Trig question

[deleted738762]

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How is the perimeter 24+ 6radical2. I know the hypotenuse equals 6radical2. But how is it 24? If the 4 lines of the square equal to 24, then what about the base of the triangle? One line isn't accounted for.

The 4 sides of the square would equate to 24, then the hypotenuse would give 6radical2, but what about the base of the triangle. Am I missing something? I thought it was 30 + 6radical 2.

Problem is attached.

 

[kimball]

I think your mistake is counting the 4th side of the square that is attached to the triangle, since its perimeter you are only adding the hypotenuse of the triangle + the base (6) + the three sides of the square so you would have (6+6+6=18) for the square + 6 for the base and 6 radical2 for the hypotenuse
= 24+ 6radical2

 

[deleted738762]

I think your mistake is counting the 4th side of the square that is attached to the triangle, since its perimeter you are only adding the hypotenuse of the triangle + the base (6) + the three sides of the square so you would have (6+6+6=18) for the square + 6 for the base and 6 radical2 for the hypotenuse
= 24+ 6radical2

ahhhhh I completely missed that. Thanks.

 

[lxx]

ahhhhh I completely missed that. Thanks.

Also, since the right triangle also has a 45 degree angle, it's an isosceles triangle (180 - 90 - 45 = 45, two angles are equal), so it's base is equal in length to its height, which is one of the square's sides. That's why the base of the triangle is 6 units in length.