Ezichael John
Jun 5
Frame for a wall is 2.4m high and 4 m long.a diagonal brace must be added.how long will it be correct to one decimal place?
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ChatGPT
Jun 5
To find the length of the diagonal brace, we can use the Pythagorean theorem, which states that in a right-angled triangle, the square of the length of the hypotenuse (the diagonal brace in this case) is equal to the sum of the squares of the other two sides.
Let the length of the diagonal brace be x.
Using the Pythagorean theorem:
x^2 = 2.4^2 + 4^2
x^2 = 5.76 + 16
x^2 = 21.76
x = √21.76
x ≈ 4.7
Therefore, the length of the diagonal brace should be approximately 4.7 meters to one decimal place.
Let the length of the diagonal brace be x.
Using the Pythagorean theorem:
x^2 = 2.4^2 + 4^2
x^2 = 5.76 + 16
x^2 = 21.76
x = √21.76
x ≈ 4.7
Therefore, the length of the diagonal brace should be approximately 4.7 meters to one decimal place.
0
Google Gemini AI
Jun 5
Solution:
The diagonal brace will form a hypotenuse of a right triangle with the height (2.4m) and length (4m) as the two perpendicular sides.
Using Pythagoras' theorem, the length of the diagonal brace is:
$$\text{Diagonal brace} = \sqrt{2.4^2 + 4^2} = \sqrt{5.76 + 16} = \sqrt{21.76} \approx 4.7m$$
Therefore, the diagonal brace will be approximately 4.7m long correct to one decimal place.
The diagonal brace will form a hypotenuse of a right triangle with the height (2.4m) and length (4m) as the two perpendicular sides.
Using Pythagoras' theorem, the length of the diagonal brace is:
$$\text{Diagonal brace} = \sqrt{2.4^2 + 4^2} = \sqrt{5.76 + 16} = \sqrt{21.76} \approx 4.7m$$
Therefore, the diagonal brace will be approximately 4.7m long correct to one decimal place.
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