Exercises on the geometry and measurement of plane figures, being solutions of the theorems, problems and questions in 'Wormell's Modern geometry'.Thomas Murby, 1883 - 192 páginas |
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Página 15
... and A B = D C ; . ' . BA + A C = DC + AC . But DC + AC is greater than A D. Consequently BA + CA is greater than AD ; that is , greater than twice A M. 17. The perimeter of a triangle is greater than the Triangles . 15.
... and A B = D C ; . ' . BA + A C = DC + AC . But DC + AC is greater than A D. Consequently BA + CA is greater than AD ; that is , greater than twice A M. 17. The perimeter of a triangle is greater than the Triangles . 15.
Página 17
... twice this sum . Let A B C be the triangle and O the point within it , then we know that B A + AC is greater than BO + O C. Hence , taking in succession each side as base , B BACA is greater than BO + CO ; ВС + СА BC + BA " " 99 BO + AO ...
... twice this sum . Let A B C be the triangle and O the point within it , then we know that B A + AC is greater than BO + O C. Hence , taking in succession each side as base , B BACA is greater than BO + CO ; ВС + СА BC + BA " " 99 BO + AO ...
Página 22
... twice the other , the hypothenuse shall be double of the shorter side . In the triangle ADC let △ ADC be right , and let DCA2 DAC . ( See fig . of Example 6 ) . Continue CD to B , making DB DC ; then △ ABD = △ A CD , △ B = ≤ C ...
... twice the other , the hypothenuse shall be double of the shorter side . In the triangle ADC let △ ADC be right , and let DCA2 DAC . ( See fig . of Example 6 ) . Continue CD to B , making DB DC ; then △ ABD = △ A CD , △ B = ≤ C ...
Página 30
... twice the sum of the diagonals . A B AD + DC > AC AB + BC > AC DA + AB > DB DC + CB > D B. By addition ... twice the sum of the sides is greater than twice the sum of the diagonals . Again , AB < AO + OB BC CO + OB DC DO + CO AD DO + AO ...
... twice the sum of the diagonals . A B AD + DC > AC AB + BC > AC DA + AB > DB DC + CB > D B. By addition ... twice the sum of the sides is greater than twice the sum of the diagonals . Again , AB < AO + OB BC CO + OB DC DO + CO AD DO + AO ...
Página 58
... twice the given middle line , construct a triangle , A B C ( see fig . of Problem 7 ) . Through C draw DC parallel to A B , and through A draw A D parallel to B C. Join AC . Since ABCD is a parallelogram , its diagonals bisect one ...
... twice the given middle line , construct a triangle , A B C ( see fig . of Problem 7 ) . Through C draw DC parallel to A B , and through A draw A D parallel to B C. Join AC . Since ABCD is a parallelogram , its diagonals bisect one ...
Términos y frases comunes
A B and C D ABCD adjacent angles bisect the angle centre chord circum circumference common tangents Consequently Construct a triangle DAB+ Decagon describe a circle diagonals diameter equal to half equilateral triangle Find the area given angle given circle given radius given side given straight line gonals greater Hence hypothenuse inscribed circle isosceles triangle joining the middle LADC line joining line parallel middle line middle point opposite angles parallel to A B parallelogram pendicular pentagon perimeter perpendicular bisector perpendicular to A B point of intersection polygon quadrilateral R₁ radii radius a circle rectangle regular polygon required circle respectively equal rhombus right angle right-angled triangle square tangent touch a given triangle A B D triangle required vertex
Pasajes populares
Página 91 - If a straight line be divided into any two parts, the square on the whole line is equal to the squares on the two parts, together with twice the rectangle contained by the parts.
Página 98 - Three times the sum of the squares on the sides of a triangle is equal to four times the sum of the squares of the lines joining the middle point of each side with the opposite angles.
Página 53 - To prove that the exterior angle of a triangle is equal to the sum of the two interior opposite angles (see fig.
Página 40 - Thus, of all straight lines drawn from a given point to a given straight line, that which is perpendicular to the given line is a minimum.
Página 78 - To construct a circle which shall pass through two given points and touch a given straight line.
Página 90 - Two triangles are equal when they have two sides and the included angle of the one, respectively equal to two sides and the included angle of the other.
Página 159 - A field in the form of a right-angled triangle is to be divided between two persons, by a fence made from the right angle meeting the hypothenuse perpendicularly, at the distance of 880 links from one end ; required the area of each person's share, the length of the division-fence being 660 links. Ans. 2a. 3r.
Página 9 - If from any point within a triangle, two straight lines be drawn to the extremities of either side, their sum will be less than Hie sum of the two other sides of the triangle.
Página 30 - Any line drawn through the point of intersection of the diagonals of a parallelogram divides it into two equal quadrilaterals.
Página 46 - Ft, and to Dd. 23. The common intersection of the three lines divides each into two parts, one of which is double of the other, and this point is the vertex of three triangles which have lines drawn from it to the bisection of the bases. Apply Euc. n. 12, 13. 24. Apply Theorem 3, p.