresponding altitude to a second side of the triangle whose length is 18 inches.

ans. Since the area of a triangle remains the same whether you are using 12 as a base with its corresponding height or 18 as a base with its height, set the two areas equal to each other. 1/2(12)(9) = 1/2(18)(h), 54=9h, h=6 inches.

3. The measures of the consecutive sides of an isosceles trapezoid are in a ratio of 8:5:2:5. The perimeter of the trapezoid is 140 inches. Find the area of the trapezoid.

ans. Place x's with the ratios as a number must have been reduced out if the perimeter is to add to 140. Therefore 2x+5x+5x+8x=140 or 20x=140 and x=7. Put the x back into the algebraic expressions to find the length of each side. 2(7)=14 5(7)=35 and 8(7)=56. Drop perpendiculars and from a rectangle and two triangles as shown.

Use the Pythagorean theorem on the triangles and find the height. 35²= 21²+ x² or 1225 = 441 + x² 1225 - 441 = x² 784=x² x=28 From here it is a matter of substituting the numbers into the formula for the area of a trapezoid. Area = 1/2 h (b₁ + b₂) or 1/2(28) (56 +14) = 14 (70) which is 980 in²

4. The area of an isosceles trapezoid is 72 cm² . The perimeter of the trapezoid is 38 cm. If the legs are 10 cm long, find the height of the trapezoid.

ans. Since the perimeter is the addition of all the sides, b₁ + b₂ +10 +10 = 38. Subtracting 20 from each side leaves b₁ + b₂ =18. The formula for a trapezoid is Area = 1/2 h (b₁ + b₂). Replace the area with 72 cm² and b₁ + b₂ with 18. 72 = 1/2(h) (18), 9h=72 , h = 8 cm.

5. A rhombus has a perimeter of 88 cm and an angle of 120^o . Find its area.

ans. Since a rhombus has 4 equal sides 4x=88cm or x = 22. If a perpendicular is dropped to represent the height, a 30-60-90 triangle is formed.

Now that you have the base and height of the rhombus use the formula for finding the area of a parallelogram A = bh since a rhombus is a parallelogram. 22 x 11 = 242

6. Find the area of an equilateral triangle whose side length is 16.

ans. Use the formula for finding the area of an equilateral triangle.

Omit 7 and 8 on your review sheet.

9. Find the area of triangle XYZ if the coordinates are as follows. X=(-4,10) Y= (12,3) and Z = (-1,0)

ans. Fence in the triangle with a rectangle and find its area . The base of the rectangle can be found by subtracting the smallest x value from the largest x value. The height is found by subtracing the smallest y value from the largest y value. The area of the rectangle is 160 sq. units. From the rectangle subtract off the areas of each of the right triangles.

Area of TriangleI = 1/2 (3)(10)=15 Area of tri II = 1/2(16)(7)=56 Area of tri III = 1/2(13)(3)=19.5, Rect - Tri I -TriII - TriIII = 160-15-56-19.5= 69.5 sq units.

11. Find the area of the trapezoid.

ans. Drop a perpendicular to form a 45-45-90 triangle. Find the height of the trapezoid by dividing by root 2

Use the area of a trapezoid = 1/2 h (b₁ + b₂) to get 1/2(3)(13+10) = 69/2 = 34.5

12. Find the area of the square.

ans. The area of a square can be found by using the formula for an area of a kite. A= 1/2d₁d₂ Since the diagonals of a square are congruent.

13. Find the area of the rhombus if AC= 18 and AD=41.

Draw in the other diagonal which divides the original into two congruent segments. Use the pythagorean theorem on the right triangles formed to find the length of DE and BE. 41²=9²+x² x = 40

Use the formula for a kite. A= 1/2d₁d₂ A = 1/2 (80)(18) A = 720 sq units.

14. Find the area of the triangle.

ans. drop a perpendicular and use the 30-60-90 triangle to find the height. The side opposite a 30 degree angle = 1/2 hypotenuse. So the height is 3 . The area of a triangle is 1/2 bh. Area = 1/2(4)(3) = 6 sq. units.

