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2Let there be a right-angled triangle, ΑΒΓ having as right the angle at Β. And let ΑΒ be 3 units, and ΒΓ 4 units. To find the area of the triangle and the hypotenuse. (diagram 2) Let right-angled parallelogram ΑΒΓΔ be filled out, where the area, as was shown above, is 12 units. But triangle ΑΒΓ is half of parallelogram ΑΒΓΔ. And so, the area of triangle ABG will be 6 units; (diagram 3) and since the angle at B is right, the squares from ΑΒ, ΒΓ are equal to the square from ΑΓ. And the squares from ΑΒ, ΒΓ are 25 units. And that from ΑΓ is, therefore, 25 units. ΑΓ itself is, therefore, 5 units.

(diagram 1) 3. Let there be an isosceles triangle, ΑΒΓ, having ΑΒ equal to ΑΓ, and each of the equal units 10, while ΒΓ to ΑΓ and each of the 10 equal units1of units 12. To find its area. (diagram 2) Let an altitude, ΑΔ, be drawn to ΒΓ. (diagram 3) And through Α, let a parallel to ΒΓ be drawn, i.e., ΕΖ, while through Β, Γ let parallels to ΑΔ, be drawn, i.e., ΒΕ, ΓΖ. Therefore, parallelogram ΒΓΕΖ is double triangle ΑΒΓ. For it has the same base as it and is in the same parallels. And since it is isosceles and ΑΔ was drawn perpendicular, ΒΔ is equal to ΔΓ. And ΒΓ is of 12 units. Therefore, ΒΔ is of 6 units. But ΑΒ is of 10 units. (diagram 4) Therefore, ΑΔ is of 8 units, since the square from ΑΒ is equal to those from ΒΔ, ΔΑ, so that also ΒΕ will be of 8 units. But ΒΓ is of 12 units. Therefore, the area of parallelogram ΒΓΕΖ is of 96 units. Thus the area of triangle ΑΒΓ is of 48 units. And this is the method. Take half of 12. It becomes 5. And 10 to themselves. They become 100. Take away 6 to themselves, which are 36. They become remainders 64. The side of this becomes 8. Altitude ΑΔ will be so much. And 12 against 8. They become 96. The half of these. They become 48. The area of the triangle is of so many units. 1This is clearly a dittography (i.e., where the copyist writes the same phrase twice).

Of non-isosceles triangles it is required to observe the angles so that we see the altitudes from the drawn angles to the sides, whether they fall within the angles or outside. And so, let there be a given triangle, ΑΒΓ, having each side of given portions (μοιρῶν). And it is required to observe, as it happens, the angle at Α, whether right or obtuse or acute. And so, if the square from ΒΓ is equal to the squares from ΒΑ, ΑΓ, it is clear that the angle at Α is right, while if it is smaller, acute, and if larger, it is clear that the angle at Α is obtuse. Let it be supposed, in fact, that the square from ΒΓ is smaller than the squares from ΒΑ, ΑΓ. Therefore, the angle at Α is acute. For if it is not acute, it is either right or obtuse. And so, it is not right. For the square from ΒΓ would have to be equal to the squares from ΓΑ, ΑΒ. But it is not. Therefore, the angle at Α is not right. However, it is not obtuse either. For the square from ΒΓ would have to be larger than the squares from ΓΑ, ΑΒ. But it is not. Therefore, it is not obtuse. But it was shown that it is not right either. Therefore, it is acute. Similarly, in fact, we will also conclude that if the square from ΒΓ is larger than the squares from ΒΑ, ΑΓ the angle at Α is obtuse.

(diagram 1) Let there be an acute-angled triangle, ΑΒΓ, having side ΑΒ of 13 units, ΒΓ of 14 units, and ΑΓ of 15 units. To find its area. It is obvious that the angle at Β is acute. For the square from ΑΓ is smaller than the squares from ΑΒ, ΒΔ. (diagram 2) Let a perpendicular, ΑΔ be drawn to ΒΓ. Therefore, that from ΑΓ is smaller than the squares from ΑΒ, ΒΓ by twice the rectangle enclosed by ΓΒ, ΒΔ, as has been shown (El. ii 13). And those from ΑΒ, ΒΓ are of 365 units, that from ΑΓ of 225 units. Therefore, twice the rectangle enclosed by ΓΒ, ΒΔ is of 140 units. Therefore, once that by ΓΒ, ΒΔ will be of 70 units. And ΒΓ is of 14 units. Therefore, ΒΔ will be of 5 units. And since that from ΑΒ is equal to those from ΑΔ, ΔΒ, and that from ΑΒ is of 169 units, while that from ΒΔ is of 25 units, therefore, a remainder, that from ΑΔ, will be of 144 units. Therefore, this ΑΔ will be of12 units. But ΒΓ is also of 14 units. Therefore, that by ΒΓ, ΑΔ will be of 168 units. And it will be double triangle ΑΒΓ. Therefore, triangle ΑΒΓ will be of 84 units.

The procedure will be like this: 13 times themselves. They become 169. And 14 times themselves. they become 196. And 15 times themselves. They become 225. From these take away 224. They become remainders 140. Half of these. They become 70. Apply along 14. They become 5. And 13 times themselves. They become 169. That from which 5 times themselves were taken away. Remainders 144. The side of these become 12. The perpendicular will be of so much. Multiply these times 14. They become 168. Half of these is 84. The area will be of so much.

