a photo measuring 7.5 cm by 5 cm is enlarged, so that the larger side becomes 18 cm. what does the shorter side become? in what ratio is the area increased?
Answers
Answer:
The length of the shorter side is 12 cm and the ratio area increased by 5.76 : 1.
Step-by-step explanation:
Consider the provided information.
The area of rectangle is = a × b, where a and b are the sides.
A photo measuring 7.5 CM by 5 cm
The area of the photo is:
7.5 × 5 = 37.5 cm²
Let x be the constant ratio
Ratio of Length : Width = 7.5x : 5x
Now the area of the largest side becomes 18cm.
7.5x = 18
x = 18/7.5
x = 2.4
Thus the width is 5x = 5 × 2.4 = 12 cm
The new length and width is 18 cm and 12 cm respectively.
Now the new area is: 18 × 12 = 216 cm²
The ratio of area increase is 216 : 37.5 which is also equals to 5.76 : 1.
Step-by-step explanation:
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Answer:
A photograph measuring 5.5 cm by 9 cm is enlarged in the ratio of 7:5. What is the new length and width of the picture?
The 7:5 or 7/5 isn’t the ratio between the width and height of the photograph. It is actually an increase to such measurements.
Let’s try this: 7/5 = 1.4
This is 40% increase of its dimensions’ measurements.
Thus,
5.5 * 1.4 = 7.7 cm
9 * 1.4 = 12.6 cm
The new width and height will be:
7.7 cm by 12.6 cm
An increase with the photograph’s width and height while retaining their original ratio.
5.5/9 = 7.7/12.6
0.6111111
The 7:5 or 7/5 isn’t the ratio between the width and height of the photograph. It is actually an increase to such measurements.
Let’s try this: 7/5 = 1.4
This is 40% increase of its dimensions’ measurements.
Thus,
5.5 * 1.4 = 7.7 cm
9 * 1.4 = 12.6 cm
The new width and height will be:
7.7 cm by 12.6 cm
An increase with the photograph’s width and height while retaining their original ratio.
5.5/9 = 7.7/12.6
0.6111111