Find the area of
the trapezoid using
and (median)(height)
Find the area of
the trapezoid using
and (median)(height)
1.
2.
The area is
2. The area of a trapezoid can also be found by multiplying the median times the height. The trapezoids area is (12)(17) =204
Find the areas of kites and rhombuses given specific information.
3. KLMN is
a kite 4. RHOM is a rhombus
3. To find the area of a kite,
the formula is
.
Since diagonals of a kite bisect each other then KM is
split into two segments of 8 each. Use the Pythagorean Theorem on the right
triangles to find the lengths of the segments that make up LN. The top triangle
has a third side length of 6 and the bottom triangle has a third side length of
15. Therefore the length of LN = 21. Using the above formula
4. A rhombus is a type of kite,
so draw the other diagonal in. The length of HM can be found by using the
Pythagorean theorem of the right triangles formed when the diagonals intersect.
HM = 18 , so the formula
=
5. Find the area of a triangle given the coordinates of its 3 vertices.
Fence the triangle in by drawing a big rectangle around the figure so that C is at the bottom left corner of the rectangle, A is on
the top and B is on the far right side of the rectangle. The rectangle's dimensions will be 11 for the base and 10 for the height giving it an area of 110. From this subtract the area of the three right triangles formed at the corners. Top left triangle has
dimensions of 10 x 5 and an area of 25, the bottom right triangle with dimensions of 11 x 1 has an area of 5.5 and the top right triangle has dimensions of 6 x 9 with an area of 27. 110 - (25 + 27 + 5.5) = 52.5
6. Each of the following is a regular polygon with given lengths. Find their areas.
6. A) Drop the apothem on the square to the side and a 45-45-90 triangle will be formed. The side of this triangle will be 4 which is half the length of the side of the square. A square with a side length of 8 has an area of 64.
b) I changed the side length to
be 8 on the octagon to make it a little easier. Divide the octagon up as
in the diagram below. The triangles formed are 45-45-90 triangles so the
side lengths are
. Use the formula,
C) If the apothem is 15, when
you draw the radius in a 30-60-90 triangle gets formed and the base of the
triangle is 1/2 the side length of the hexagon. To get back to the short
leg, divide by 
. The base is
and the side length is
The area can be found by taking 1/2 the
apothem times the perimeter. 
D) Drawing in the apothem of
the equilateral triangle, the length is 1/2 of 10 as it is the short leg of a
30-60-90 triangles. The base of the 30-60-90 is
and the side of the equilateral triangle is
twice this or 
. The area then becomes
7. Find the area of the sector shown.
To find the area of the sector use
the formula
8. Find the
areas of the arc segment.
Area of the sector - area of the
triangle. 
9. Find the areas of each shaded region.
9. The area of the equilateral
triangle - 1/2 circle
B) The area of a sector with a
central angle of 90 + the area of a 45-45-90 triangle with a leg of 5.
C) Shaded region is between the hexagon and the circle -
Area of the circle- area of the hexagon.
D) Area of two 45-45-90 triangles with side lengths
of 
and two sector areas with central angles
of 90 and radii of .
The area shaded is the region outside of the circle
bounded by the circle and the two tangents. Draw in CE forming two
30-60-90 triangles . The legs of the triangles are 10 and 10
.
The area of the region is the area of the two triangle - the area of the sector
with a central angle of 120 and radius of 10.
The area of the annulus is the subtraction of the smaller
circle with radius of 4 from the larger with radius of 6.
