If the side of an equilateral triangle is increased by 34%, then by what percentage will its area increase?
79.56%
This problem asks us to find the percentage increase in the area of an equilateral triangle when its side length is increased by a specific percentage. To solve this, we need to understand the formula for the area of an equilateral triangle and how changes in the side length affect the area.
The area of an equilateral triangle is calculated using the formula:
\(A = \frac{\sqrt{3}}{4} s^2\)
Where:
Notice that the area is directly proportional to the square of the side length (\(s^2\)). This means if the side length changes, the area will change by the square of the factor by which the side changes.
Let the original side length be \(s_1\). We are told that the side is increased by 34%.
So, the new side length is 1.34 times the original side length.
Using the area formula:
Substitute \(s_2 = 1.34 s_1\) into the formula for \(A_2\):
\(A_2 = \frac{\sqrt{3}}{4} (1.34 s_1)^2 = \frac{\sqrt{3}}{4} (1.34^2) s_1^2\)
We can see that the new area \(A_2\) is \(1.34^2\) times the original area \(A_1\).
\(A_2 = (1.34^2) A_1\)
Let's calculate \(1.34^2\):
\(1.34^2 = 1.34 \times 1.34 = 1.7956\)
So, \(A_2 = 1.7956 A_1\). This means the new area is 1.7956 times the original area.
To find the percentage increase in area, we use the formula:
Percentage Increase = \(\frac{\text{New Value} - \text{Original Value}}{\text{Original Value}} \times 100\%\)
In this case:
Percentage Increase in Area = \(\frac{A_2 - A_1}{A_1} \times 100\%\)
Substitute \(A_2 = 1.7956 A_1\):
Percentage Increase in Area = \(\frac{1.7956 A_1 - A_1}{A_1} \times 100\%\)
Percentage Increase in Area = \(\frac{(1.7956 - 1) A_1}{A_1} \times 100\%\)
Percentage Increase in Area = \((1.7956 - 1) \times 100\%\)
Percentage Increase in Area = \(0.7956 \times 100\%\)
Percentage Increase in Area = \(79.56\%\)
Therefore, the area of the equilateral triangle will increase by 79.56% when its side is increased by 34%.
| Concept | Formula/Value |
|---|---|
| Original Side | \(s_1\) |
| Percentage Increase in Side | 34% |
| New Side (\(s_2\)) | \(s_1 \times (1 + 0.34) = 1.34 s_1\) |
| Original Area (\(A_1\)) | \(\frac{\sqrt{3}}{4} s_1^2\) |
| New Area (\(A_2\)) | \(\frac{\sqrt{3}}{4} s_2^2 = \frac{\sqrt{3}}{4} (1.34 s_1)^2 = 1.34^2 \times \frac{\sqrt{3}}{4} s_1^2 = 1.7956 A_1\) |
| Percentage Increase in Area | \(\frac{A_2 - A_1}{A_1} \times 100\%\) |
| Property | Formula |
|---|---|
| Side Length | \(s\) |
| Area | \(\frac{\sqrt{3}}{4} s^2\) |
| Perimeter | \(3s\) |
| Height | \(\frac{\sqrt{3}}{2} s\) |
The concept of percentage change is widely used in various fields. When a quantity changes, the percentage change tells us the magnitude of the change relative to the original quantity. The formula is always:
Percentage Change = \(\frac{\text{Change in Value}}{\text{Original Value}} \times 100\%\)
If the new value is greater than the original value, it's a percentage increase. If the new value is less than the original value, it's a percentage decrease.
In this problem, increasing the side length by a factor \(k\) (here \(k=1.34\)) causes the area (which depends on \(s^2\)) to increase by a factor of \(k^2\) (here \(1.34^2 = 1.7956\)). The percentage increase is then \((k^2 - 1) \times 100\%\).
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