the sides being _n_, (_n_^2 - 1)/2, and (_n_^2 + 1)/2. If for _n_ we put 3, we have 3, 4, 5. If we take the various odd numbers, we have _n_ = 1, 3, 5, 7, 9, ..., (_n_^2 - 1)/2 = 0, 4, 12, 24, 40, ..., (_n_^2 + 1)/2 = 1, 5, 13, 25, 41, .... Of course _n_ may be even, giving fractional values. Thus, for _n_ = 2 we have for the three sides, 2, 1 1/2, 2 1/2. Other formulas are also known. Plato's, for example, is as follows: (2_n_)^2 + (_n_^2 - 1)^2 = (_n_^2 + 1)^2. If 2_n_ = 2, 4, 6, 8, 10, ..., then _n_^2 - 1 = 0, 3, 8, 15, 24, ..., and _n_^2 + 1 = 2, 5, 10, 17, 26, .... This formula evidently comes from that of Pythagoras by doubling the sides of the squares.[81] THEOREM. _In any triangle the square of the side opposite an acute angle is equal to the sum of the squares of the other two sides diminished by twice the product of one of those sides by the projection of the other upon that side._ THEOREM. _A similar statement for the obtuse triangle._ These two propositions are usually proved by the help of the Pythagorean Theorem. Some writers, however, actually construct the squares and give a proof similar to the one in that proposition. This plan goes back at least to Gregoire de St. Vincent (1647). [Illustration] It should be observed that _a_^2 = _b_^2 + _c_^2 - 2_b'c_. If [L]_A_ = 90 deg., then _b'_ = 0, and this becomes _a_^2 = _b_^2 + _c_^2. If [L]_A_ is obtuse, then _b'_ passes through 0 and becomes negative, and _a_^2 = _b_^2 + _c_^2 + 2_b'c_. Thus we have three propositions in one. [Illustration] At the close of Book IV many geometries give as an exercise, and some give as a regular proposition, the celebrated problem that bears the name of Heron of Alexandria, namely, to compute the area of a triangle in terms of its sides. The result is the important formula Area = [sqrt](_s_(_s_ - _a_)(_s_ - _b_)(_s_ - _c_)), where _a_, _b_, and _c_ are the sides, and _s_ is the semiperimeter 1/2(_a_ + _b_ + _c_). As a practical application the class may be able to find a triangular piece of land, as here shown, and to measure the sides. If the piece is clear, the result may be checked by measuring the altitude and applying the formula _a_ = 1/2_bh_. It may be stated to the class that Heron's formula is only a special case of the more general one developed about 640 A.D., by a famous Hindu mathematician, Brahmagupta. This formula gives the area of an inscribed quadrilateral as [sqrt]((_s_ - _a_)(_s_ - _b_)(_s_ - _c_)(_s_ - _d_)), where _a_, _b_, _c_, and _d_ are the sides and _s_ is the semiperimeter. If _d_ = 0, the quadrilateral becomes a triangle and we have Heron's formula.[82]