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 σελίδες |
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Σελίδα 70
... inscribed in a circle . Show that the straight line which bisects A and the straight line drawn from the centre perpendicular to BC meet on the circumference . Let the perpendicular bisector of B C meet the E bisector of the angle A in ...
... inscribed in a circle . Show that the straight line which bisects A and the straight line drawn from the centre perpendicular to BC meet on the circumference . Let the perpendicular bisector of B C meet the E bisector of the angle A in ...
Σελίδα 72
... circle , .. A and BOC are supplementary : FOEL BOC . .. Take away the common BOE ; ../ FOB = COE . Again , the quadrilateral B FOD , having two opposite angles B FO , B DO right angles , may be inscribed in a circle , and ../ FOB = 4FDB ...
... circle , .. A and BOC are supplementary : FOEL BOC . .. Take away the common BOE ; ../ FOB = COE . Again , the quadrilateral B FOD , having two opposite angles B FO , B DO right angles , may be inscribed in a circle , and ../ FOB = 4FDB ...
Σελίδα 77
Richard Wormell. CHAPTER X. POLYGONS . 84. The centres of the circles inscribed in and circumscribed about an ... circle , and because FO , E O , DO bisect at right angles the sides A B , A C , B C , O is the centre of the circum- scribed ...
Richard Wormell. CHAPTER X. POLYGONS . 84. The centres of the circles inscribed in and circumscribed about an ... circle , and because FO , E O , DO bisect at right angles the sides A B , A C , B C , O is the centre of the circum- scribed ...
Σελίδα 78
... inscribed circle . △ DBA + △ DAB = half ( ≤ ABC + Z BAC Therefore angle . = half a right angle . BDA = three halves of a right Hence , because △ BDA is constant , the locus of D is the arc of a circle . Because B DA is the ...
... inscribed circle . △ DBA + △ DAB = half ( ≤ ABC + Z BAC Therefore angle . = half a right angle . BDA = three halves of a right Hence , because △ BDA is constant , the locus of D is the arc of a circle . Because B DA is the ...
Σελίδα 79
... inscribed is equal to the excess of the sum of the two sides over the hypothenuse . Let A B C be the right - angled triangle and O the centre of the inscribed circle ... inscribed circle ; .. the diameter of the inscribed circle the excess of ...
... inscribed is equal to the excess of the sum of the two sides over the hypothenuse . Let A B C be the right - angled triangle and O the centre of the inscribed circle ... inscribed circle ; .. the diameter of the inscribed circle the excess of ...
Συχνά εμφανιζόμενοι όροι και φράσεις
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
Δημοφιλή αποσπάσματα
Σελίδα 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.
Σελίδα 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.
Σελίδα 53 - To prove that the exterior angle of a triangle is equal to the sum of the two interior opposite angles (see fig.
Σελίδα 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.
Σελίδα 78 - To construct a circle which shall pass through two given points and touch a given straight line.
Σελίδα 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.
Σελίδα 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.
Σελίδα 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.
Σελίδα 30 - Any line drawn through the point of intersection of the diagonals of a parallelogram divides it into two equal quadrilaterals.
Σελίδα 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.