The procedure will be this. 13 times themselves become 169. And 11 times themselves become 121. And 20 times themselves become 400. Add 169 and 121. They become 290. Take these from 400. Remainders 110. Have of these become 55. Apply along 110. They become 5. And 13 times themselves become 169. Take 5 times themselves. Remainders 144. The side of these become 12. These times 11 become 132. Half of these is 66. The area of the triangle will be of so much.

There is a general procedure for finding the area given three sides of any triangle whatever and without the altitude. For example, let the sides of the triangle be of 7, 8, 9 units. Compose 7 and 8 and 9. They become 24. Take half of these. They become 12. Take away the 7 units. Remainders 5. Again take away the 8 from the 12. Remainders 4. And further the 9. Remainders 3. Make 12 times 5. They become 60. These times 4. They become 240. These times 3. They become 720. Take the side of these and it will be the area of the triangle.

Here is the geometrical demonstration of this: the sides of a triangle being given, to find the area. And so, it is possible by drawing a single altitude of it and providing its magnitude to find the area of the triangle, but let it be required to provide the area without the altitude.

(diagram 1) Let the given triangle be ΑΒΓ, and let each of ΑΒ, ΒΓ, ΓΑ be given. To find the area. (diagram 2) Let a circle, ΔΕΖ, whose center is Η, be inscribed in the triangle (El. iv 4), and let ΑΗ, ΒΗ, ΓΗ, ΔΗ, ΕΗ, ΖΗ be joined. Therefore, that by ΒΓ, ΕΗ is double triangle ΒΗΓ, while that by ΓΑ, ΖΗ is double triangle ΑΓΗ, and that by ΓΑ, ΖΗ is double triangle ΑΓΗ. Therefore, that by the perimeter of triangle ΑΒΓ and ΕΗ, that is that from the center of circle ΔΕΖ, is double triangle ΑΒΓ. (diagram 3) Let ΓΒ be extended, and let ΒΘ lie equal to ΑΔ. Therefore, ΓΒΘ is half the perimeter of triangle ΑΒΓ, due to the fact that ΑΔ is equal to ΑΖ and ΔΒ to Β and ΖΓ to ΓΕ. Therefore, that by ΓΘ, ΕΗ is equal to triangle ΑΒΓ. But that by ΓΘ, ΕΗ is a side of that from ΓΘ times that from ΕΗ. Therefore, the area of triangle ΑΒΓ times itself will become equal to that from ΘΓ times that from ΕΗ. (diagram 4) Let ΗΛ be drawn at right angles to ΓΗ, ΒΛ to ΓΒ. And let ΓΛ be joined. And so, since each of ΓΗΛ, ΓΒΛ is right, therefore, quadrilateral ΓΗΒΛ is in a circle (see theorem A below). Therefore, the angles enclosed by ΓΗΒ, ΓΛΒ are equal to two right angles (see theorem B below). (diagram 5) But those by ΓΗΒ, ΑΗΔ are also equal to two right angles since the angles at Η are bisected by ΑΗ, ΒΗ, ΓΗ and those by ΓΗΒ, ΑΗΔ are equal to those by ΑΗΓ, ΔΗΒ, and all of them are equal to four right angles (see theorem C below). (diagram 6) Therefore,the angle by ΑΗΔ is equal to the angle by ΓΛΒ. (diagram 7) But a right-angle, by ΑΔΗ, is also equal to a right-angle, by ΓΒΛ. Therefore, triangle ΑΗΔ is similar to triangle ΓΒΛ. As, therefore, ΒΓ is to ΒΛ, ΑΔ is to ΔΗ, that is ΒΘ to ΕΗ. And, alternando, as ΓΒ is to ΒΘ, ΒΛ is to ΕΗ (diagram 8) that is ΒΚ to ΚΕ, since ΒΛ is a parallel to ΕΗ. And, componendo, as ΓΘ is to ΒΘ, so is ΒΕ to ΕΚ. Thus, as that from ΓΘ is to that by ΓΘ, ΘΒ, so too is that by ΒΕΓ to that by ΓΕΚ, that is that from ΕΗ. For a perpendicular, ΕΗ, has been drawn in a rectangle from the right-angle to the base. Thus that from ΓΘ times that from ΕΗ, whose side was the area οf triangle ΑΒΓ, will be equal to that by ΓΘΒ times ΓΕΒ. And each of ΓΘ, ΘΒ, ΒΕ, ΓΕ is given. For ΓΘ is half of the perimeter of triangle ΑΒΓ, while ΒΘ is the excess by which half of the perimeter exceeds ΓΒ, and ΒΕ the excess by which half of the perimeter exceeds ΑΓ, and ΕΓ the excess by which half of the perimeter exceeds ΑΒ, since really ΕΓ is equal to ΓΖ, and ΒΘ to ΑΖ, since it is also equal to ΑΔ. Therefore, the area of triangle ΑΒΓ is also given.

It will be synthesized, in fact, in this way. Let ΑΒ be of 13 units, ΒΓ of 14 units, and ΑΓ of 15 units. Add 13 and 14 and 15. And they become 42. half of which become 21. Remove the 13. Remainders 8. Then the 14. Remainders 7. And further the 15. Remainders 6. 21 times 8, and what come-to-be times 7, and further what come-to-be times 6. They are combined 7056. The side of these of so much as 84 will be the area of the triangle.

Jrg G. Meyer learned all about luxurious interior design from the ground up. He started out as a carpenter and wood technician, going on to work as an architect and project manager, and he is now well-known at metrica for his comprehensive understanding of materials, expert craftsmanship and his knack for avoiding unpleasant surprises through reliable construction processes and planning cost certainty.

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