Question

A postcard has an area of 24 square inches. Two enlargements of the postcard have areas of 54 square inches and 96 square inches. In each postcard, the length is 1 and 1/2 times the width. What could be the dimensions of the postcard or the enlargements.

Answer 1

Answer:

Step-by-step explanation:

Given:

Initial area, Ai = 24 in^2

First enlargement, A1 = 54 in^2

Second enlargement, A2 = 96 in^2

Length, L = 3/2 × B

Area of a rectangle = L × B

Ai = 3/2B × B

24 = 3/2 B^2

B = sqrt(16)

= 4 in

A1 = 3/2B × B

54 = 3/2 B^2

B = sqrt(36)

= 6 in

A2 = 3/2B × B

96 = 3/2 B^2

B = sqrt(64)

= 8 in

For the first enlargement,

2in : 3in

Second enlargement,

1in : 4in

Answer 2

Answer:

original is 4 & 6.

for 54 sq inch= 6 & 9 and for 96 sq inch=8 & 12

Step-by-step explanation:

Given: The length is 1.5 times the width.

so the length is, l = 1.5w ----(a)

lw = 24  =1.5w(w)

lw = 54= 1.5w(w)

lw = 96 =1.5w(w)

Further simplifying it,

1.5$w^{2}$=24

1.5$w^{2}$=54

1.5$w^{2}$=96

so,

$w^{2}$ = $\frac{24}{1.5}$=16

$w^{2}$= $\frac{54}{1.5} =$36

$w^{2}$= $\frac{96}{1.5} =$64

taking the square root, we get:

w = 4

w = 6

w = 8

By putting the above values in eq (a), we can find their corresponding lengths:

l = 1.5(4) = 6

l = 1.5(6) = 9

l = 1.5(8) = 12

So a few lengths could be:

(l, w)

(6,4)

(9,6)

(12,8)