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 3
... Chords and Secants 2. - Angles in a Circle XI . - PROBLEMS ON THE CIRCLE ... 64 64 ... 68 ... ... ... 77 80 ... ... XII . - AREAS ... ... SEC . 300 ... ... ... 93 XIV . - PROPORTION ... ... 1. - Equality of Areas 2. - Theorem of ...
... Chords and Secants 2. - Angles in a Circle XI . - PROBLEMS ON THE CIRCLE ... 64 64 ... 68 ... ... ... 77 80 ... ... XII . - AREAS ... ... SEC . 300 ... ... ... 93 XIV . - PROPORTION ... ... 1. - Equality of Areas 2. - Theorem of ...
Página 63
... one of the given angles ; at E make △ CED half another . From C draw CA and C B. making = < DCA = 4 CDA and △ BCE = / CEB . Then A B C shall be the triangle required . CHAPTER IX . THE CIRCLE . SECTION I. - Chords Problems . 63.
... one of the given angles ; at E make △ CED half another . From C draw CA and C B. making = < DCA = 4 CDA and △ BCE = / CEB . Then A B C shall be the triangle required . CHAPTER IX . THE CIRCLE . SECTION I. - Chords Problems . 63.
Página 64
Richard Wormell. CHAPTER IX . THE CIRCLE . SECTION I. - Chords and Secants . 65. If two chords in a circle bisect each other , they both pass through the centre . Let AB and CD be chords bisecting each other in O , then shall O be the ...
Richard Wormell. CHAPTER IX . THE CIRCLE . SECTION I. - Chords and Secants . 65. If two chords in a circle bisect each other , they both pass through the centre . Let AB and CD be chords bisecting each other in O , then shall O be the ...
Página 65
... chords ; then M N is perpen- dicular to all . First let A B be any one of the chords , and let it cut M N in C ; then , by Theorem XXIII . , A B is bisected in C. Conversely , let C be any point M A B C N in M N , and AB the chord which ...
... chords ; then M N is perpen- dicular to all . First let A B be any one of the chords , and let it cut M N in C ; then , by Theorem XXIII . , A B is bisected in C. Conversely , let C be any point M A B C N in M N , and AB the chord which ...
Página 66
... chords intersect within the circle : prove that the parts of the one are respectively equal to the parts of the other ... chord through C , and OD the perpendicular on it from B the centre . ZODC > 20CD ; ... OC > O D. .. EFA B. 70. Two ...
... chords intersect within the circle : prove that the parts of the one are respectively equal to the parts of the other ... chord through C , and OD the perpendicular on it from B the centre . ZODC > 20CD ; ... OC > O D. .. EFA B. 70. Two ...
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.