Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

# MATH PROBLEM ?

A plain mat is placed around a picture measuring 28cm by 36cm so that the width of the mat is equal on all sides. The area of the mat is 3/4 of the area of the picture. What is the width of the mat, to the nearest millimetre??

Relevance
Lv 7
1 month ago

• Philip
Lv 6
1 month ago

Linear units = [cm].

Picture area = P = 28*36 = 1008.

Mat area = M = (3/4)P = 756.

M+P = (7/4)P = 49*36 = 1764.

Put mat width = w. Now the figure of M+P has

2 strips 36*w & 2 strips (28+2w)*w equivalent

to 1 strip 72*w & 1 strip (56+2w)*w equivalent

to 1 strip (128+2w)*w whose area = 2w(56+w)

1764. Then w^2+56w-882 = 0 and 2w = -56+D,

where D^2=56^2+4*1*882=6664 and D =

81.63332653. Then w = 12.81666326 [cm] =

128 mm to nearest mm.

• 1 month ago

let w be the width

A of picture is 28•36

A of mat+picture = (28+2w)(36+2w)

A of mat = (28+2w)(36+2w) – 28•36

(28+2w)(36+2w) – 28•36 = (4/3)28•36

(14+w)(18+w) – 7•36 = (1/3)28•36

(14+w)(18+w) – 252 = (12)28

252 + 64w + w² – 252 = 336

w² + 64w – 336 = 0

w = –32 + 4√85 = 4.88 cm

check

A of picture is 28•36 = 1008

total area = (28+9.76)(36+9.76) = (37.76)(45.76) = 1728

A of mat = 1728 – 1008 = 720

3/4 of picture area = 756

close, need to check the math...