Elements of GeometryHilliard, Gray,, 1841 - 235 páginas |
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Página 2
... isosceles ( fig . 8 ) , when two only of its sides are equal , and scalene ( fig . 9 ) , when no two of its sides are equal . 16. A right - angled triangle is that which has one right angle . The side opposite to the right angle is ...
... isosceles ( fig . 8 ) , when two only of its sides are equal , and scalene ( fig . 9 ) , when no two of its sides are equal . 16. A right - angled triangle is that which has one right angle . The side opposite to the right angle is ...
Página 9
... isosceles triangle , the angles opposite to the equal sides are equal . = Demonstration . Let the side AB AC ( fig . 28 ) , then will Fig 28 . the angle C be equal to B. Draw the straight line AD from the vertex A to the point D , the ...
... isosceles triangle , the angles opposite to the equal sides are equal . = Demonstration . Let the side AB AC ( fig . 28 ) , then will Fig 28 . the angle C be equal to B. Draw the straight line AD from the vertex A to the point D , the ...
Página 10
... isosceles triangle to the middle of the base , is perpendicular to that base , and divides the vertical angle into two equal parts . In a triangle that is not isosceles , any one of its sides may be taken indifferently for a base ; and ...
... isosceles triangle to the middle of the base , is perpendicular to that base , and divides the vertical angle into two equal parts . In a triangle that is not isosceles , any one of its sides may be taken indifferently for a base ; and ...
Página 35
... isosceles , and the angle BAC ABC ; = the triangle CAD is also isosceles , and the angle CAD = ADC ; hence BAC + CAD , or BAD≈ ABD + ADB . But , if the two angles B and D of the triangle ABD are together equal to the third BAD , the ...
... isosceles , and the angle BAC ABC ; = the triangle CAD is also isosceles , and the angle CAD = ADC ; hence BAC + CAD , or BAD≈ ABD + ADB . But , if the two angles B and D of the triangle ABD are together equal to the third BAD , the ...
Página 53
... isosceles , we have AC = AB + BC = 2AB ; therefore the square described upon the diagonal AC is double of the square described upon the side AB . This property may be rendered sensible by drawing , through the points A and C , parallels ...
... isosceles , we have AC = AB + BC = 2AB ; therefore the square described upon the diagonal AC is double of the square described upon the side AB . This property may be rendered sensible by drawing , through the points A and C , parallels ...
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Términos y frases comunes
ABC fig adjacent angles altitude angle ACB angle BAC base ABCD bisect centre chord circ circular sector circumference circumscribed common cone consequently construction convex surface Corollary cube cylinder Demonstration diagonals diameter draw drawn equal angles equiangular equilateral equivalent faces figure formed four right angles frustum GEOM given point gles greater hence homologous sides hypothenuse inclination intersection isosceles triangle join less Let ABC let fall Let us suppose line AC mean proportional measure the half meet multiplied number of sides oblique lines opposite parallelogram parallelopiped perimeter perpendicular plane MN polyedron prism produced proposition radii radius ratio rectangle regular polygon right angles Scholium sector segment semicircle semicircumference side BC similar solid angle sphere spherical polygons spherical triangle square described straight line tangent THEOREM third three angles triangle ABC triangular prism triangular pyramids vertex vertices whence
Pasajes populares
Página 67 - The square of the hypothenuse is equal to the sum of the squares of the other two sides ; as, 5033 402+302.
Página 9 - If two triangles have the three sides of the one equal to the three sides of the other, each to each, the triangles are congruent.
Página 65 - The areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles. A D A' Hyp. In triangles ABC and A'B'C', To prove AABC A A'B'C' A'B' x A'C ' Proof. Draw the altitudes BD and B'D'.
Página 160 - ABC (fig. 224) be any spherical triangle ; produce the sides AB, AC, till they meet again in D. The arcs ABD, ACD, will be...
Página 168 - In any spherical triangle, the greater side is opposite the greater angle ; and conversely, the greater angle is opposite the greater side.
Página 157 - CIRCLE is a plane figure bounded by a curved line, all the points of which are equally distant from a point within called the centre; as the figure ADB E.
Página 8 - Any side of a triangle is less than the sum of the other two sides...
Página 82 - The perimeters of two regular polygons of the same number of sides, are to each other as their homologous sides, and their areas are to each other as the squares of those sides (Prop.
Página 29 - Two equal chords are equally distant from the centre ; and of two unequal chords, the less is at the greater distance from the centre.
Página 182 - CD, &c., taken together, make up the perimeter of the prism's base : hence the sum of these rectangles, or the convex surface of the prism, is equal to the perimeter of its base multiplied by its altitude.