Make sure you can also do Number 28 on page 522 and #9 on page 515.

11.5 Areas of Regular Polygons

Remember - Regular Polygons by definition have all congruent sides and congruent angles.

All regular Polygons can be inscribed in a circle.

The center of the circle is referred to as the center of the polygon

The radius of the polygon is any segment drawn from the center to a vertex of the polygon,

The apothem of the polygon is the segment drawn from the center perpendicular to any of the sides.

Notice that the radius bisects the angle to which it is drawn.

To find an angle of a regular polygon remember the formula Angle = (n-2)(180)/n

When all the radii are drawn in, the polygon is divided into a number of triangles = to the number of sides on the figure

To find the area of the polygon we could find the area of each triangle and then multiply this times the number of triangles formed. Since the base of each triangle is a side of the polygon and the height is the apothem of the formula can be derived

Area of a Regular polygon = 1/2 (apothem)(perimeter)

1/2 aP

Below are some examples of finding the areas of regular polygons. Notice the prevalence of the 30-60-90 and 45-45-90

Be sure to check out the pentagon in the review below and be able to find the area of an octagon.

If the radius of a regular octagon is given 8 triangles may be formed by drawing all the radii in . The central angles are each 45 degrees and if a height is dropped in each triangle it will equal 1/2 the radius times the square root of 2. Find the area of one triangle and multiply this by 8.

11.6 Areas of Circles, Sectors and Arc Segments.

Formulas

Area of a circle = r²

Circumference of a circle = 2r

If the circumference of a circle is given and the question asks for the area, solve for the radius and substitute it in to the circles area formula.

Ex: If the circumference of a circle is 12^, what is the area of the circle.

12= 2r r = 6 Area of a circle = 6²= 36

If the area of a circle is given and you are asked for the circumference, solve for the radius and substitute in to the circumference formula

Ex. If the area of a circle is 49^, what is the circumference of the circle

r²=49 r = 7 C =2r = 27 = 14

Sector of a circle is a region of a circle bounded by two radii of the circle and their intercepted arc.

examples:

An arc segment is a region of a circle bounded by a chord of the circle and its corresponding arc.

Ex. 1 Ex. 2 Ex. 3

ans. Ex. 1

ans. Ex. 2

ans. Ex. 3

Miscellaneous shaded regions.

Find the area of each shaded region.

Ans. Find the area of the equilateral triangle and subtract off the three sector areas.
Each sector has a central angle of 60 degrees due to the angles of the equilateral triangle. 3 – 60 degree angles = 180 degrees or half of a circle with a radius of 6.

11.7 Ratios of areas

If two figures are similar, the ratio of their areas compares as the square of ratio of their sides.

Example. Find the area of triangle DEF is ABC has an area of 36 cm²

ans. The ratio of the sides is 9/6 or 3/2. The areas must compare as the square of 3/2 or 9/4. Set up a proportion 9/4 = 36/x Cross multiply 9x = 144, x = 16

Sometimes figures have one dimension equal but are changed in some other manner such as a height that is doubled. When this occurs you must set up the formula for finding the area and represent the areas with the dimensions given.

Example. Given two triangles with the same height but one's base is twice as big as the other, find the ratio of their areas.

ans. 1/2(2y)x/ 1/2(x)y = 2/1 The area of the first triangle is twice the area of the second.

Sometimes the similar figures must be subtracted from the entire area in order to find the ratio of two given areas as shown in the example below.

Ex. Find the ratio of area I to II

ans. Triangle BEF ~Triangle BAC by AA~. The left side of the small triangle corresponds to the left side of the larger triangle and their ratio is 9/12 or 3/4. Since areas compare as squares of scale factors the areas are 9 to 16. Since the top part is 9 parts of the whole, their are 7 of the 16 parts left for the bottom area. Therefore , the ratio of area I to II is 9/7

11.8 Hero’s and Brahmagupta’s Formulas

Hero’s formula is used to find the area of a triangle when only the lengths of the 3 sides are known.

