round you answer to the two decimal places
2007-06-09 14:35:48 · 11 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
7.77
2007-06-09 14:38:14 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The hypotenuse in a 45-45-90 triangle is the side (they are the same) * sqrt(2).
Thus, if the hypotenuse is 11, the side must be 11 / sqrt(2) or 11 * sqrt(2) / 2
2007-06-09 14:40:13 · answer #2 · answered by MathProf 4 · 0⤊ 0⤋
a^2+b^2=c^2 Pythagorean theorem now, because you realize that the triangle is a 40 5-40 5-ninety, the legs could have equivalent lengths 8^2+8^2=128 c^2=128, so the dimensions of the hypotenuse is the sq. root of 128
2016-11-27 21:30:07 · answer #3 · answered by waltman 3 · 0⤊ 0⤋
the ratio between the legs of a 45-45-90 triangle and its hypothenuse is 1:sqrt(2). If the hyp is 11 then the leg can be found using the ratio:
1/ sqrt(2) = leg/ 11
leg = 11/sqrt(2) = 11/1.414 = 7.779
2007-06-09 14:47:40 · answer #4 · answered by ignoramus_the_great 7 · 0⤊ 0⤋
You should use a model triangle for this. Use your ratios; you know that the legs in the model triangle are 1 and the hypothenuse is square root of 2. If the hypothenuse of this triangle is 11, you'd set up this ratio, and solve for x
11/(square root of 2) = x/1
2007-06-09 14:40:50 · answer #5 · answered by jlange 2 · 0⤊ 0⤋
sin 45° = y / 11
y = 11.sin 45°
y = 7.78 (to 2 dec. places)
2007-06-10 03:35:36 · answer #6 · answered by Como 7 · 0⤊ 0⤋
11 = x sqrt(2)
x =11/sqrt(2)
x = 7.78
2007-06-09 14:40:13 · answer #7 · answered by 7 · 0⤊ 0⤋
x sq rt 2 = 11
x = 11 / sq rt 2 = 7.78
2007-06-09 14:39:25 · answer #8 · answered by richardwptljc 6 · 0⤊ 0⤋
cos45 = x/11
x = 11cos45
x = 7.78 (to 2 decimal places)
2007-06-09 15:02:31 · answer #9 · answered by Kemmy 6 · 0⤊ 0⤋
a^2+b^2=c^2
a=b
2a^2=c^2
2a^2=121
a^2=60.5
a=sqrt 60.5 = 7.78
2007-06-09 14:39:11 · answer #10 · answered by Master Maverick 6 · 0⤊ 0⤋
a^2+b^2=c^2
where a=b and c=11
2a^2=121
a^2=60.5
a=b=sqrt 60.5
2007-06-09 14:43:45 · answer #11 · answered by gordonmorrison 6 · 0⤊ 0⤋