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 18
... produce AD and B C to meet in H. Then the triangles AC D HCD will have the right ACD = < HCD , since angles at D equal each of them is equal to common ; therefore AC = BC F , and the side DC HC . Similarly , if M is any other position ...
... produce AD and B C to meet in H. Then the triangles AC D HCD will have the right ACD = < HCD , since angles at D equal each of them is equal to common ; therefore AC = BC F , and the side DC HC . Similarly , if M is any other position ...
Página 19
... produced to meet E F in a point N , then the dis- tances of the point N from A and B make the same angle with E F , but their sum is then greater than the sum of any other distances , A M , B M. Similarly , if the points A and B in ...
... produced to meet E F in a point N , then the dis- tances of the point N from A and B make the same angle with E F , but their sum is then greater than the sum of any other distances , A M , B M. Similarly , if the points A and B in ...
Página 21
... produced far enough . Let E G be the perpendicular to AE B A B at the point E , then E G and CD are parallel . And since we can have only one straight line through a given point parallel to another straight line , E F cannot be parallel ...
... produced far enough . Let E G be the perpendicular to AE B A B at the point E , then E G and CD are parallel . And since we can have only one straight line through a given point parallel to another straight line , E F cannot be parallel ...
Página 22
Richard Wormell. AN , they will , if produced , intersect in some point N. Because M N intersects A N , one of two parallels , it also intersects PQ , the other . Let R be the point of intersection of M N and P Q. Because PE and M N are ...
Richard Wormell. AN , they will , if produced , intersect in some point N. Because M N intersects A N , one of two parallels , it also intersects PQ , the other . Let R be the point of intersection of M N and P Q. Because PE and M N are ...
Página 32
... produced in F , then BCFE is a parallelogram ; .. CF BE = AE . Also FDC ZDAE . LEDA , and D C F = Consequently triangles A D E , C D F are equal , and A D C D. = 43. The line joining the middle points of two sides of a triangle is equal ...
... produced in F , then BCFE is a parallelogram ; .. CF BE = AE . Also FDC ZDAE . LEDA , and D C F = Consequently triangles A D E , C D F are equal , and A D C D. = 43. The line joining the middle points of two sides of a triangle is equal ...
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.