(5+2x)(8+2x) - 5*8 = 30
now solve for x
A picture measuring 5cm by 8cm is mounted on a frame so as to leave a uniform margin of width X cm all round .if the area of the margin is 30cm² ,find x
3 answers
The margin has a rectangle at the top and bottom its length = 5 cm
Its area = 5 ∙ x
That means margin has two rectangles of area 5 x
The margin also has a rectangle on the left and right sides its length = 8 - x - x = 8 - 2 x
Its area = ( 8 - 2 x ) ∙ x
That means margin has two rectangles of area ( 8 - 2 x ) ∙ x
Total area of margin is:
2 ∙ 5 x + 2 ∙ ( 8 - 2 x ) ∙ x = 30 cm²
10 x + 2 ∙ ( 8 x - 2 x² ) = 30
10 x + 16 x - 4 x² = 30
- 4 x² + 26 x = 30
Now you must solve this equation.
Multiply both sides by - 1
4 x² - 26 x = - 30
Add 30 to both sides
4 x² - 26 x + 30 = 0
The solutions are:
x = 3 / 2 = 1.5 cm
and
x = 5 cm
The margin cannot be 5 cm, because in this case the length of the rectangle at the top and bottom of the picture would be:
5 - 2 x = 5 - 2 ∙ 5 = 5 - 10 = - 5 cm
This is impossible because the length of the rectangle cannot be negative.
So the solution is:
x = 1.5 cm
Its area = 5 ∙ x
That means margin has two rectangles of area 5 x
The margin also has a rectangle on the left and right sides its length = 8 - x - x = 8 - 2 x
Its area = ( 8 - 2 x ) ∙ x
That means margin has two rectangles of area ( 8 - 2 x ) ∙ x
Total area of margin is:
2 ∙ 5 x + 2 ∙ ( 8 - 2 x ) ∙ x = 30 cm²
10 x + 2 ∙ ( 8 x - 2 x² ) = 30
10 x + 16 x - 4 x² = 30
- 4 x² + 26 x = 30
Now you must solve this equation.
Multiply both sides by - 1
4 x² - 26 x = - 30
Add 30 to both sides
4 x² - 26 x + 30 = 0
The solutions are:
x = 3 / 2 = 1.5 cm
and
x = 5 cm
The margin cannot be 5 cm, because in this case the length of the rectangle at the top and bottom of the picture would be:
5 - 2 x = 5 - 2 ∙ 5 = 5 - 10 = - 5 cm
This is impossible because the length of the rectangle cannot be negative.
So the solution is:
x = 1.5 cm
Yes
1 answer