can anyone solve these word problems?

Question

1. The length of a picture without its border is twice its width. If the border is 2cm wide and border's area is 208 square cm. what are the dimensions of the picture alone?

2. The perimeter of a rectangle is 20 meters. If the length is increased by 2 meters and the width is increased by 3 meters, the area is increased by 30 square meters. What is the length of the rectangle?

Answer 1

Question 1
========
l = 2 * w
b = 2

Note that the length and width of the picture plus border are (l+4) and (w+4) respectively - the border is on each side.

Area of picture + border minus area of picture = 208
(l+4) * (w+4) - (l * w) = 208
(2w + 4) * (w + 4) - (2w * w) = 208
2w^2 + 8w + 4w + 16 - 2w^2 = 208
12w + 16 = 208
12w = 192

Solution:
w = 16
l = 2w = 32.

Checking:
Area of picture alone = w * l = 16 x 32 = 512

Area of picture plus border
= (w + 4) * (l + 4) = 20 x 36 = 720 ( = 512 + 208). QED.

Question 2
========
Perimeter = 2l + 2w = 20
l + w = 10
w = 10 - l

New area = old area + 30
(l + 2) * (w + 3) = lw + 30
(l + 2) * (10 - l + 3) = l(10-l) + 30
(l + 2) (13 - l) = 10l - l^2 + 30
13l + 26 - l^2 - 2l = 10l - l^2 + 30
11l = 10l + 4

Solution: l = 4

Checking: w = 10 - l = 10 - 4 = 6
Old area = w * l = 6 * 4 = 24
New area = (l + 2) * (w + 3)
= 6 * 9 = 54 (= old area + 30). QED

Answer 2

Q1
If the length is x cm, the width is 2x cm. Adding a border of 2 cms all round, the new rectangle with borders would be of size x+2+2 and 2x+2+2 or x+4 and 2x+4. The border's area would be the difference between the areas of the two rectangles. Hence
(x+4)(2x+4) less 2x*x=208 or 12x+16=208, giving x=16. The dimensions therefore would be 16 and 32 cms.
Q2
Let the sides be x and y, x being the length.So, the perimeter is 2* (x+y)=20 or x+y=10 The new rectangle would thus be of dimensions x+2 and y+3. It is given that the difference of the two areas is 30 sq.mts. That is, (x+2)(y+3) less x*y= 30; that is,
3x+2y+6=30 or 3x+2y=24. Substituting x=10-y, we get
3*(10-y)+2y=24 which gives 30-y=24or y=6, and so x=4. The length of the rectangle would thus be 4 meters.

Answer 3

1) l = 2w; (l + 4cm)(w + 4 cm) - (l*w) = 208 sq. cm
(2w+ 4)(w+4) - 2w^2 = 208
2w^2 + 12w + 16 - 2w^2 = 208
3w + 4 = 52
3w = 48
w = 12
l = 24

The PIcture is 24cm X 12 cm.

2) 2L+2W = 20; (L+2)(W+3) = L*W +30
L+W = 10, W = 10 - L;
(l+2)(13-l) = 10L - L^2 + 30
-L^2 + 11L + 26 = -L^2 +10L + 30
L = 4

The length of the rectangle is 4 meters

Answer 4

problem 1...

x = width

(x+2)(2x+2) - x(2x) = 208cm
(2x^2+4x+2x+4) - 2x^2 = 208cm
6x+4 = 208cm
6x = 204cm
x = 34
so width = 34 cm

and length = 2x or 68 cm
dimension of picture alone is 34x68

reds got me again... I should have added 4 not 2. This answer is wrong

Answer 5

question 2, answer is 16.