9"A"- 27/04/09

Today in class we started of by correcting the h.w which was pg. 539 # 11-26.  We  asked to solve some on the bord so I'll give an example of 2 of them. 

19) A= 95, find the value of x.  

The formula for the area of a trapezoid is   so we substitute and get 

   then multiply 95 x 2 and the answer is divided by 2 and we get 10.

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21) A rhombus has an angle measure of 120, and its longer diagonal has a lenght of 10 inches. Find the AREA of the rhombus.

 After drawing all the given information we can apply SOH CAH TOA to find the area of 1 triangle.  Tan 60º= 5/x

      

  after we simplify we get   after this we just have to substitute in the formula for area of a triangle and times the answer by 4 to get the area of the rhombus.

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After finishing with the h.w we continued to AREAS OF REGULAR POLYGONS. Mrs. Saatkamp gave us a print out with 3 figures: 1 hexagon, 1 pentagon and 1 octagon. We had to draw the diagonals in order to split the shape in triangles. With this we were going to come up a formula that works to find the are of all regular polygons.

Keep in mind-

The base of 1 triangle is 1 side of the polygon.

The hight of a triangle is a apothem.

An apothem is 1/3 of the hight of the polygon.

Hexagon: All 5 triangles are equilateral. 

To find the area of the polygon without drawing it and the triangles inside it we need to get one of the triangles and apply the formula for area of a triangle BUT replace hight by apothem and base by side. To finish the formula should be multiply by the total number of sides in the polygon.

 The angle of 1 of the triagles is = interior angle/2

Formulas: 

-Area of a regular polygon      p= perimeter

a= apothem      p= perimeter

-Area of a circle = r^2π 

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We have no h.w for tomorrow yey!!

see you in class :]

Malena