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 7
... perpendicular to AB ; let a line , C D , revolve about the point C , and let it first coincide with C A. At first the angle A CD is zero , and therefore less than the given angle , and it gradually increases as CD revolves until CD ...
... perpendicular to AB ; let a line , C D , revolve about the point C , and let it first coincide with C A. At first the angle A CD is zero , and therefore less than the given angle , and it gradually increases as CD revolves until CD ...
Página 9
... perpendicular bisector of a straight line to the extremities of the straight line are equal to one another . Let A be a point in the line A D which bisects . BC and is at right angles to BC at the point D ; then shall CA = BA . The ...
... perpendicular bisector of a straight line to the extremities of the straight line are equal to one another . Let A be a point in the line A D which bisects . BC and is at right angles to BC at the point D ; then shall CA = BA . The ...
Página 10
... perpendicular to A B , then there cannot be another perpendicular from C to A B. If possible let CE be per- pendicular to A B. A E D F B Then CED and CDB will be equal , because they are right angles ; that is to say , the exterior ...
... perpendicular to A B , then there cannot be another perpendicular from C to A B. If possible let CE be per- pendicular to A B. A E D F B Then CED and CDB will be equal , because they are right angles ; that is to say , the exterior ...
Página 12
... perpendicular to B C. 11. The perpendicular drawn to the base of an isosceles triangle from its vertex will bisect the base as well as the angle at the vertex . Let A B C be the isosceles triangle and AD the perpendicular from the ...
... perpendicular to B C. 11. The perpendicular drawn to the base of an isosceles triangle from its vertex will bisect the base as well as the angle at the vertex . Let A B C be the isosceles triangle and AD the perpendicular from the ...
Página 13
... perpendicular at the middle point , D , of B C. Consequently the perpendicular to BC at the middle point is the bisector of BAC . 13. When two isosceles triangles stand on the same base , the straight line which passes through their ...
... perpendicular at the middle point , D , of B C. Consequently the perpendicular to BC at the middle point is the bisector of BAC . 13. When two isosceles triangles stand on the same base , the straight line which passes through their ...
Términos y frases comunes
A B C D ABCD base centre chord circum circumference Consequently Construct a triangle contained Continue describe describe a circle diagonals diameter difference distance divided draw drawn equal Example extremities figure Find the area formed four given angle given circle given length given point given radius given side given straight line greater half height Hence hypothenuse inscribed circle isosceles triangle less Let A B C line joining lines drawn locus meet middle line middle point parallel parallelogram pass pentagon perimeter perpendicular point of intersection polygon problem produced Prove quadrilateral radii rect rectangle right angle sides Similarly square Suppose Take triangle triangle required twice vertex vertical yards
Pasajes populares
Página 93 - 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 100 - 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 54 - 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 80 - 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 161 - 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.