Respuesta :
Given:
2 parallelograms with an area of 9 1/3 yd²
height of each parallelogram is 1 1/3 yd
Area of parallelogram = base * height
We need to divide the combined area into two to get each parallelogram's base.
9 1/3 = ((9*3)+1)/3 = 28/3
28/3 ÷ 2 = 28/3 * 1/2 = 28/6 yd² or 4 4/6 yd² ⇒ 4 2/3 yd²
Area of each parallelogram is 4 2/3 yd²
4 2/3 yd² = base * 1 1/3 yd
14/3 yd² ÷ 4/3 yd = base
14/3 yd² x 3/4 yd = base
14*3 / 3*4 = base
42 / 12 = base
3 6/12 yd = base
or 3 1/2 yd = base
a) the base of each parallelogram is 3 1/2 yards
b) we can assume that the two parallelograms form a rectangle.
area of a rectangle is length times width.
length is 3 1/2 yds * 2 = 7 yds
width is 3 1/2 yds
Area of rectangle = 7 yds * 3 1/2 yds
Area = 7 yd * 7/2 yd
Area = 7*7 / 2 yd²
Area = 49 / 2 yd²
Area = 24 1/2 yd²
2 parallelograms with an area of 9 1/3 yd²
height of each parallelogram is 1 1/3 yd
Area of parallelogram = base * height
We need to divide the combined area into two to get each parallelogram's base.
9 1/3 = ((9*3)+1)/3 = 28/3
28/3 ÷ 2 = 28/3 * 1/2 = 28/6 yd² or 4 4/6 yd² ⇒ 4 2/3 yd²
Area of each parallelogram is 4 2/3 yd²
4 2/3 yd² = base * 1 1/3 yd
14/3 yd² ÷ 4/3 yd = base
14/3 yd² x 3/4 yd = base
14*3 / 3*4 = base
42 / 12 = base
3 6/12 yd = base
or 3 1/2 yd = base
a) the base of each parallelogram is 3 1/2 yards
b) we can assume that the two parallelograms form a rectangle.
area of a rectangle is length times width.
length is 3 1/2 yds * 2 = 7 yds
width is 3 1/2 yds
Area of rectangle = 7 yds * 3 1/2 yds
Area = 7 yd * 7/2 yd
Area = 7*7 / 2 yd²
Area = 49 / 2 yd²
Area = 24 1/2 yd²
Answer:
Base of each parallelogram is [tex]3\frac{2}{4}yards[/tex]
The area of the smallest rectangle of wall that the mirror could fit on [tex]24\frac{1}{2} yards^{2}[/tex]
Step-by-step explanation:
Given : The parallelograms have a combined area of 9 1/3 square yards. The height of each parallelogram is 1 1/3 yards.
To Find:a) how long is the base of each parallelogram? b) what is the area of the smallest rectangle of wall that the mirror could fit on?
Solution :
Since The parallelograms have a combined area of 9 1/3 square yards = [tex]\frac{28}{3} yards^{2}[/tex]
Since this is the area of two combined congruent parallelograms
So, area of each parallelogram = [tex]\frac{\frac{28}{3}}{2}=\frac{14}{3}[/tex]
Thus area of each parallelogram is [tex]\frac{14}{3} yards^{2}[/tex]
To Calculate base of each parallelogram
Formula of area of parallelogram = Base * Height
Since height of each parallelogram is [tex]1\frac{1}{3} =\frac{4}{3}[/tex]
Thus [tex]\frac{14}{3} = Base *=\frac{4}{3}[/tex]
[tex]\frac{14*3}{3*4} = Base[/tex]
[tex]\frac{14}{4} = Base[/tex]
[tex]\frac{7}{2}yards = Base[/tex]
[tex]3\frac{1}{2}yards = Base[/tex]
hence the base of each parallelogram is [tex]3\frac{1}{2}yards[/tex]
b) we can assume that the two parallelograms form a rectangle.
So, in this case length will increase since two parallelograms are combines so length of the resultant will be twice the length of each parallelogram
area of a rectangle is length times width.
length is 3 1/2 yards * 2 = 7 yards
width is 3 1/2 yards
Area of rectangle = 7 yards * 3 1/2 yards
Area = 7 yd * 7/2 yd
Area = 7*7 / 2 yd²
Area = 49 / 2 yd²
Area = [tex]24\frac{1}{2} yards^{2}[/tex]
The area of the smallest rectangle of wall that the mirror could fit on [tex]24\frac{1}{2} yards^{2}[/tex]
