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3 cube *(243)-2 by 3 *9-1 by 3 |
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Answer» Answer: Steps: 1) DRAW a line SEGMENT AB of length 9.3 units. 2) Extend the line by 1 unit more such that BC=1 unit . 3) Find the midpoint of AC. 4) Draw a line BD perpendicular to AB and LET it intersect the semicircle at point D. 5) Draw an arc DE such that BE=BD. Therefore, BE= 9.3
unitstex] \begin{lgathered}\tt \implies \: (25 - X) {}^{2} = (x) {}^{2} + (5) {}^{2} \\ \tt \implies625 - 50 + x {}^{2} = x {}^{2} + 25 \\ \tt \implies - 50xtex] \begin{lgathered}\tt \implies \: (25 - x) {}^{2} = (x) {}^{2} + (5) {}^{2} \\ \tt \implies625 - 50 + x {}^{2} = x {}^{2} + 25 \\ \tt \implies - 50xtex] \begin{lgathered}\tt \implies \: (25 - x) {}^{2} = (x) {}^{2} + (5) {}^{2} \\ \tt \implies625 - 50 + x {}^{2} = x {}^{2} + 25 \\ \tt \implies - 50xtex] \begin{lgathered}\tt \implies \: (25 - x) {}^{2} = (x) {}^{2} + (5) {}^{2} \\ \tt \implies625 - 50 + x {}^{2} = x {}^{2} + 25 \\ \tt \implies - 50xtex] \begin{lgathered}\tt \implies \: (25 - x) {}^{2} = (x) {}^{2} + (5) {}^{2} \\ \tt \implies625 - 50 + x {}^{2} = x {}^{2} + 25 \\ \tt \implies - 50x |
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