= a2 × (55° 22′) ÷ 2 = ·4831a2 .
Therefore the area of the added portion, T′,
= ·4831 {(a + x)2 − a2}.
And this, by hypothesis,
= T = ·3927a2 .
We get, accordingly, since a = 1,
x2 + 2x = ·3927 ⁄ ·4831 = ·810,
and, solving,
x + 1 = √(1·81) = 1·345, or x = 0·345.
Working out x′ in the same way, we arrive at the approximate value, x′ + 1 = 1·517. {369}