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weeeeeb [17]
3 years ago
11

Find the diagonal of the square whose side are of the given measure GIVEN=5

Mathematics
1 answer:
cestrela7 [59]3 years ago
5 0
Here to find the diagonal, you would simply split the square diagonally into 2 triangles and use the Pythagorean theorem which is used on triangles:
a² + b² = c²
a and b are both short sides and c is the diagonal side. So since it's a square, a and b would both be 5 and you would solce for c which is the diagonal:
a² + b² = c²
5² + 5² = c²              square the 5s
25 + 25 = c²            add the 25s
50 = c²                    square root both numbers to get rid of Cs 2nd power
√50 = √c²                 
√50 = c                  you can reduce the square root if you want but √50 should be fine.
√(25 × 2) = c
√(5×5×2)
5√2 = c                  this is the reduced root
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\large\underline{\sf{Solution-}}

<u>Let us assume that:</u>

\sf \longmapsto x =  \sqrt{6 +  \sqrt{6 +  \sqrt{6 + ... \infty } } }

We can also write it as:

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By splitting the middle term:

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<u>Therefore:</u>

\longmapsto\begin{cases} \sf (x+ 2) =0 \\ \sf (x - 3) = 0 \end{cases}

\sf \longmapsto x =  - 2,3

<u>But x cannot be negative. </u>

\sf \longmapsto x = 3

Therefore, the value of the expression is 3.

\large\underline{\sf{Verification-}}

Given:

\sf\longmapsto x=3

We can also write it as:

\sf\longmapsto x = \sqrt{9}

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\sf\longmapsto x = \sqrt{6 + \sqrt{9}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+3}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{9}}}

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+3}}}

This pattern will continue.

\sf\longmapsto x = \sqrt{6 + \sqrt{6+\sqrt{6+...\infty}}}

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