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Sagot :
To solve this problem, let's follow a step-by-step approach:
1. Understanding the Relationship:
The diagonal of a square divides it into two right-angled triangles. The diagonal acts as the hypotenuse of these triangles. According to the Pythagorean theorem, for a right-angled triangle with sides [tex]\(a\)[/tex] and [tex]\(b\)[/tex], and hypotenuse [tex]\(c\)[/tex], the relationship is given by:
[tex]\[ c^2 = a^2 + b^2 \][/tex]
In the case of a square, both [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are the sides of the square, and [tex]\(c\)[/tex] is the diagonal.
2. Applying the Pythagorean Theorem:
Let the side of the square be [tex]\(a\)[/tex], and the diagonal be [tex]\(d\)[/tex]. We can write:
[tex]\[ d = \sqrt{a^2 + a^2} = \sqrt{2a^2} \][/tex]
Since [tex]\(d = 11.3\)[/tex] meters, we have:
[tex]\[ 11.3 = \sqrt{2a^2} \][/tex]
3. Solving for the Side Length:
Squaring both sides of the equation:
[tex]\[ 11.3^2 = 2a^2 \][/tex]
This simplifies to:
[tex]\[ 127.69 = 2a^2 \][/tex]
Dividing both sides by 2:
[tex]\[ a^2 = \frac{127.69}{2} = 63.845 \][/tex]
Taking the square root of both sides:
[tex]\[ a = \sqrt{63.845} \approx 7.990306627407987 \, \text{meters} \][/tex]
4. Calculating the Perimeter:
The perimeter of a square is given by four times the side length:
[tex]\[ \text{Perimeter} = 4a = 4 \times 7.990306627407987 \approx 31.96122650963195 \, \text{meters} \][/tex]
Therefore, the approximate perimeter of the square, given its diagonal is 11.3 meters, is [tex]\(31.96122650963195\)[/tex] meters.
1. Understanding the Relationship:
The diagonal of a square divides it into two right-angled triangles. The diagonal acts as the hypotenuse of these triangles. According to the Pythagorean theorem, for a right-angled triangle with sides [tex]\(a\)[/tex] and [tex]\(b\)[/tex], and hypotenuse [tex]\(c\)[/tex], the relationship is given by:
[tex]\[ c^2 = a^2 + b^2 \][/tex]
In the case of a square, both [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are the sides of the square, and [tex]\(c\)[/tex] is the diagonal.
2. Applying the Pythagorean Theorem:
Let the side of the square be [tex]\(a\)[/tex], and the diagonal be [tex]\(d\)[/tex]. We can write:
[tex]\[ d = \sqrt{a^2 + a^2} = \sqrt{2a^2} \][/tex]
Since [tex]\(d = 11.3\)[/tex] meters, we have:
[tex]\[ 11.3 = \sqrt{2a^2} \][/tex]
3. Solving for the Side Length:
Squaring both sides of the equation:
[tex]\[ 11.3^2 = 2a^2 \][/tex]
This simplifies to:
[tex]\[ 127.69 = 2a^2 \][/tex]
Dividing both sides by 2:
[tex]\[ a^2 = \frac{127.69}{2} = 63.845 \][/tex]
Taking the square root of both sides:
[tex]\[ a = \sqrt{63.845} \approx 7.990306627407987 \, \text{meters} \][/tex]
4. Calculating the Perimeter:
The perimeter of a square is given by four times the side length:
[tex]\[ \text{Perimeter} = 4a = 4 \times 7.990306627407987 \approx 31.96122650963195 \, \text{meters} \][/tex]
Therefore, the approximate perimeter of the square, given its diagonal is 11.3 meters, is [tex]\(31.96122650963195\)[/tex] meters.
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