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 11
... Hence each of them is a right angle . 10. The straight line which is drawn from the vertex of an isosceles triangle so as to bisect the base will also bisect the angle at the vertex and be per- pendicular to the base . Here the ...
... Hence each of them is a right angle . 10. The straight line which is drawn from the vertex of an isosceles triangle so as to bisect the base will also bisect the angle at the vertex and be per- pendicular to the base . Here the ...
Σελίδα 12
... Hence the perpen- dicular from A to B C is also the line from A which bisects the base . Consequently BD CD and 4 BAD = CAD ( Example 10 ) . 12. The perpendicular to the base of an isosceles triangle drawn from the middle point of the ...
... Hence the perpen- dicular from A to B C is also the line from A which bisects the base . Consequently BD CD and 4 BAD = CAD ( Example 10 ) . 12. The perpendicular to the base of an isosceles triangle drawn from the middle point of the ...
Σελίδα 17
... Hence , taking in succession each side as base , B BACA is greater than BO + CO ; ВС + СА BC + BA " " 99 BO + AO ; 99 CO + AO . By addition , therefore , 2 ( A B + BC + CA ) is greater than 2 ( BO + AO + CO . ) Consequently the sum of ...
... Hence , taking in succession each side as base , B BACA is greater than BO + CO ; ВС + СА BC + BA " " 99 BO + AO ; 99 CO + AO . By addition , therefore , 2 ( A B + BC + CA ) is greater than 2 ( BO + AO + CO . ) Consequently the sum of ...
Σελίδα 19
... Hence , by subtracting M H , we see that A H is greater than A M - MH ; ... AC B C is greater than A M - B M. - Note . - In Example 20 , if the points A B be joined and the line A B be produced to meet E F in a point N , then the dis ...
... Hence , by subtracting M H , we see that A H is greater than A M - MH ; ... AC B C is greater than A M - B M. - Note . - In Example 20 , if the points A B be joined and the line A B be produced to meet E F in a point N , then the dis ...
Σελίδα 27
... Hence the triangles are equal in all respects , and △ ADB = 4 C B D and △ ABD = 4CDB ; But A D B and C BD are alternate angles , there- fore A D and C B are parallel , and C D B and A B D are alternate angles , therefore CD and AB are ...
... Hence the triangles are equal in all respects , and △ ADB = 4 C B D and △ ABD = 4CDB ; But A D B and C BD are alternate angles , there- fore A D and C B are parallel , and C D B and A B D are alternate angles , therefore CD and AB are ...
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.