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Problem. 6. In any Right angled Triangle either the base or cathetus
& the alternate segment of the Hyp: being given to find the
other segment --
Let C = 45 Cathetus.
And B alternate segment = 48. [Diagrams to represent these quantities]
1 Then B:e::e:A.
2 ba = ee.
1 47. Euc: 3 bb - aa = ee
4 bb - aa = ba \frac{RR}{4} = cc + \frac{bb}{4.}
45 5 aa + ba = cc 4RR = 16cc + 4bb
C. \square 6 aa + ba + \frac{bb}{4} = cc + \frac{bb}{4} = \frac{RR}{4} RR = 4cc + bb
a + \frac{b}{2} = \frac{R}{2} R = 2c + b
a = \frac{R-b}{2.} R = \sqrt{4cc + bb}
a = \frac{\sqrt{4cc+bb} - b}{2} = 27 a = \frac{\sqrt{4cc + bb} - b}{2}

Geometrical Construction
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aa = ba + cc. Make y^e Coefficient b | 45 48 6
& the Root of y^e resolvend c in to a Rect^d par: | 45 48
Upon the middle points of the side B | --- 2 --- 3
describe a semicircle which shall [Diagram] | 225 384
pass. thro. &. V. (ampleating its | 180 192
Diameter as in y^e annexed. scheme. vl. 13 | ---- ---- [Uncertainty regarding the reading of 'vl.']
Then will either part of the diametre a be the | 4|2025|506 2304
required segment For \angle L b d is a \Rangle^e. & a:b::b:R. | |200 | 2
&. a:c::c:b+a & aa + ba = cc. | 4
---------------------------------------------------------------- ----
| 8100
Problem. given. b. the base |
d alternate segment of Hyp. | 8100
Then b:d::c:a | 2304
AB - dc. [diagram] | -----
bb - dd= cc - aa | 10404(102
a = \frac{dc}{b} | 202 ---
\frac{dc}{b} \times \frac{dc}{b} | 1404
cc = \frac{dc^2}{bb} = bb - dd | 404
bbcc - dc^2 = bbbb - ddbb. | ---
bbbb-ddbb = bb |
cc - ddcc = bb - dd | 102
---------------- | 48
bb | ---
| 2)54(27.
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AnnaBeddoes