Answer:
Let the base of the parallelogram is denoted by 'b'
Height of the parallelogram is denoted by 'h'.
and the area is given by 'A'.
We know that the area of a parallelogram is given as:
[tex]Area=Base\times Height\\\\i.e.\\\\A=bh[/tex]
1)
We have:
[tex]b=15\dfrac{3}{4}\ \text{ft.}=\dfrac{63}{4}\ \text{ft.}[/tex]
[tex]A=147\ \text{ft^2}[/tex]
Hence, the height of parallelogram is:
[tex]147=\dfrac{63}{4}\times h\\\\h=\dfrac{147\times 4}{63}\\\\h=\dfrac{588}{63}\\\\h=9\dfrac{21}{63}\ \text{ft.}[/tex]
Hence, height of parallelogram is:
[tex]h=9\dfrac{21}{63}\ \text{ft.}[/tex]
2)
We have:
[tex]h=11\dfrac{1}{4}=\dfrac{45}{4}\ \text{ft.}[/tex]
[tex]A=140\dfrac{5}{8}=\dfrac{1125}{8}\ \text{ft^2}[/tex]
[tex]\dfrac{1125}{8}=\dfrac{45}{4}\times b\\\\b=12\dfrac{1}{2}\ \text{ft}[/tex]
Hence, base of parallelogram is:
[tex]b=12\dfrac{1}{2}\ \text{ft}[/tex]
3)
We have:
[tex]b=10\dfrac{1}{4}\ \text{ft}=\dfrac{41}{4}[/tex]
[tex]A=151\dfrac{3}{16}\ \text{ft^2}=\dfrac{2419}{16}[/tex]
[tex]\dfrac{2419}{16}=\dfrac{41}{4}\times h\\\\h=\dfrac{59}{4}\\\\\\h=14\dfrac{3}{4}\ \text{ft}[/tex]
Hence, the height of parallelogram is:
[tex]h=14\dfrac{3}{4}\ \text{ft}[/tex]
