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 62
... height of the triangle = BG the given height , and ABC the given angle . = 30. Construct a triangle , having given the sum of two sides and the angles . Take BD equal to the sum of the two given sides ( see preceding fig . ) ; at D make ...
... height of the triangle = BG the given height , and ABC the given angle . = 30. Construct a triangle , having given the sum of two sides and the angles . Take BD equal to the sum of the two given sides ( see preceding fig . ) ; at D make ...
Página 89
... height AD be given . Con- struct a semicircle on A B , and describe an arc from A with radius equal to AD , the given height intersecting the semi - circumference in D. At A make BAC = the given angle , and through D draw B C , then ...
... height AD be given . Con- struct a semicircle on A B , and describe an arc from A with radius equal to AD , the given height intersecting the semi - circumference in D. At A make BAC = the given angle , and through D draw B C , then ...
Página 90
... height ( as in Problem 5 ) . From A , with radius A B , describe an arc cutting C B in B , then A B C is the triangle required . = 52. Construct a triangle , having given the radius of the inscribed circle , one angle , and the height ...
... height ( as in Problem 5 ) . From A , with radius A B , describe an arc cutting C B in B , then A B C is the triangle required . = 52. Construct a triangle , having given the radius of the inscribed circle , one angle , and the height ...
Página 90
... height ( as in Problem 5 ) . From A , with radius A B , describe an arc cutting C B in B , then A B C is the triangle required . = 52. Construct a triangle , having given the radius of the inscribed circle , one angle , and the height ...
... height ( as in Problem 5 ) . From A , with radius A B , describe an arc cutting C B in B , then A B C is the triangle required . = 52. Construct a triangle , having given the radius of the inscribed circle , one angle , and the height ...
Página 119
... height of the required rectangle . Proof . The rectangle on B HF parallelogram HB on the base HF , and the rectangle on B D = parallelogram F D on the base BD . But the parallelogram H B is equal to the parallelogram FD ; therefore rect ...
... height of the required rectangle . Proof . The rectangle on B HF parallelogram HB on the base HF , and the rectangle on B D = parallelogram F D on the base BD . But the parallelogram H B is equal to the parallelogram FD ; therefore rect ...
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.