Asked by Punkie
I am to find the area of a square. It is cut in half and the line is measured at 6 square root 2. It also says that the two triangles are equallateral triangles. I know that the equation for a equalateral triangle is
A = the square root of 3 times s squared / 4
How would I solve this so that I can have the area of the square?
A = the square root of 3 times s squared / 4
How would I solve this so that I can have the area of the square?
Answers
Answered by
MathMate
Using one single cut, I can cut a square into two isosceles triangles, but not two equilateral triangles.
Could you please check your question?
Could you please check your question?
Answered by
Punkie
Oh, I guess I assumed that the triangle was and equilateral triangle because the only other information it gave me was that it had an angle of 45 degrees in the corner that was split.
Answered by
MathMate
So that makes two isosceles right triangles. The hypotenuse (the longest side) is the cut, right? Draw a little diagram to confirm that.
The area of an isosceles right triangle (i.e. 45°,90°45°) is equal to the hypotenuse²/4. The hypotenuse is the cut, or the side that joins the two 45° angles.
Take it from here, and post your answer for verification if you wish.
The area of an isosceles right triangle (i.e. 45°,90°45°) is equal to the hypotenuse²/4. The hypotenuse is the cut, or the side that joins the two 45° angles.
Take it from here, and post your answer for verification if you wish.
Answered by
Punkie
A = 6 square root 2 to the power of 2 / 4
A = 12/4
A = 3
Then to find the area of the square I would multiply 3 time 2 to equal 6
A = 12/4
A = 3
Then to find the area of the square I would multiply 3 time 2 to equal 6
Answered by
MathMate
A = <b>(</b>6 square root 2<b>)</b> to the power of 2 / 4
=6²*(√2)²/4
=?
=6²*(√2)²/4
=?
Answered by
Punkie
Oh ok I didn't square both the 6 and the square root 2...that makes sense. Thank you.
Answered by
MathMate
You're welcome!
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