How do you find the Area of a parallelogram with a 45 degree angle?

Question

Find the area of the parallelogram ABCD in Figure 4, if m Ang A = 45 deg, AB = 10 sqrt(3), and AD = 6


Find the area of the trapezoid in Figure 5 if m Ang A = m Ang D = 45 deg, AB = 4 sqrt(2), and BC = 4.

figures 4 and 5:
http://s20.photobucket.com/albums/b231/Catalainya/?action=view¤t=geom_293.345.jpg

Answer 1

1. To get the height:
it is AD *sin 45 deg = 6*sqrt2 / 2 = 3 sqrt2
... area = 30 sqrt 6

2. height: AB (sin45) = 4 sqrt 2 * sqrt 2 / 2 = 4.
... area of trapezoid: 1/2(b1 + b2)h = 1/2(AD + 4)*4
... Now, to get AD, it is 4 + 2*(AB*cos45) = 4 + 4sqrt2

... area of trapezoid: 2(8 + 4 sqrt2) = 16 + 8 sqrt2.


d:

Answer 2

You need the base and the height, so draw a line from D perpendicular to AB. It makes a 45 45 90 degree triangle, meaning the sides are sqrt(2)/2 times the hypotenuse. Since the hypotenuse is 6, the other sides are 3sqrt(2).
base*height = 10sqrt(3)*3sqrt(2) = 30sqrt(6)

I do not know about the next one because there is not enough information... People are assuming that angle A is 45, but it could be anything (between 0 and 90 including 90)

I got 16[sqrt(2)sin(m)+cos(m)+sin(m)]
Heres how...
Draw a line from B perp to AD, we will call this line h.
sin(m)=h/(4sqrt2) solve for h
h=4sqrt(2)sin(m) now we have the height.

If you draw a line from C perp to AD, it forms a 45 45 90 triangle (we will call that new intercetion E). That means the two sides of the triangle are the same, making ED the same as the height, or 4sqrt(2)sin(m) .

New all we need is the other side of the first triangle we formed (from A to the line though B perp to AD). Lets call it x.
cos(m) = x/4sqrt2
x=4sqrt(2)cos(m)

b1 = 4
b2 = 4 + 4sqrt(2)cos(m) + 4sqrt(2)sin(m)
h = 4sqrt(2)sin(m)

(1/2)[4sqrt(2)sin(m)][4 + 4 + 4sqrt(2)cos(m) + 4sqrt(2)sin(m)]
[2sqrt(2)sin(m)][8 + 4sqrt(2)cos(m) + 4sqrt(2)sin(m)]
[2sqrt(2)sin(m)]{8 + 4sqrt(2) [cos(m) + sin(m)]}
16sqrt(2)sin(m) + 16[cos(m) + sin(m)]
16[sqrt(2)sin(m) + cos(m) + sin(m)]

Answer 3

Both questions require use of the relation of sides for a 45/45/90 right triangle. For this triangle, the ratio of the hypotenuse to either side is sqrt(2):1.

1. Draw DE perpendicular to AB. AD is the hypotenuse of the 45/45/90 right triangle. DE is the height of the parallelogram.
Area = DE * AB
2. Draw BE perpendicular to AD. AB is hypotenuse for 45/45/90 right triangle. BE is the height of the trapezoid. 2* BE+BC= base of trapezoid. BC = top.
A = 0.5 * (base+top) * height

Answer 4

4.) Construct an altitude DE making 90 degrees with AB

sin45 = DE/AD
DE = AD(sin45)
DE = 6/(sq.rt(2))
DE = 4.24

AREA = base * height
area = DE * AB
area =10(sq.rt(3)) * 4.24
area of the parallelogram = 73.44 sq.units.

5.) construct an altitude BE to AD,

sin45 = BE/AB
BE = 4(sq.rt(2)/(root(2)
BE = 4

Area = 1/2 (height)(sum of the length of parallel sides)
Area = 1/2 (BE)(BC + AD)

AD = 2cos(45)(AB) + BC
AD = 12 approx.

Area = 1/2 (4)(12 + 4)
Area = 1/2 ( 4)(16)
area of the trapezoid = 32 sq. units

Answer 5

The first answer is correct this is totally independent of the angle, the width perpendicular to the length and the length, take the product.

Answer 6

6*sin 45 = x , 6*cos 45 = y
(0.5) x * y = z
(z *2) + (10root3 * x) = answer

4root2*sin 45 = a
4root2*cos45 = b
(4 * a) + (a *b) = answer

Answer 7

parralellograms are the same as a rectangle for area = l * w
the trapezoid is best broken into parts to get the area. seperate it into simple triangles and rectangles and add them up.