Plane Geometry |
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... drawn , while he is proving the theorem about the base angles of an isosceles triangle . The real question is how far shall this sort of practice be allowed . The prevailing view is that it should be practiced no longer than necessary ...
... drawn , while he is proving the theorem about the base angles of an isosceles triangle . The real question is how far shall this sort of practice be allowed . The prevailing view is that it should be practiced no longer than necessary ...
Page 13
... draw an angle , first draw a straight line , which is to be one side of the angle , as BC in the figure , 4. If the angle is to be 50 ° , place the protractor so that the side BC falls along the 50 ° mark , then draw the side BA along ...
... draw an angle , first draw a straight line , which is to be one side of the angle , as BC in the figure , 4. If the angle is to be 50 ° , place the protractor so that the side BC falls along the 50 ° mark , then draw the side BA along ...
Page 13
... draw an angle , first draw a straight line , which is to be one side of the angle , as BC in the figure , § 4. If the angle is to be 50 ° , place the protractor so that the side BC falls along the 50 ° mark , then draw the side BA along ...
... draw an angle , first draw a straight line , which is to be one side of the angle , as BC in the figure , § 4. If the angle is to be 50 ° , place the protractor so that the side BC falls along the 50 ° mark , then draw the side BA along ...
Page 13
... draw an angle , first draw a straight line , which is to be one side of the angle , as BC in the figure , § 4. If the angle is to be 50 ° , place the protractor so that the side BC falls along the 50 ° mark , then draw the side BA along ...
... draw an angle , first draw a straight line , which is to be one side of the angle , as BC in the figure , § 4. If the angle is to be 50 ° , place the protractor so that the side BC falls along the 50 ° mark , then draw the side BA along ...
Page 13
... draw an angle , first draw a straight line , which is to be one side of the angle , as BC in the figure , § 4. If the angle is to be 50 ° , place the protractor so that the side BC falls along the 50 ° mark , then draw the side BA along ...
... draw an angle , first draw a straight line , which is to be one side of the angle , as BC in the figure , § 4. If the angle is to be 50 ° , place the protractor so that the side BC falls along the 50 ° mark , then draw the side BA along ...
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Common terms and phrases
AABC ABCD acute angle ADEF adjacent adjacent angles altitude angle equal angles are equal base bisects chord circumference circumscribed circle construct a regular Construct a square corresponding sides decagon diagonals diameter distance divided drawing an angle drawn equal respectively equal sides equiangular polygon equilateral polygon equilateral triangle EXERCISES figure Find the area Find the length geometry given circle given line given point hypotenuse inch inscribed angle interior angles intersecting isosceles trapezoid isosceles triangle median middle point number of degrees number of sides obtuse parallel lines parallelogram pentagon perimeter perpendicular bisector points equidistant prime numbers Proof protractor Prove quadrilateral radii radius ratio rectangle regular hexagon regular inscribed polygon regular octagon regular polygon rhombus right angle right triangle secant segment semicircle Show sides equal supplementary tangent Theorem transversal triangle ABC vertex vertex angle vertices
Popular passages
Page 153 - In any triangle, the square of a 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 and the projection of the other side upon it.
Page 189 - The straight line joining the middle points of two sides of a triangle is parallel to the third side, and equal to half of it.
Page 139 - The area of a parallelogram is equal to the product of its base and its height: A = bx h.
Page 233 - ... as the squares of their radii, or as the squares of their...
Page 80 - ... the third side of the first is greater than the third side of the second.
Page 148 - ... they have an angle of one equal to an angle of the other and the including sides are proportional; (c) their sides are respectively proportional.
Page 94 - Theorem. In the same circle or in equal circles, equal chords are equidistant from the center; and of two unequal chords the greater is nearer the center. Given two equal © M, M ' , with chords AB = A'B', AE > A'B', and OC, OD, O'C' ±'s from center 0 to AB, AE, and from center O
Page 149 - The square described on the hypotenuse of a right triangle is equivalent to the sum of the squares on the other two sides.
Page 135 - The area of a rectangle is equal to the product of its base and altitude.
Page 148 - Two triangles which have an angle of one equal to the supplement of an angle of the other are to each other as the products of the sides including the supplementary angles.