Question

an architect designed three different parallelograms shaped brick patios. write the missing dimension in the table

Answer 1

An architect designed three different parallelograms shaped brick patios. The formula for the area of a parallelogram is A = bh where A is the area, b is the base and h is the height. For the first brick, the height is 28/3 ft. For the second brick, the base is 25/2 ft and for the third brick, the height is 59/4 ft.

Answer 2

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:

$Area=Base\times Height\\\\i.e.\\\\A=bh$

1)

We have:

$b=15\dfrac{3}{4}\ \text{ft.}=\dfrac{63}{4}\ \text{ft.}$

$A=147\ \text{ft^2}$

Hence, the height of parallelogram is:

$147=\dfrac{63}{4}\times h\\\\h=\dfrac{147\times 4}{63}\\\\h=\dfrac{588}{63}\\\\h=9\dfrac{21}{63}\ \text{ft.}$

Hence, height of parallelogram is:

$h=9\dfrac{21}{63}\ \text{ft.}$

2)

We have:

$h=11\dfrac{1}{4}=\dfrac{45}{4}\ \text{ft.}$

$A=140\dfrac{5}{8}=\dfrac{1125}{8}\ \text{ft^2}$

$\dfrac{1125}{8}=\dfrac{45}{4}\times b\\\\b=12\dfrac{1}{2}\ \text{ft}$

Hence, base of parallelogram is:

$b=12\dfrac{1}{2}\ \text{ft}$

3)

We have:

$b=10\dfrac{1}{4}\ \text{ft}=\dfrac{41}{4}$

$A=151\dfrac{3}{16}\ \text{ft^2}=\dfrac{2419}{16}$

$\dfrac{2419}{16}=\dfrac{41}{4}\times h\\\\h=\dfrac{59}{4}\\\\\\h=14\dfrac{3}{4}\ \text{ft}$

Hence, the height of parallelogram is:

$h=14\dfrac{3}{4}\ \text{ft}$