To use the formula, the semi-perimeter of the triangle must be found. The semi-perimeter is the addition of the sides divided by 2.

The formula :

Example:

Brahmagupta's formula is used to find the area of any quadrilateral that can be inscribed in circle. (cyclic quadrilaterals).

To find the area of an inscribed quadrilateral, use the following formula.

s = semiperimeter (half the perimeter) and a,b,c,and d are sides of the quadrilateral.

Example: If an inscribed quadrilateral has side lengths of 10,8,10 and 20, find the area of the quadrilateral.

Review 11.5-8

1. If the circumference of a circle is 36, find the area of the circle.

2. Find the area of a parallelogram with sides of 6 cm and 10 cm and an included angle of 45 degrees.

Drop a perpendicular forming a 45-45-90 triangle. The side becomes the hypotenuse and the perpendicular is the hypotenuse divided by the square root of 2. Use the formula A = bh

3. Find the length of the diagonal of a square whose area = 24 sq. cm

Since the Area of a square = s² , then s² =24 and The diagonal of a square is the hypotenuse of a 45-45-90 and can be found by multiplying the side times The diagonal then equals .

Other method A = 1/2 d² = 24 d² =48 d =

4. If two similar trapezoids have corresponding bases of 8 and 12 cm, and thee area of the smaller trapezoid is 100 cm² , find the area of the larger trapezoid.

ans. Ratio of the sides is 8/12 or 2/3. The ratio of the areas is the square of the ratio of the sides or 4/9. Set up a proportion

5. Find the ratio of area I to area II if the figure is a trapezoid.

ans. The triangles are similar by AA~ and the ratio of the sides is 8/18 which reduces to 4/9. The rule is the areas compare as the square of the scale factor. Area I/AreaII = 16/81

6. Find the area of an inscribed quadrilateral with side lengths of 5, 3, 5, and 9 cm.

ans. Use Brahmagupta's formula where s= semiperimeter (half of the perimeter) and a,b,c,d are the lengths of the sides. The semiperimeter = 11

7. Find the area of a regular hexagon with side apothem of length 9 cm.

ans. find the measure of one angle of a regular hexagon using the formula (n-2)180/n. The angle (4)(180)/6=120 and when the radius is drawn to a vertex it bisects the angle forming a 30-60-90 triangle with the apothem and half of the side. To get back to the side opposite the 30 degree angle, divide 9 by the square root of 3

, this is half of the side length so one side of the hexagon measures and the perimeter of the hexagon = The formula for the area of a regular polygon is A = 1/2 aP where a = apothem and P = perimeter of the polygon.

8. Find the area of a regular pentagon with side length of 10 cm.

ans. 172 - To solve, find an angle of the pentagon using the formula (n-2)180/n where n = number of sides (5) The angle = 108. Draw in the apothem and radius of the pentagon. The radius will bisect the angle. Since each side is 10 and the apothem bisects the side to which it is drawn, you can use the trigonometric function Tan 54 = a/5 to find the measure of the apothem. The apothem = 6.88. The formula for finding the measure of a regular polygon is A=1/2aP A = 1/2(6.88)50 = 172

Find the areas of each of the shaded regions.

Find the area the larger circle and subtract the area of the smaller circle.

10.

Introduce the segment OK to the center of the circle. Find the area of the sector using the formula to this add the area of the right triangle.

11 Given an equilateral triangle with an inscribed circle of radius 4, find the area of the yellow shaded region.

Find the area of the equilateral triangle and subtract off the area of the circle. Since the triangle formed when the radius is drawn in is a 30-60-90 triangle the side across from the 60 is the short leg times The side of the triangle measure 8 and the area of the triangle can be found by A=1/2aP or A=1/2(4)(24)= 48. From this subtract the area of the circle

12.

Draw in the square formed by joining the centers of the circles. The sides of the squares will be equal to two radii of the smaller circles or 6. The diagonal of the square will be the side. The diameter of the large circle will be 6 + 6 and the radius of the big circle will be 3+3 Find the area of the large circle and subtract off the 4 circles of radii 3.