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To determine which equations are equivalent to the formula for the area of a rhombus, [tex]\( A = \frac{1}{2} d_1 d_2 \)[/tex], we need to manipulate it to isolate [tex]\( d_1 \)[/tex] and [tex]\( d_2 \)[/tex] while keeping the relationships consistent.
1. Starting with the original formula:
[tex]\[ A = \frac{1}{2} d_1 d_2 \][/tex]
2. Isolate [tex]\( d_1 \)[/tex]:
First, multiply both sides of the equation by 2 to eliminate the fraction:
[tex]\[ 2A = d_1 d_2 \][/tex]
Then, divide both sides by [tex]\( d_2 \)[/tex] to solve for [tex]\( d_1 \)[/tex]:
[tex]\[ d_1 = \frac{2A}{d_2} \][/tex]
3. Isolate [tex]\( d_2 \)[/tex]:
Similarly, from [tex]\( 2A = d_1 d_2 \)[/tex], divide both sides by [tex]\( d_1 \)[/tex] to solve for [tex]\( d_2 \)[/tex]:
[tex]\[ d_2 = \frac{2A}{d_1} \][/tex]
Now, let's check the provided equations:
- [tex]\( d_1 = 2 A d_2 \)[/tex]:
[tex]\[ \text{This implies } d_1 = 2A \cdot d_2, \text{ which is incorrect.} \][/tex]
- [tex]\( d_1 = \frac{2 A}{d_2} \)[/tex]:
[tex]\[ \text{This is correct, as derived.} \][/tex]
- [tex]\( d_2 = \frac{d_1}{2 A} \)[/tex]:
[tex]\[ \text{Rearranging } d_2 = \frac{2 A}{d_1} \text{ does not yield this. This is incorrect.} \][/tex]
- [tex]\( d_2 = \frac{2 A}{d_1} \)[/tex]:
[tex]\[ \text{This is correct, as derived.} \][/tex]
- [tex]\( d_2 = 2 A d_1 \)[/tex]:
[tex]\[ \text{This implies } d_2 = 2A \cdot d_1, \text{ which is incorrect.} \][/tex]
Thus, the two correct equivalent equations are:
- [tex]\( d_1 = \frac{2 A}{d_2} \)[/tex]
- [tex]\( d_2 = \frac{2 A}{d_1} \)[/tex]
1. Starting with the original formula:
[tex]\[ A = \frac{1}{2} d_1 d_2 \][/tex]
2. Isolate [tex]\( d_1 \)[/tex]:
First, multiply both sides of the equation by 2 to eliminate the fraction:
[tex]\[ 2A = d_1 d_2 \][/tex]
Then, divide both sides by [tex]\( d_2 \)[/tex] to solve for [tex]\( d_1 \)[/tex]:
[tex]\[ d_1 = \frac{2A}{d_2} \][/tex]
3. Isolate [tex]\( d_2 \)[/tex]:
Similarly, from [tex]\( 2A = d_1 d_2 \)[/tex], divide both sides by [tex]\( d_1 \)[/tex] to solve for [tex]\( d_2 \)[/tex]:
[tex]\[ d_2 = \frac{2A}{d_1} \][/tex]
Now, let's check the provided equations:
- [tex]\( d_1 = 2 A d_2 \)[/tex]:
[tex]\[ \text{This implies } d_1 = 2A \cdot d_2, \text{ which is incorrect.} \][/tex]
- [tex]\( d_1 = \frac{2 A}{d_2} \)[/tex]:
[tex]\[ \text{This is correct, as derived.} \][/tex]
- [tex]\( d_2 = \frac{d_1}{2 A} \)[/tex]:
[tex]\[ \text{Rearranging } d_2 = \frac{2 A}{d_1} \text{ does not yield this. This is incorrect.} \][/tex]
- [tex]\( d_2 = \frac{2 A}{d_1} \)[/tex]:
[tex]\[ \text{This is correct, as derived.} \][/tex]
- [tex]\( d_2 = 2 A d_1 \)[/tex]:
[tex]\[ \text{This implies } d_2 = 2A \cdot d_1, \text{ which is incorrect.} \][/tex]
Thus, the two correct equivalent equations are:
- [tex]\( d_1 = \frac{2 A}{d_2} \)[/tex]
- [tex]\( d_2 = \frac{2 A}{d_1} \)[/tex]
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